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Lecture 1 f:VARIANTS OF SIMPLEX METHOD

Jeff Chak-Fu WONG

Department of Mathematics

Chinese University of Hong Kong

MAT581 SSMathematics for Logistics

Produced by Jeff Chak-Fu WONG 1

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Example 1 Use two phase simplex method to

Miximise z = 2x 1 x 2

subject tox 1 + x2 2

x 1 + x2 4

x 1 , x 2 0

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Solution:

Convert the LPP into maximisation by using

min z = max ( z),

the LPP becomes

Maximise ( z) = 2 x 1 + x2

subject to

x 1 + x2 2

x 1 + x2 4x 1 , x 2 0

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By introducing slack variable s 2 , surplus variable s 1 , articial variableA1 , the standard form of LPP is

max ( z) = 2 x 1 + x2

subject to

x 1 + x2 s1 + 0 s 2 + A1 = 2

x 1 + x2 + 0 s 1 + s2 = 4

x 1 , x 2 , s 1 , s 2 , A 1 0

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The initial basis matrix is:

B = sA 1 s2 = I 2 .

An obvious initial basis feasible solution is s2 = 4 , A 1 = 2 , i.e.,

X B = B 1

b = I 2 b = [2 4]T

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y 1 = B 1

a 1 = I 2 [1 1]T = [1 1]T

y 2 = B 1

a 2 = I 2 [1 1]T = [1 1]T

y 1 = B 1

s 1 = I 2 [ 1 0]T = [ 1 0]T

y 2 = B 1

s 2 = I 2 [0 1]T = [0 1]T

y A 1 = B 1

s A 1 = I 2 [1 0]T = [1 0]T

z1 c1 = C B x 1 c1 = [ 1 0]1 1]T 0 = 1

z2 c2 = C B x 2 c2 = [ 1 0][1 1]T 0 = 1

z3 c3 = C B s 1 c3 = [ 1 0][ 1 0]T

0 = 1z4 c4 = C B s 2 c4 = [ 1 0][0 1]T 0 = 0

z5 c5 = C B y A 1 c5 = [ 1 0][0 1]T ( 1) = 0

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Phase I The objective function of the auxiliary LPP ismax ( z ) = A1 .

Starting Tableau:Tableau 6.1.1

cj : (0 0 0 0 -1)

y 1 y 2 y 3 y 4 y A 1

C B X B x1

x2

s1

s2

A1

Min. Ratio-1 A1 2 1 1 -1 0 1 2/1=2

0 s2 4 1 1 0 1 0 4/1=4

zj cj -1 -1 1 0 0

From Tableau 6.1.1 we observe that

the non-basic variable x1 enters into the basis;

the basic variable A1 leaves the basis.

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cj : (0 0 0 0 -1)

C B X B x1 x2 s1 s2 A1 Min. Ratio-1 A1 2 1 1 -1 0 1 2/1=2

0 s2 4 1 1 0 1 0 4/1=4 R2 R 1

zj cj

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cj : (0 0 0 0 -1)

C B X B x1 x2 s1 s2 A1

0 x1 2 1 1 -1 0 1

0 s2 2 0 0 0 1 1

zj cj

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The new basis matrix is:

B = sA 1 s2 = I 2 B = a 1 s2 .

Find the inverse of B , i.e., B 1 , we have

B I 2 I 2 B 1 .

That is,

1 0 1 0

1 1 0 1 1 0 1 0

0 1 1 1 .

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y 1 = B 1

a 1 = [1 0]T

y 2 = B 1

a 2 = [1 0]T

y 3 = B 1 s 1 = [ 1 1]T

y 4 = B 1

s 2 = [0 1]T

z1 c1 = C B y 1 c1 = [0 0][1 0]T 0 = 0

z2 c2 = C B y 2 c2 = [0 0][1 0]T 0 = 0

z3 c3 = C B y 3 c3 = [0 0][ 1 1]T

0 = 0z4 c4 = C B y 4 c4 = [0 0][0 1]T 0 = 0

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First iteration:

Tableau 6.1.2cj : (0 0 0 0)

C B X B x1 x2 s1 s2

0 x1 2 1 1 -1 0

0 s2 2 0 0 1 1

zj cj z (X B ) = 0 0 0 0 0

Since all zj cj 0 and no articial variable appears in theoptimum basis, the current basic feasible solution is optimal to theauxiliary LPP and we proceed to Phase II .

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Phase II Assign the actual costs to the original variables. The newobjective function then becomes

Maximise ( z) = 2 x1

+ x2

+ 0 s1

+ 0 s2

The initial basic feasible solution for this phase is the one obtained atthe end of Phase I .

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The initial basis matrix is:

B = B = a 1 s2 .

An obvious initial basis feasible solution is x1 = 2 , s 2 = 2 , i.e.,

X B = B 1

b = [2 2]T

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y 1 = [1 0]T

y 2 = [1 0]T

y 3 = [ 1 1]T

y 4 = [0 1]T

z1 c1 = C B y 1 c1 = [2 0][1 0]T ( 2) = 0

z2 c2 = C B y 2 c2 = [2 0][1 0]T ( 1) = 1

z3 c3 = C B y 3 c3 = [2 0][ 1 1]T

0 = 2z4 c4 = C B y 4 c4 = [2 0][0 1]T 0 = 0

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The iterative simplex tableaus for this phase is:

Initial iteration:Tableau 6.1.3cj : (2 1 0 0)

y 1 y 2 y 3 y 4

C B X B x1 x2 s1 s2 Min. Ratio2 x1 2 1 1 -1 0

0 s2 2 0 0 1 1 2/1=2

z

j cj z(X B ) = 4 0 1 -2 0From Tableau 6.1.3, we observe that s1 enters into the basis and s 2leaves the basis.

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y 1 y 2 y 3 y 4


0 s2 2 0 0 1 1 2/1=2

z


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y 1 y 2 y 3 y 4


0 s2 2 0 0 1 1 2/1=2

z


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Tableau 6.1.3cj : (2 1 0 0)

C B X B x1 x2 s1 s2 Min. Ratio

2 x1 2 1 1 -1 0 R1 + R2

0 s2 2 0 0 1 1 2/1=2

zj cj z(X B ) = 4 0 1 -2 0 R3 + 2 R 2

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Tableau 6.1.3cj : (2 1 0 0)


2 x1 4 1 1 0 10 s1 2 0 0 1 1

zj cj z(X B ) = 8 0 1 0 2

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First iteration:Tableau 6.1.4

cj : (2 1 0 0)

C B X B x1 x2 s1 s2

2 x1 4 1 1 0 1

0 s1 2 0 0 1 1

z

j cj z(X B ) = 8 0 1 0 2

Since all zj cj 0, the current basic feasible solution is optimal.

The optimal solution is

x 1 = 4 , x 2 = 0 with Maximum of ( z) = 8

That is, x1 = 4 , x 2 = 0 with minimum of z = 8

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VARIANTS OF SIMPLEX METHOD

ARIANTS OF SIMPLEX METHOD 23

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In this section we discuss some complications and variations which

are very often encountered during simplex procedure.The following are some of them:

1. Degeneracy and cycling

2. Unbounded solution

3. Multiple solution or alternative optimum solution

4. No feasible solution

5. Unrestricted variables.

ARIANTS OF SIMPLEX METHOD 24

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1. DEGENERACY (TIE FOR MINIMUM RATIO)

. D EGENERACY (TIE FOR MINIMUM RATIO) 25

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But sometimes this ratio may not be unique

(i.e. more than one variable is eligible to leave the basis)

or

at the very rst iteration, the value of one or more basic variablesin X B column becomes zero ,

this causes the problem of degeneracy.


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If the minimum ratio is zero, for two or more basic variables,degeneracy may result and the simplex routine to cycleindenitely.

That is, the solution which we have obtained in one iterationmay repeat again after few iterations and therefore, nooptimum solution may be obtained.

This concept is known as cycling or circling .


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Step 1: First nd out the rows for which the minimum non-negativeratio is the same (tie), for example, suppose there is a tie betweenrst and second row.

Step 2: Now, rearrange the columns of the usual simplex table sothat the columns forming the original unit matrix come rst.

Step 3: Find the minimum of the ratio

elements of rst column of unit matrixcorresponding elements of key column

Only for the rows for which the minimum ratio was not unique(i.e. for tied rows, in our example for rst and second rows).

(i) If this minimum is attained for second row (say) then this row willdetermine the key element by intersecting with the key column.

(ii) If this minimum is also not unique, then go to the next step.

ETHOD TO RESOLVE DEGENERACY (TIE) 30

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Step 4: Compute the minimum of the ratioelements of second column of unit matrix

corresponding elements of key column

only for the rows for which minimum ratio is not unique in Step 3

.If the minimum is also not unique, then go to next step.

Step 5: Next compute the minimum of the ratio

elements of third column of unit matrixcorresponding elements of key column

If this minimum is still not unique, then go on repeating the

above procedure till the unique minimum ratio is obtained toresolve the degeneracy.

After the resolution of this tie, simplex method is applied toobtain the optimum solution.


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Remark 1 If we have a tie for an articial variable and some other

variable we can choose the articial variable to leave the basis.


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Example 2 Solve the LPP

Maximise z = 3 x 1 + 9 x 2

subject to

x 1 + 4 x 2 8

x 1 + 2 x 2 4

x 1 , x 2 0


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Solution:

Introduce the slack variables s1

, s2

0, the LPP becomesMaximise z = 3 x 1 + 9 x 2 + 0 s 1 + 0 s 2

subject to

x 1 + 4 x 2 + s1 = 8x 1 + 2 x 2 + + s2 = 4

x 1 , x 2 , s 1 , s 2 0

An obvious initial basic feasible solution is s1 = 8 , s 2 = 4 .


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Tableau 6.2.1cj : (3 9 0 0)


0 s1

8 1 4 1 00 s2 4 1 2 0 1

zj cj z(X B ) -3 -9 0 0

From Tableau 6.2.1, we observe that the non-basic variable x2enters into the basis.

Since the minimum ratio is 2 for both the slack variables s1 ands 2 , there is a tie for the variable to leave the basis.

This is an indication for the existence of degeneracy in the givenLPP.

Rearrange the columns of the simplex table so that the initial

identity matrix appears rst.


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Tableau 6.2.1cj : (3 9 0 0)


0 s1

8 1 4 1 00 s2 4 1 2 0 1

zj cj z(X B ) -3 -9 0 0




Rearrange the columns of the simplex table so that the initialidentity matrix appears rst.


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Tableau 6.2.1cj : (3 9 0 0)


0 s1

8 1 4 1 0 8/4= 20 s2 4 1 2 0 1 4/2= 2

zj cj z(X B ) -3 -9 0 0






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Tableau 6.2.1cj : (3 9 0 0)


0 s1

8 1 4 1 0 8/4= 20 s2 4 1 2 0 1 4/2= 2

zj cj z(X B ) -3 -9 0 0






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Now, using the Step 3 of the procedure for resolving degeneracy,we nd

min elements of rst column


= min 14

, 02

= 0 .

which occurs for the second row. Hence, s2 must leave the basisand the pivot element is 2.

Tableau 6.2.2cj : (0 0 3 9)

C B X B s1 s2 x1 x2 Min. Ratio0 s1 8 1 0 1 4 1/4

0 s2 4 0 1 1 2 0/2=0

zj cj z(X B ) = 0 0 0 -3 -9


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Now, using the Step 3 of the procedure for resolving degeneracy,we nd

min elements of rst column


= min 14

, 02

= 0 .

which occurs for the second row. Hence, s2 must leave the basisand the pivot element is 2.

Tableau 6.2.2cj : (0 0 3 9)

C B X B s1 s2 x1 x2 Min. Ratio0 s1 8 1 0 1 4 1/4

0 s2 4 0 1 1 2 0/2=0

zj cj z(X B ) = 0 0 0 -3 -9


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Tableau 6.2.2cj : (0 0 3 9)

C B X B s1 s2 x1 x2 Min. Ratio

0 s1 8 1 0 1 4 1/4

0 s2 4 0 1 1 2 0/2=0 12 R 2

zj cj z(X B ) = 0 0 0 -3 -9

Tableau 6.2.2cj : (0 0 3 9)

C B X B s1 s2 x1 x2 Min. Ratio

0 s1 0 1 -2 -1 0 1/4 R1 4 R 2

0 s2 2 0 1/2 1/2 1 0/2=0

zj cj z(X B ) = 18 0 9/2 3/2 0 R3 + 9 R 2


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cj : (0 0 3 9)

C B X B s1 s2 x1 x2

0 s1 0 1 -2 -1 0

9 x2 2 0 1/2 1/2 1

zj cj z(X B ) = 18 0 9/2 3/2 0

Since all zj cj 0, an optimum solution has been reached.

Hence, the optimum basic feasible solution is

x 1 = 0 , x 2 = 2 with maximum of z = 18


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2. Unbounded solutions: The case of unbounded solutions occur when the feasible

region is unbounded such that the value of the objectivefunction can be increased indenitely.

It is not necessary, however, that an unbounded feasible regionwill give rise to an unbounded value of the objective function.

The following example will illustrate these points.

. Unbounded solutions: 44

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UNBOUNDED SOLUTION SPACE BUT BOUNDED OPTIMAL SOLUTION

Example 3 Solve the following LPP

Maximise z = 3 x 1 x 2

subject to

x 1 x 2 10

x 1 20

x 1 , x 2 0

NBOUNDED SOLUTION SPACE BUT BOUNDED OPTIMAL SOLUTION 45

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Solution:

By introducing the slack variables s1 , s 2 0, the LPP becomes

Maximise z = 3 x 1 x 2 + 0 s 1 + 0 s 2

subject to

x 1 x2 + s1 = 10

x 1 + s2 = 20

x 1 , x 2 , s 1 , s 2 0



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Tableau 6.3.3c j : (3 -1 0 0)

y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio

3 x 1 10 1 -1 1 00 s 2 10 0 1 -1 1

z j c j z(X B ) = 30 0 -2 3 0

Second Iteration

Tableau 6.3.4c j : (3 -1 0 0)


3 x 1 10 1 -1 1 0

0 s 2 10 0 1 -1 1 10/1 = 10

z j c j z(X B ) = 30 0 -2 3 0


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Tableau 6.3.5 c j : (3 -1 0 0)y 1 y 2 y s 1 y s 2

C B X B x 1 x 2 s 1 s 2 Min. Ratio

3 x 1 10 1 -1 1 0 R 1 + R 20 s 2 10 0 1 -1 1 10/1 = 10

z j c j z(X B ) = 30 0 -2 3 0 R 3 + 2 R 2

Tableau 6.3.6c j : (3 -1 0 0)


3 x 1 20 1 0 0 1

-1 x 2 10 0 1 -1 1

zj

cj

z(X

B ) = 50 0 0 1 2

The optimum solution is:

x 1 = 20 , x 2 = 10 with maximum of z = 50


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The graphical solution of the LPP is given in Figure 1.

0

x = 20

x - x = 10

B(20,10)

A(10,0)

15105 20

5

10

x1

1

1 2

x2

15

(a)

0

x = 20

x - x = 10

B(20,10)

A(10,0)

15105 20

5

10

x1

1

1 2

x2

15

(b)

0

x = 20

x - x = 10

B(20,10)

A(10,0)

15105 20

5

10

x1

1

1 2

x2

15

(c)

Figure 1:

We observe that the solution space is unbounded. But the optimalsolution occurs at the vertex (20, 10) .


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BOUNDED SOLUTION SPACE AND UNBOUNDED O PTIMAL SOLUTION


Maximise z = 2 x 1 + x2

subject tox 1 x 2 10

2x 1 x 2 40

x 1 0, x 2 0

NBOUNDED SOLUTION SPACE AND UNBOUNDED O PTIMAL SOLUTION 51

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Solution:

Introducing the slack variables s 1 , s 2 0 the LPP in standard form is

Maximise z = 2 x 1 + x2 + 0 s 1 + 0 s 2

subject to

x 1 x2 + s1 = 10

2x 1 x2 + s2 = 40

x 1 , x 2 , s 1 , s 2 0

An obvious IBFS is s1 = 10 , s 2 = 40 .


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Starting Table:Tableau 6.4.1

c j : (2 1 0 0)


0 s 1 10 1 -1 1 0

0 s 2 40 2 -1 0 1

z j c j z(X B ) = 0 -2 -1 0 0

Tableau 6.4.2c j : (2 1 0 0)


0 s 1 10 1 -1 1 0 10

1 = 100 s 2 40 2 -1 0 1 402 = 20

z j c j z(X B ) = 0 -2 -1 0 0


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Tableau 6.4.3c j : (2 1 0 0)

y 1 y 2 y s 1 y s 2

C B X B x 1 x 2 s 1 s 2 Min. Ratio0 s 1 10 1 -1 1 0

0 s 2 40 2 -1 0 1 R 2 2 R 1

z j c j z(X B ) = 0 -2 -1 0 0 R 3 + 2 R 1

Tableau 6.4.4c j : (2 1 0 0)


2 x 1 10 1 -1 1 0

0 s 2 20 0 1 -2 1

z j c j z(X B ) = 20 0 -3 2 0


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Second iteration:

Table 6.4.8c j : (2 1 0 0)


2 x 1 30 1 0 -1 1

1 x 2 20 0 1 -2 1

z j c j z(X B ) = 80 0 0 -4 3

From Table 6.4.8 we see that s 1 column is the pivotal column,but there is no positive element in that column.

Hence, there exists an unbounded solution to the given LPP. If we solve the LPP by graphical method we can see that the

feasible region is unbounded. See Figure 2.


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0x1

x2

5 10 20

10

20

30

40

10

20

P(30,20)

x - x = 101 2

2 x - x = 401 2

30

Figure 2:


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3. M ULTIPLE SOLUTIONS OR ALTERNATE O PTIMAL SOLUTIONS While solving the LPP by simplex method, in the optimum simplex

table, if the net evaluation zj cj = 0 for all non-basic variables,then the problem is said to have a unique optimal solution.

On the other hand, if the net evaluation zj cj = 0 for at leastone non-basic variable, then the problem is said to have analternative or innite number of solutions.

. M ULTIPLE SOLUTIONS OR A LTERNATE O PTIMAL SOLUTIONS 60

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In a graphical method if the optimal solution occurs at a vertexof the solution space, then the problem said to havea unique optimal solution.

If the optimum solution occurs on an edge of the solutionspace, then the problem is said to have an alternative or innitenumber of solutions .

. M ULTIPLE SOLUTIONS OR A LTERNATE O PTIMAL SOLUTIONS 61

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MULTIPLE SOLUTIONS OR ALTERNATE O PTIMAL SOLUTIONS


Maximise z = x1 + 12

x 2

subject to2x 1 + x2 4

x 1 + 2 x 2 3

x 1 , x 2 0

ULTIPLE SOLUTIONS OR A LTERNATE O PTIMAL SOLUTIONS 62

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Solution:

By introducing the slack variables s1 , s 2 0, the LPP becomes

Maximise z = x1 + 12

x 2 + 0 s 1 + 0 s 2

subject to

2x 1 + x2 + s1 = 4

x 1 + 2 x 2 + s2 = 3

x 1 , x 2 , s 1 , s 2 0



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First Iteration

Tableau 6.5.1c j : (1

12 0 0)


0 s 1 4 2 1 1 0

0 s 2 3 1 2 0 1

z j c j z(X B ) = 0 -1 12 0 0


12 0 0)


0 s 1 4 2 1 1 0 42 = 2 12 R 1

0 s 2 3 1 2 0 1 31 = 4

z j c j z(X B ) = 0 -1 12 0 0


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Tableau 6.5.3

c j : (1 1

2 0 0)


0 s 1 2 1 1/2 1/2 0 42 = 20 s 2 3 1 2 0 1 31 = 4 R 2 R 1

z j c j z(X B ) = 0 -1 12 0 0 R 3 + R 1

Tableau 6.5.4c j : (1 12 0 0)


1 x 1 2 1 1/2 1/2 0

0 s 2 1 0 3/2 -1/2 1

z j c j z(X B ) = 2 0 0 1/2 0


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12 0 0)


1 x 1 2 1 1/2 1/2 0 21 / 2 = 4

0 s 2 1 0 3/2 -1/2 1 13 / 2 = 2 / 3 2

3 R 2

z j c j z(X B ) = 2 0 0 1/2 0

Tableau 6.5.6

c j : (1 12 0 0)


1 x 1 2 1 1/2 1/2 0 2

1 / 2 = 4 R 1 1

2 R 20 s 2 23 0 1 -1/3 2/3

13 / 2 = 2 / 3

z j c j z(X B ) = 2 0 0 1/2 0


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Tableau 6.5.7

c j : (1 12 0 0)

y 1 y 2 y s 1 y s 2

C B X B x 1 x 2 s 1 s 2 Min. Ratio1 x 1 53 1 0 2/3 -1/312 x 2

23 0 1 -1/3 2/3

z j c j z(X B ) = 2 0 0 1/2 0


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NO FEASIBLE SOLUTION OR NON - EXISTING FEASIBLE SOLUTION

O FEASIBLE SOLUTION OR N ON -EXISTING FEASIBLE SOLUTION 68

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In an LPP, where there is no point in the solution space satisfyingall the constraints, then the problem is said to have no feasiblesolution .

In simplex method, if there exists at least one articial variable inthe basis at positive level, and even though optimalityconditions are satised which is the indication of non-feasiblesolution.


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C.f. Example 2 in Lecture Note 1d.


subject to

2x 1 + x2 23x 1 + 4 x 2 12

x 1 , x 2 0.


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cj : (3 2 0 0 M )C B X B x1 x2 s1 s2 A1

2 x2 2 2 1 1 0 0

M A 1 4 -5 0 -4 -1 1

zj cj z(X B ) = 4M + 4 5M + 1 0 4M + 2 M 0

Here, the coefcient of M in each z j c j 0 , and an articial variableA 1 appears in the basis at non-zero level.

Thus, the given LPP does not possess any feasible solution.

We can say the LPP possess a pseudo-optimal solution.


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UNRESTRICTED VARIABLE

NRESTRICTED VARIABLE 72

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In an LPP, if any variable is unrestricted (it can have positivevalue or negative value or zero value) it can be expressed asthe difference between two non-negative variables.

The problem can be converted into an equivalent one involvingonly non-negative variables.


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Example 6 Solve the LPP


subject to

x 1 + 2 x 2 4x 1 + x2 6

x 1 + 3 x 2 9

and x1 , x 2 are unrestricted.


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Tableau 6.6.1c j : (2 -2 3 -3 0 0 0)

y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3C B X B x

1 x

1 x

2 x

2 s 1 s 2 s 3 Min Ratio

0 s1

4 -1 1 2 -2 1 0 00 s 2 6 1 -1 1 -1 0 1 0

0 s 3 9 1 -1 3 -3 0 0 1

z j c j z(X B ) = 0 -2 2 -3 3 0 0 0

Tableau 6.6.2 c j : (2 -2 3 -3 0 0 0)y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3

C B X B x

1 x

1 x

2 x


0 s 1 4 -1 1 2 -2 1 0 0 4/2 = 2

0 s 2 6 1 -1 1 -1 0 1 0 6/1 = 60 s 3 9 1 -1 3 -3 0 0 1 9/3 = 3

z j c j z(X B ) = 0 -2 2 -3 3 0 0 0


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Tableau 6.6.3c j : (2 -2 3 -3 0 0 0)


1 x

1 x

2 x


0 s 1 2 -1/2 1/2 1 -1 1/2 0 0 4/2 = 2 12 R 10 s 2 6 1 -1 1 -1 0 1 0 6/1 = 6

0 s 3 9 1 -1 3 -3 0 0 1 9/3 = 3

z j c j z(X B ) = 0 -2 2 -3 3 0 0 0

Tableau 6.6.4c j : (2 -2 3 -3 0 0 0)


1 x

1 x

2 x


3 x 2 2 -1/2 1/2 1 -1 1/2 0 0 4/2 = 2

0 s 2 4 3/2 -3/2 0 0 -1/2 1 0 R 2 R 10 s 3 3 5/2 -5/2 0 0 -3/2 0 1 R 3 3 R 1

z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0 R 4 + 3 R 1


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Tableau 6.6.5c j : (2 -2 3 -3 0 0 0)


1 x

1 x

2 x


3 x 2 2 -1/2 1/2 1 -1 1/2 0 0

0 s 2 4 3/2 -3/2 0 0 -1/2 1 0

0 s 3 3 5/2 -5/2 0 0 -3/2 0 1

z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0


C B X B x

1 x

1 x

2 x


3 x 2 2 -1/2 1/2 1 -1 1/2 0 0

0 s 2 4 3/2 -3/2 0 0 -1/2 1 0 4/(3/2) = 2.6667

0 s 3 3 5/2 -5/2 0 0 -3/2 0 1 3/(5/2) = 1.2 25 R 3

z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0


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Tableau 6.6.7c j : (2 -2 3 -3 0 0 0)


1 x

1 x

2 x

2 s 1 s 2 s 3 M. R.

3 x 2 2 -1/2 1/2 1 -1 1/2 0 0 R 1 + 1

2 R 30 s 2 4 3/2 -3/2 0 0 -1/2 1 0 R 2 32 R 3

0 s 3 6/5 1 -1 0 0 -3/5 0 2/5

z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0 R 4 + 7

2 R 3


C B X B x

1 x

1 x

2 x

2 s 1 s 2 s 3 M. R.

3 x 2 13/5 0 0 1 -1 1/5 0 1/5

0 s 2 11/5 0 0 0 0 2/5 1 -3/5

2 x 1 6/5 1 -1 0 0 -3/5 0 2/5

z j c j z(X B ) = 51 / 5 0 0 0 0 -3/5 0 7/5


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If we solve the LPP by ordinary simplex method, the optimumsolution is

x 1 = x 1 x 1 = 9 / 2 0 = 9 / 2

x 2 = x 2 x 2 = 3 / 2 0 = 3 / 2with maximum of z = 27 / 2


5.1 lecture_1f

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