big-m two phase methods
DESCRIPTION
OptimisationTRANSCRIPT
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In the standard simplex table
bull The coefficients of the basic variables in the z-rowshould be zero
bull Th r iv i i n in h n r in m rix
1
must be identity coefficients
bull In case it is not we have to use elementary row
operations to keep it in the standard simplex formbefore we apply the simplex method
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Contents
bull Artificial Variables Techniques
bull Big M Method
bull Two Phase Method
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If in a starting simplex tableau we donrsquot have an
identity sub matrix (ie an obvious starting BFS) then
we introduce artificial variables to have a starting BFS
This is known as artificial variable technique
There are two methods to find the starting BFS and
so ve t e prob em ndash
(1) Big M Method
(2) Two-Phase Method
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Big M Method
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Suppose that ith constraint equation doesnt have aslack variable ie
ith constraint in the original LPP is either ge or = type
Then an artificial variable Ri is added to form the ithunit vector column
Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal
zero in the optimum solution
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Artificial variable objective coefficient
= - M (Sum of artificial variables) in a maximization
problem
= sum o t e art c a var a es n a m n m zat onproblem
where M is a very large positive number
Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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Tabular form
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Since the z column does not change in the iterations it may be
deleted
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The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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Apply the usual simplex method for solvingminimization problem
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Selecting pivot row and pivot column for a minimization problem
Minratio
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61
---
Select the most positive element of the z row
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Minratio
Second iteration
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Final iteration
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Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
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1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
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1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
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1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
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Phase - II
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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Standard simplex table
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Applying the usual simplex method for solving aminimization problem
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Minratio
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At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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Form a simplex table
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Is this standard simplex table
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z ndashrow = z row + 6(x2 row) + 5(x3 row)
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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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Contents
bull Artificial Variables Techniques
bull Big M Method
bull Two Phase Method
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If in a starting simplex tableau we donrsquot have an
identity sub matrix (ie an obvious starting BFS) then
we introduce artificial variables to have a starting BFS
This is known as artificial variable technique
There are two methods to find the starting BFS and
so ve t e prob em ndash
(1) Big M Method
(2) Two-Phase Method
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Big M Method
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Suppose that ith constraint equation doesnt have aslack variable ie
ith constraint in the original LPP is either ge or = type
Then an artificial variable Ri is added to form the ithunit vector column
Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal
zero in the optimum solution
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Artificial variable objective coefficient
= - M (Sum of artificial variables) in a maximization
problem
= sum o t e art c a var a es n a m n m zat onproblem
where M is a very large positive number
Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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983106983137983155983145983139 983162 983160983089
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983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
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Since the z column does not change in the iterations it may be
deleted
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983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
983160983090
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983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983106983137983155983145983139 983160983089
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983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
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102
61
---
Select the most positive element of the z row
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983106983137983155
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983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
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983123983090
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983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
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Applying the usual simplex method for solving aminimization problem
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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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If in a starting simplex tableau we donrsquot have an
identity sub matrix (ie an obvious starting BFS) then
we introduce artificial variables to have a starting BFS
This is known as artificial variable technique
There are two methods to find the starting BFS and
so ve t e prob em ndash
(1) Big M Method
(2) Two-Phase Method
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Big M Method
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Suppose that ith constraint equation doesnt have aslack variable ie
ith constraint in the original LPP is either ge or = type
Then an artificial variable Ri is added to form the ithunit vector column
Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal
zero in the optimum solution
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Artificial variable objective coefficient
= - M (Sum of artificial variables) in a maximization
problem
= sum o t e art c a var a es n a m n m zat onproblem
where M is a very large positive number
Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
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Since the z column does not change in the iterations it may be
deleted
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983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
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983123983091
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The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
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983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
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102
61
---
Select the most positive element of the z row
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983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
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983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
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983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
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form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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Standard simplex table
983089
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Applying the usual simplex method for solving aminimization problem
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Minratio
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983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
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7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Big M Method
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7172019 Big-M Two Phase Methods
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Suppose that ith constraint equation doesnt have aslack variable ie
ith constraint in the original LPP is either ge or = type
Then an artificial variable Ri is added to form the ithunit vector column
Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal
zero in the optimum solution
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Artificial variable objective coefficient
= - M (Sum of artificial variables) in a maximization
problem
= sum o t e art c a var a es n a m n m zat onproblem
where M is a very large positive number
Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983162 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
983088 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983088 983085983091 983090 983088 983088 983088 983088 983089 983094
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
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983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Suppose that ith constraint equation doesnt have aslack variable ie
ith constraint in the original LPP is either ge or = type
Then an artificial variable Ri is added to form the ithunit vector column
Since the artificial variables are not part of theoriginal LPP they are assigned a very high penalty inthe objective function and thus forcing them to equal
zero in the optimum solution
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Artificial variable objective coefficient
= - M (Sum of artificial variables) in a maximization
problem
= sum o t e art c a var a es n a m n m zat onproblem
where M is a very large positive number
Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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983106983137983155983145983139 983162 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
983088 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983088 983085983091 983090 983088 983088 983088 983088 983089 983094
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
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983123983090
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983122983090
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983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
983160983089
983160983090
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983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Artificial variable objective coefficient
= - M (Sum of artificial variables) in a maximization
problem
= sum o t e art c a var a es n a m n m zat onproblem
where M is a very large positive number
Mathematically M rarrinfinrarrinfinrarrinfinrarrinfin
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
7
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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983106983137983155983145983139 983162 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
983088 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983088 983085983091 983090 983088 983088 983088 983088 983089 983094
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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7172019 Big-M Two Phase Methods
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Example 1
Min z = 4x1 + 3x2
Subject to
2x1 + x2 ge 10x1 + x2 ge 6
-3x + 2x le 6
x1ge0 x2ge0 r t ng z as equat on an a ngslack and surplus variables to the
constraints
z ndash 4x1 ndash 3x2 = 0
2x1 +x2 ndash S1 = 10x1 + x2 - S2 = 6
-3x1 + 2x2 + S3 = 6
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7172019 Big-M Two Phase Methods
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
9
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983162 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
983088 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983088 983085983091 983090 983088 983088 983088 983088 983089 983094
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7172019 Big-M Two Phase Methods
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
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983122983090
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form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
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Applying the usual simplex method for solving aminimization problem
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983085
Minratio
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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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Form a simplex table
983106983137983155983145983139 983160983089
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983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
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7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Addition of artificial variables to obtain a starting BFS
Artificial variables
R1 and R2 added tothe constraints
2x1 + x2 ndash S1 +R1 = 10
x1 + x2 ndash S2 +R2 = 6- 3x1 + 2x2 + S3 = 6
These artificial variables are penalized in the objectivefunction by assigning some very large value (say M) to
them in the objective function
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983162 983160983089
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983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
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983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
983160983090
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983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983106983137983155983145983139 983160983089
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983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
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102
61
---
Select the most positive element of the z row
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983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
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983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
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983123983090
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983122983090
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983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
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983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
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983155983090
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983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
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983155983090
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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
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983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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z = 4x1 + 3x2 + MR1 + MR2
2x1 + x2 ndash s1 +R1 = 10
x1 + x2 ndash s2 +R2 = 6- 3x1 + 2x2 + s3 = 6
Objective function
constraints
Where M is a very large penalty assigned onartificial variables R1 and R2
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983162 983160983089
983160983090
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983155983151983148983157983156983145983151983150
983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
983088 983089 983089 983088 983085983089 983088 983089 983088 983094
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7172019 Big-M Two Phase Methods
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
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983123983090
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983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
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983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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7172019 Big-M Two Phase Methods
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
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983155983090
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983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
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983155983090
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983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
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983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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7172019 Big-M Two Phase Methods
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 1051
983106983137983155983145983139 983162 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983089 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122 983088 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Tabular form
983122983090
983088 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983088 983085983091 983090 983088 983088 983088 983088 983089 983094
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7172019 Big-M Two Phase Methods
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
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983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
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983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
983160983090
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983155983090
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983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
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983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
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983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Since the z column does not change in the iterations it may be
deleted
983106983137983155983145983139 983160983089
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983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983085983091 983088 983088 983085 983117 983085 983117 983088 983088
983122983089
983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090 983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
The initial BFS as obtained from the table is
R1 = 10 R2 = 6 and z = 0 Is this tablecorrect
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
983160983089
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983123983090
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983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
983160983090
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983123983090
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983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
7172019 Big-M Two Phase Methods
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
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983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
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form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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Standard simplex table
983089
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Applying the usual simplex method for solving aminimization problem
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Minratio
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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
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983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
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983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
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At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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Form a simplex table
983106983137983155983145983139 983160983089
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983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983106983137983155983145983139 983160983089
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983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
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Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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But if R1 = 10 R2 = 6 then the value ofobjective function should be z = 16 M sincewe have defined the objective function as
z = 4x1 + 3x2 + MR1 + MR2
Wh hi in n i n r
This is because of the fact that the elementsof the basic variable have a non zero
coefficient in the z-row
The starting table now becomes
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983106983137983155983145983139 983160983089
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983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983106983137983155983145983139 983160983089
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983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
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102
61
---
Select the most positive element of the z row
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983162 983088 983085983090983083983117
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983085983092983085983117
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983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
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512
112
2172
983089983087983090
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983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
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Final iteration
983160983090
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Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
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TWO PHASE METHOD
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
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983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
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983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
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7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
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7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
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7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
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983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122983089 983090 983089 983085983089 983088 983089 983088 983088 983089983088
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091 983085983091 983090 983088 983088 983088 983088 983089 983094
Apply the usual simplex method for solvingminimization problem
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983106983137983155983145983139 983160983089
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983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
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102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
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983162 983088 983085983090983083983117
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983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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983106983137983155983145983139 983160983089
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983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
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7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
7172019 Big-M Two Phase Methods
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
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7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
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983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
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983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
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983155983090
983122983089
983122983090
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983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
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983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983085983092 983083 983091983117 983085983091983083983090983117 983085983117 983085983117 983088 983088 983088 983089983094983117
983122 983090 983089 983085983089 983088 983089 983088 983088 983089983088
Selecting pivot row and pivot column for a minimization problem
Minratio
983122983090
983089 983089 983088 983085983089 983088 983089 983088 983094
983123983091
983085983091 983090 983088 983088 983088 983088 983089 983094
102
61
---
Select the most positive element of the z row
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983106983137983155
983145983139
983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
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Tutorial - 3
22
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
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7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
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7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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983106983137983155
983145983139
983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983085983090983083983117
983090
983085983092983085983117
983090
983085983117 983092 991251 983091983117
983090
983088 983088 983090983088 983083983117
Minratio
Second iteration
983160983089
983089 983089 983090 983085983089 983090 983088 983089 983088 983088 983093
983122983090
983088 983089983087983090 983089983087983090 983085983089 983088 983089 983088 983089
983123983091 983088 983095983087983090 983085983091983087983090 983088 983088 983088 983089 983090983089
512
112
2172
983089983087983090
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
16
7172019 Big-M Two Phase Methods
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
20
7172019 Big-M Two Phase Methods
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
7172019 Big-M Two Phase Methods
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 1651
983106983137983155983145983139 983160983089
983160983090
983123983089
983123983090
983122983089
983122983090
983123983091
983155983151983148983157983156983145983151983150
983162 983088 983088 983085983089 983085983090 983089 991251 983117 983080983090983085983117983081983087983090 983088 983090983090
983160983089
983089 983088 983085983089 983088 983089 983088 983088 983092
Final iteration
983160983090
983088 983089 983089 983085983089 983088 983089 983088 983090
983123983091
983088 983088 983085983089983091983087983090 983088 983088 983088 983089 983089983092
Solution x1 = 4 x2 = 2 z = 22
16
7172019 Big-M Two Phase Methods
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Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 1851
Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
20
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2151
If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2251
Tutorial - 3
22
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 1751
Consider the following LPP
Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
17
7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
7172019 Big-M Two Phase Methods
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
20
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2151
If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2251
Tutorial - 3
22
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
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Tutorial - 4
48
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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Introducing surplus and artificial variables s2 R1 and R2 the
LPP is modified as follows
Maximize
Subject to
1 2 3 1 2
2 3 5 z x x x M R M R= + minus minus minus
1 2 3 1 7 x x x R+ + + =
Now we solve the above LPP by the Simplex method
1 2 3 2 2
1 2 3 2 1 2
2 5 1 0 0
x x x s R
x x x s R R
minus + minus + =
ge
18
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
7172019 Big-M Two Phase Methods
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
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7172019 Big-M Two Phase Methods
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
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Tutorial - 3
22
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol
z 1 -2 -3 5 0 M M 0
R1 0 1 1 1 0 1 0 7
z 1 0 -8 - 6 - -1 - 0 1 + 10 -
x1 0 1 -52 12 -12 0 12 5
7M2 M2 M2 3M2 2MR1 0 0 72 12 12 1 -12 2
x2 0 0 1 17 17 27 -17 47
z 1 0 0 507 17 167 + -17 1027M +M
x1 0 1 0 67 -17 57 17 457
7172019 Big-M Two Phase Methods
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The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
20
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2151
If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2251
Tutorial - 3
22
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2951
Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3051
Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2051
The optimum (Maximum) value of
z = 1027
and it occurs at
x1 = 457 x2 = 47 x3 = 0
20
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2151
If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2251
Tutorial - 3
22
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2551
(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
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Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
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Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
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7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
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In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
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Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
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(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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If in any iteration there is a tie for leaving variablebetween an artificial variable and other variable
(decision surplus or slack) we must prefer the
artificial variable to leave the basis
into the basis
If in the final optimal tableau an artificial variable ispresent in the basis at a non-zero level this means our
original problem has no feasible solution
21
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Tutorial - 3
22
7172019 Big-M Two Phase Methods
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(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
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(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
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TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
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Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
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Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
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Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2251
Tutorial - 3
22
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2451
(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2551
(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2951
Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3051
Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2351
(1) Solve the following LPPs using Simplex method
(i) Max
subject to
1 21 2 1 5 z x x= minus minus
1 2
1 2
1 2
4 3 1 22 5 1 0
0
x x x x
x x
+ le
+ le
ge
(ii) Max
subject to
23
1 22 3 z x x= minus
1 2
1 2
1 2
1 2
2
2 2
2
0 u n r e s t r i c e d
x x
x x
x x
x x
minus + le
minus le
minus minus le
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2451
(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2551
(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2451
(iii) Max
subject to
1 2 33 2 5 z x x x= + +
1 2 3
1 3
1 2
1 2 3
2 4 3 0
3 2 4 6 0
4 4 2 0
0
x x x
x x
x x
x x x
+ + le
minus minus ge minus
+ le
ge
(iv) Minsubject to
24
1 2 36 2 6 z x x x= minus minus minus
1 2 3
1 2 3
1 2 3
1 2 3
2 3 1 4
4 4 1 0 4 6
2 2 4 3 7
2 1 3
x x x
x x x
x x x
x x x
minus + le
minus + + le
+ minus le
ge ge ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2551
(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3051
Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
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s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2551
(2) Solve the following LPPs using Big-M method
(i) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 23 3
4
x x
x x
x
+ ge
+ le
le
(ii) Min
subject to
1 2 0 x x ge
25
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 5
2 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
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Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
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983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
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z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
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Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
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(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2651
TWO PHASE METHOD
26
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
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7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
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Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
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This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2751
Why use two phase method when Big M - methodis already there
One major disadvantage of Big M - method is that itis computationally inefficient
Mathematically it is said that M should be a verylarge number but sometimes very large values of Mmay lead to a wrong result
27
7172019 Big-M Two Phase Methods
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Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
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r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
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r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2851
Phase - I
983127983141 983139983151983150983155983156983154983157983139983156 983137983150 983137983157983160983145983148983145983137983154983161 983116983120983120 983080983137983148983159983137983161983155 983137 983149983145983150983145983149983145983162983137983156983145983151983150
983152983154983151983138983148983141983149983081983086
983124983144983141 983151983138 983141983139983156983145983158983141 983157983150983139983156983145983151983150 983145983155 983155983157983149 983151 983137983154983156983145 983145983139983145983137983148 983158983137983154983145983137983138983148983141983155
983159983145983156983144 983156983144983141 983151983154983145983143983145983150983137983148 983139983151983150983155983156983154983137983145983150983156983155
983124983144983141 983151983152983156983145983149983137983148 983155983151983148983157983156983145983151983150 983151983142 983156983144983141 983137983157983160983145983148983145983137983154983161 983152983154983151983138983148983141983149 983159983145983148983148 983143983145983158983141
983137 983155983156983137983154983156983145983150983143 983106983110983123 983142983151983154 983156983144983141 983151983154983145983143983145983150983137983148 983152983154983151983138983148983141983149983086
28
7172019 Big-M Two Phase Methods
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Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
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Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 2951
Phase - II
983127983141 983157983155983141 983156983144983141 983138983137983155983145983139 983142983141983137983155983145983138983148983141 983155983151983148983157983156983145983151983150 983137983158983137983145983148983137983138983148983141 983142983154983151983149 983156983144983141
983085
983152983154983151983138983148983141983149983086
29
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3051
Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3051
Example
Min z = 4x1 + 6x2 + 5x3
Subject to2x1 + 4x2 + 3x3 ge 32x + 2x + 4x ge 28
x1 x2 x3 ge 0
30
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3151
s1 s2 ndash surplus variableR1 R2 ndash artificial variables
Min z = 4x1 + 6x2 + 5x3
Subject to
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Artificial variables are added to get an identity sub matrix
31
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3251
Phase I (construct an auxiliary LPP)
construct a new objective function in which thecoefficients of artificial variables are 1 and thecoefficients of all other variables are zero
Min r = R1 + R2New objective function
2x1 + 4x2 + 3x3 ndash s1 + R1 = 32x1 + 2x2 + 4x3 ndash s2 + R2 = 28x1 x2 x3 ge 0
Constraints
32
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3351
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983122983089
983090 983092 983091 983085983089 983088 983089 983088 983091983090
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
form a new r row (similar to the manner in which new z- row is
formed in Big M method)
ThusNew r-row = r-row + R1 row + R2 row
33
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3451
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Standard simplex table
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096
34
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3551
Applying the usual simplex method for solving aminimization problem
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091 983094 983095 983085983089 983085983089 983088 983088 983094983088
983085
Minratio
983089
983122983090
983089 983090 983092 983088 983085983089 983088 983089 983090983096 284983092
35
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3651
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983122983089
983122983090
983155983151983148983157983156983145983151983150
983154 983091983087983092 983093983087983090 983088 983085983089 983095983087983092 983088 983085983095983087983092 983089983089
983122983089
983093983087983092 983093983087983090 983088 983085983089 983091983087983092 983089 983085983091983087983092 983089983089983093983087983090
983160983091
983089983087983092 983089983087983090 983089 983088 983085983089983087983092 983088 983089983087983092 983095
36
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3751
983106983137983155983145983139 983160983089 983160983090 983160983091 983155983089 983155983090 983122983089 983122983090 983155983151983148983157983156983145983151983150
983154 983088 983088 983088 983088 983088 983085983089 983085983089 983088
983160983090
983089983087983090 983089 983088 983085983090983087983093 983091983087983089983088 983090983087983093 983085983091983087983089983088 983090983090983087983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983085983089983087983093 983090983087983093 983090983092983087983093
At this point both the artificial variables have completed theirpurpose and hence their columns can be removed from thetable
37
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3851
Phase II
We will use the BFS obtained in Phase-I as a starting
BFS of the given problem and use simplex method tofind the optimal solution
38
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 3951
Form a simplex table
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085983092 983085983094 983085983093 983088 983088 983088
983160 983089 983090 983089 983088 983085983090 983093 983091 983089983088 983090983090 983093
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Is this standard simplex table
39
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4051
983106983137983155983145983139 983160983089
983160983090
983160983091 983155983089
983155983090
983155983151983148983157983156983145983151983150
983162 983085 983089 983088 983088 983085983089983090983087983093 983085983090 983090983093983090983087983093
z ndashrow = z row + 6(x2 row) + 5(x3 row)
983090
983160983091
983088 983088 983089 983089983087983093 983085983090983087983093 983090983092983087983093
Since all the conditions for optimality (for a minimization problem) arebeing satisfied Phase two ends at this point
40
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4151
Solve the following LPP using Two-Phase method
Minimize
Sub ect to
321 352 x x x z ++=
1 2 3
1 2 3
1 2 3
2 4 5 0
0
x x x
x x x
x x x
minus +
+ + =
ge
41
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4251
Phase I
Minimize
Subject to
1 2r R R= +
1 2 3 1 12 20 x x x s Rminus + minus + =
=1 2 3 2
1 2 3 1 1 2 0 x x x s R R ge
42
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4351
r 1 0 0 0 0 -1 -1 0
Basic r x1 x2 x3 s1 R1 R2 Sol
3 2 2 -1 0 0 70
R1 0 1 -2 1 -1 1 0 20
R2 0 2 4 1 0 0 1 50
x1 0 1 -2 1 -1 1 0 20
r 1 0 8 -1 2 -3 0 10
R2 0 0 8 -1 2 -2 1 10
43
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4451
r 1 0 8 -1 2 -3 0 10
Basic r x1 x2 x3 s1 R1 R2 Sol
x1 0 1 -2 1 -1 1 0 20
R2 0 0 8 -1 2 -2 1 10
x1 0 1 0 34 -12 12 14 452
r 1 0 0 0 0 -1 -1 0
x2 0 0 1 -18 14 -14 18 54
44
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4551
This is optimal tableau Thus Phase I has ended and wenow start Phase II with the starting BFS as the optimal
solution of Phase I
Phase II
subject to the same constraints as given in the original
problem
321
45
Basic z x1 x2 x3 s1 R1 R2 Sol
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4651
z 1 -2 -5 -3 0 0
1 2 3 1 1 2
x1 0 1 0 34 -12 452
x2 0 0 1 -18 14 54
- -
0 0 -178 14 2054
x1 0 1 2 12 0 25
s1 0 0 4 -12 1 5
Thus the optimal solution is x1 = 25 x2 = 0 x3 = 0 and the
optimal z = Min z = 5046
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4751
In Phase I we always minimize the sum of the
artificial variables
If in the optimal table of phase I an artificial variablesrema ns at non-zero eve n t e as s t en t e pro em
has no feasible solution
47
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4851
Tutorial - 4
48
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 4951
Solve the following LPPs using Two-Phase Method
(i) Maximize
Subject to
1 2 32 3 5 z x x x= + minus
1 2 3 7 x x x+ + =
1 2 3
1 2 3
2 5 10
0
x x x
x x x
minus + ge
ge
49
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5051
(ii) Maximize
Subject to
321 422 x x x z ++=
1 2 32 2 x x x+ + le
1 2 3
1 2 3 0
x x x
x x x
+ +
ge
50
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge
7172019 Big-M Two Phase Methods
httpslidepdfcomreaderfullbig-m-two-phase-methods 5151
(iii) Max
subject to
1 23 z x x= minus +
1 2
1 2
1
2 2
3 34
0
x x
x x
x
x x
+ ge
+ le
le
ge
(iv) Min
subject to
51
1 2 39 2 z x x x= + +
1 2 3
1 2 3
1 2 3
4 2 52 3 4
0
x x x
x x x
x x x
+ + ge
+ + ge
ge