chap4 duality and dual simplex.pdf
TRANSCRIPT
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CHAPTER 4
Duality of Linear Programming
4.1 INTRODUCTION
One of the interesting features of linear programming is duality. For every linear
programming problem, there is a twin linear programming problem that has a special and
unique relationship to the first one. These two problems stand as pairs, or duals of each other.
Not only is duality a rather nice theoretical relationship; it has also proved to be of immense
value in devising other operations research techniques. Furthermore, duality has a useful
economic interpretation and is widely used in economic theory. Besides being of theoretical
interest, duality is at the core of sensitivity analysis in linear programming. That, however, is
the topic of Chapter 5. Chapter 4 assumes that you have a thorough grasp of Chapter 3.
This programme can be rewritten by transposing (reversing) the rows and columns of the
algebraic statement of the problem. Inverting the programme in this way results in dual (D)
programme. A solution to the dual programme may be found in a manner similar to that used
for the primal(P). The two programmes have very closely related properties so that optimal
solution of the dual problem gives complete information about the optimal solution of the
primal problem and vice versa.
Duality is an extremely important and interesting feature of linear programming. The various
useful aspects of this property are
(i) If the primal problem contains a large number of rows (constraints) and a smaller number
of columns (variables), the computational procedure can be considerably reduced by
converting it into dual and then solving it. Hence it offers an advantage in many applications.
(ii) It gives additional information as to how the optimal solution changes as a result of the
changes in the coefficients and the formulation of the problem. This forms the basis of post
optimality or sensitivity analysis.
(iii) Duality in linear programming has certain far reaching consequences of economic nature.
This can help managers answer questions about alternative courses of action and their relative
values.
(iv) Calculation of the dual checks the accuracy of the primal solution.
(v) Duality in linear programming shows that each linear programme is equivalent to a two-
person zero-sum game for player A and player B. This indicates that fairly close relationships
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exist between linear programming and the theory of games.
(vi) Duality is not restricted to linear programming problems only but finds application in
economics, management and other fields. In economics it is used in the formulation of input
and output systems.
(vii) Economic interpretation of the dual helps the management in making future decisions.
(viii) Duality is used to solve L.P. problems (by the dual simplex method) in which the initial
solution is infeasible.
4.2 THE DUAL PROBLEM
The power generating problem was viewed in example 2.7 in Chapter 2 as a problem of
allocating scarce resources. Let us now look at this problem from an entirely different angle.
The county council, which is the largest customer of the power generating plant, is
considering making an offer to purchase the plant. In order to make such an offer, the council
needs to determine fair prices for the existing plant resources. For our purpose these resources
can be viewed as the available loading capacity, the available pulverizer capacity, and the
available capacity to emit smoke. (We shall neglect the "capacity" to emit sulfur for the
moment and reintroduce it later on.)
Theoretically, the prices of resources are not necessarily related to their average or marginal
costs, but rather to the revenues that they can produce. Economists tell us that as long as the
price offered for a resource is less than the marginal revenue product of the resource, i.e., the
revenue produced by the last unit of the resource employed, the firm has no incentive to sell
any of this resource. The marginal revenue products can also be viewed as the prices the firm
should be willing to pay for additional amounts of scarce resources. In linear programming,
these prices or marginal revenue products are called imputed values or shadow prices-
imputed because they are not actual costs or prices, but the prices or values that can be
inferred from the particular productive system in question.
The problem of finding these prices turns out to be another linear program. In our example,
the resources are used to produce steam. Hence, rather than express the prices in monetary
units, we shall express them in terms of steam equivalents. Furthermore, since the original
resource allocation problem is on a per-hour basis, the prices of the resources will be on the
basis of per-hour use. Let
y1 be the steam that can be produced by using up 1 kg of smoke emission capacity,
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y2 be the steam that can be produced by 1 ton of loading capacity,
y3 be the steam that can be produced by 1 hour of pulverizer capacity.
The objective of the problem is to find prices 321 yandy,y that minimize the council's total
cost of acquiring the resources presently owned by the firm. The cost of acquiring the smoke
capacity is 1y12 (= quantity available price); the cost of the loading capacity is 2y20 and the cost of the pulverizer capacity is 3y1 . So the objective function is as follows:
321 yy20y12Minimize ++ The prices that the firm will accept depend on what the resources can do for the firm. The
firm will insist on prices that give a return that is at least equal to the return produced by each
of the two activities in which the resources are used, namely, burning coal A and burning coal
B.
In burning 1 ton of coal A, the firm produces 24 units of steam. The resources required to
burn 1 ton of coal A are 0. 5,kg/hr of smoke emission capacity, 1 ton of loading capacity, and
1/16 hr of pulverizer capacity. At the prices 321 yandy,y , the council will pay
321 y161yy5.0 ++ , per hour for these resources. However, since the firm can already make 24
units of steam from 1 ton of coal A, the council must be willing to pay (per hour) at least 24
units of steam for these resources for the firm to find their offer acceptable, i.e.,
24y161yy5.0 321 ++
Similarly, for coal B,
20y241yy 321 ++
0yandy,y 321 Let us write out this linear program again and compare it with problem (2.7) of Chapter 2
(without the sulfur constraint):
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Original problem (allocation of resources): New problem (pricing of resources):
)4(
0x,x0x1800x120020xx
1x241x
161
12xx5.0tosubject
x20x24zMaximize
21
21
21
21
21
21
+++
+=
)B4(
0y,y,y
20y241yy
24y161yy5.0
tosubjectyy20y12zMinimize
321
321
321
321
++
++
++=
How are the problems (4-A) and (4-B) related?
Each problem is called the dual of the other problem. The relationship between them is two-
way: what applies from problem (4-A) to problem (4-B) also applies from (4-B) to (4-A).
Some algebraic manipulations are needed to show this for property 4 of DRI. In the
terminology of linear programming, we call one problem the primal and the other the dual. It
does not matter which problem is called the primal and which is called the dual. Normally,
the problem we formulate first is referred to as the primal, the other becomes the dual. In this
case, problem (4-B) is the primal, and problem (4-A) is the dual. (Note our convention of
denoting the value of the dual objective function by a z/ and the value of the primal objective
function by a lowercase z.)
DUALITY RELATION 1 (DRI)
1. Each constraint in one problem is associated with a variable in the other and vice versa,
2. The LHS coefficients of each constraint of one problem are the same as the LHS
coefficients of the corresponding variable of the other problem.
3. The RHS parameters of one problem are the objective function coefficients of the
corresponding variables in the other problem and vice versa.
4. One problem is a minimizing problem with constraints and nonnegative variables, and the other is a maximizing problem with constraints and nonnegative variables.
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MORE ON DUALITY RELATIONS
Let us define standard form (canonical form) problems as follows:
1. A standard form maximizing problem is a linear program with all constraints as inequalities and nonnegative variables.
2. A standard form minimizing problem is a linear program with all constraints as inequalities and nonnegative variables.
The dual of a standard form maximizing problem is a standard form minimizing problem and
vice versa. This is property 4 of DR1. If the primal is not in standard form, neither is the dual.
Deviations from the standard form could mean that a problem has both and constraints or equality constraints and/or some non-positive or unrestricted variables. Fortunately, any
nonstandard problem can be converted to a standard form problem by some simple algebraic
manipulations.
We will use the concept of the standard form to develop rules for finding the dual of a
nonstandard primal. The fact that all linear programming problems have a standard form
equivalent also means that statements about duality in terms of standard form problems are
completely general. Let us demonstrate these ideas with the original problem (2.7) from
Chapter 2.
Primal Problem:
)C4(
0x,x0x1800x120020xx
1x241x
161
12xx5.0tosubject
x20x24zMaximize
21
21
21
21
21
21
+++
+=
This problem is not in standard form. The sulfur constraint is a inequality. However, the
problem can easily be converted to a standard form by multiplying the sulfur constraint
through by -1.
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Standardized primal problem:
iablevardualassociated
0x,x)y(0x1800x1200
)y(1x241x
161
)y(20xx)y(12xx5.0
tosubjectx20x24zMaximize
21
321
221
321
121
21
+
+++
+=
The dual associated with this standardized primal is as follows.
Standardized dual:
0y,y,y,y
20y800y241yy
24y1200y161yy5.0
tosubjecty0yy20y12zMinimize
3321
3321
3321
3321
+++
++
++=
Compare now the original primal with the standardized dual. Properties 2 and 3 of DR1 are
not satisfied for those coefficients associated with the sulfur constraint and 3y . The standardized dual is thus not the proper dual of the original problem. The proper dual can,
however, easily be obtained by reversing the standardization operation used to get the
standardized primal. We multiply the coefficients of 3y through by -1 and define a new variable w which is the negative of 3y . This yields the following dual. Dual of the original primal:
)D4(
0y0,y,y,y
20y800y241yy
24y1200y161yy5.0
tosubjecty0yy20y12zMinimize
4
321
4321
4321
4321
++
+++
+++=
We now see that properties 2 and 3 of DRI are satisfied. But we also note that the new dual
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variable 4y is restricted to be non-positive (since 3y was nonnegative). There is no need to go through the process of first standardizing, then finding the dual, and finally unscrambling
the dual. Instead, we can go directly to the dual by using the following duality relationship.
What is the nature of the dual if the primal has equality constraints? In order to find out, we
resort to the following trick for each such constraint. We replace the equality by two
inequalities of opposite direction, i.e., one is a inequality, the other a inequality. The LHS coefficients and the RHS parameter are the same as in the original constraint. Since both
have to be satisfied simultaneously, the feasible region will be identical to the original
constraint. We have just seen how to handle a problem with mixed inequality constraints. The
dual will have two dual variables, say +iy and iy , one of which is restricted to be nonnegative
and the other to be non-positive. Both variables have, however, exactly the same coefficients
in the dual. We now undo the trick of substituting two inequality constraints for the equality
constraint. We define a new variable iy that can assume both nonnegative and non-positive
values, i.e., one that is unrestricted in sign, where =iy ( +iy - iy ). So iy replaces +iy if iy assumes a nonnegative value, and replaces iy if iy assumes a non-positive value. Again, we
can avoid actually using this trick by applying the next duality relation.
DUALITY RELATION 2 (DR2)
If the direction of the inequality constraint in one problem deviates from the standard
form, the corresponding variable in the other problem is restricted to be non-positive and
vice versa.
DUALITY RELATION 3 (DR3)
If a constraint in the one problem is a strict equality, then the corresponding variable in the
other problem has no sign restriction and vice versa.
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0x,......x,x,xbxa...xaxaxa
.................................bxa...xaxaxa
bxa...xaxaxa
xc...xcxcxczMax
n321
mnmn33m22m11m
2nn2323222121
1nn1313212111
nn332211
++++
++++++++
++++=to subject
be problem primal the let problem dual the of Definition
iablesvardualcalledarey...,yy,ywhere0y...,yy,y
cyb...yayaya.................................
cya...yayayacya...yayaya
yb...ybybybzMin
m321
m321
nnmn3n32n21n1
2m2m332222112
1m1m331221111
mm332211
++++
++++++++
++++=to subject
as.definedisproblemdualThe
Table 4.1 demonstrates the duality relations DR I through DR3 in general terms.
TABLE 4.1. Primal and dual in general form
Primal Dual
Maximize = ijj bxcz subject to
ijij bxa ijij bxa = ijij bxa ix unrestricted 0x i 0x i
Maximize = ii ybz Subject to
0yi
0yi
iy unrestricted
= ijij cya ijij cya ijij cya
Example 4.1 write the dual of the following primal LP problem.
321 xx2xzMax ++= subject to
Implies DR1
Implies DR2
Implies DR3
Implies DR3
Implies DR2
Implies DR1
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6xxx46x5xx
2xxx2
321
321
321
+++
+
0x,x,x 321 Solution since the problem is not in the canonical form, we interchange the inequality of
second constraint.
321 xx2xzMax ++= subject to
6xxx46x5xx2xxx2
321
321
321
++++
0x,x,x 321 Dual Problem: let 321 yy,y be the dual variables.
321 y6y6y2zMin ++= subject to
1yy5y2yxy
1y2y2y2
321
321
321
+++
++
0yy,y 321 Example 4.2 Find the dual of the following LPP.
321 xxx3zMax += subject to
13xx512x3xx8
8xx4
31
321
21
++
0x,x,x 321 Solution since the problem is not in the canonical form, we interchange the inequality of the
second constraint.
321 xxx3zMax += subject to
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13xx512x3xx8
8xx4
31
321
21
0x,x,x 321 xczMax T=
subject to0xBAx
( )
=
3
2
1
xxx
113z ,
=13128
b
=
605318
014A
Dual Problem: let 321 yy,y be the dual variables.
0yandcyA
ybzMinT
T
=
i.e., ( )
=
3
2
1
yyy
13128zMin
subject to
11
3
yyy
630011584
3
2
1
321 y13y12y8zMin += subject to
1y6y3y01y0yy
3y5y8y4
321
321
321
++
+
0yy,y 321
Example 4.3 write the dual of the following LPP.
22 x5x2zMin += subject to
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4x3xx6x6xx2
2xx
321
321
21
=+++
+
0x,x,x 321 Solution since the given primal problem is not in the canonical form, we interchange the
inequality of the constraint. also the third constraint is an equation. this equation can be
converted into two inequalities.
221 x5x2x0zMin ++= subject to
4x3xx4x3xx
6x6xx22x0xx
321
321
321
321
++
++
0x,x,x 321 Again rearranging the constraints, we have
221 x5x2x0zMin ++= subject to
4x3xx4x3xx
6x6xx22x0xx
321
321
321
321
++
++
0x,x,x 321 Dual Problem: since there are four constraints in the primal, we have four dual variable
namely 321 yy,y
//3
/321 y4y4y6y2zMax +=
subject to
0y,y,y2,y
5y3y3y6y0
2yyyy
0yyy2y
//3
/321
//3
/321
//3
/321
//3
/321
+
++
Let //3/33 yyy =
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)yy(4y6y2zMax //3/321 +=
2)yy(yy
0)yy(y2y//3
/321
//3
/321
+
5)yy(3y6y0 //3/321 +
finally, we have,
321 y4y6y2zMax += subject to
2yyy
0yy2y
321
321
+
5y3y6y0 321 + edunrestrictisy,0y,y 321
Example 4.4 give the dual of the following problem
21 x2xzMax += subject to
5x4x34x3x2
21
21
=++
0x1 and 2x unrestricted. Solution since the variable 2x is unrestricted, it can be expressed as //2/22 xxx = . On
reformulating the given problem, we have
)xx(2xzMax //2/21 +=
subject to
5)xx(4x3
5)xx(4x3
4)xx(3x2
//2
/21
//2
/21
//2
/21
++
0x,x,x //2/21
Since the problem is not in the canonical form we rearrange the
constraints. //2
/21 x2x2xzMax +=
subject to
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5x4x4x3
5x4x4x3
4xx3x2
//2
/21
//2
/21
//2
/21
++
+
0x,x,x //2/21
Dual Problem: Since there are three variables and three constraints, in dual we have three
variables namely //2/21 y,y,y
//2
/21 y5y5y4zMin +=
subject to
2y4y4y3
2y4y4y3
1yy3y2
//2
/21
//2
/21
//2
/21
++
+
0y,y,y //2/21
Let //2/22 y,yy = , so that the dual variable 2y is unrestricted in sign. finally the dual is )yy(5y4zMin //2
/21 +=
subject to
2)y4y(4y3
2)yy(4y3
1)yy(3y2
//2
/21
//2
/21
//2
/21
++
i.e.,
21 y5y4zMin += subject to
2y4y3
2y4y3
1y3y2
21
21
21
++
0y1 and 2y is unrestricted. ie: 21 y5y4zMin += subject to
2y4y3
2y4y3
1y3y2
21
21
21
+++
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ie: 21 y5y4zMin += subject to
2y4y3
1y3y2
21
21
=++
0y1 and 2y is unrestricted. EXAMPLE 4.5 Management analysts at a Apollo laboratory have developed the following
LP primal problem:
21 x18x23zMinimize += subject to
0x,x,x116x4x9125x6x4120x4x8
321
21
21
21
+++
This model represents a decision concerning number of hours spent by biochemists on certain
laboratory experiments (x1) and number of hours spent by biophysicists on the same series of
experiments (x2). A biochemist costs $23 per hour, while a biophysicist's salary averages $18
per hour. Both types of scientists can be used on three needed laboratory operations: test 1,
test 2, and test 3. The experiments and their times are as follows:
TABLE 4.2 Apollo laboratory data
Lab Experiment Scientists Type Minimum Test Time
Needed per day Biophysicist Biochemist
Test 1 8 4 120
Test 2 4 6 115
Test 3 9 4 116
This means that a biophysicist can complete 8,4 and 9 hours of tests 1, 2, and 3 per day.
Similarly, a biochemist can perform 4 hours of test 1, 6 hours of test 2, and 4 hours of test 3
per day.
SOLUTION I
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TABLE 4.3 The optimal Simplex Table for the lab's primal problem is
1x 2x s1 s2 s3 b Basic Variable
0 0 -1.19 0.13 1 12.13 s3 0 1 0.13 -0.25 0 13.75 x2 1 0 0.12 -0.19 0 8.13 x1 0 0 -2.06 -1.63 0 434.38 z
Optimum number of hours of biophysicists and biochemist are costs $23 per hour, while a
biophysicist's salary x1 = 8.12 hours and x2 = 13.75 hours total cost = $434.37 per day
SOLUTION II Using the Dual Problem
The primal problem is
21 x18x23zMinimize += subject to
0x,x,x116x4x9125x6x4120x4x8
321
21
21
21
+++
The Dual problem of this will be
321 y116y125y120zMaximize ++= subject to
0y,y,y18y4y6y423y9y4y8
321
321
321
++++
TABLE 4.4 The optimal Simplex Table for the lab's Dual problem is
1y 2y y3 s1 s2 b Basic Variable
1 0 1.19 0.19 -0.13 2.06 x1 0 1 0.12 -0.13 -0.13 1.63 x2 0 0 12.13 13.75 8.13 434.38 z
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The optimal solution to the dual problem is y1 = 2.07, y2 = 1.63, y3 =0 which is described
worth of Test 1 , Test 2 and Test 3 per hour.
Note that optimal values of x1 = 8.12 hours and x2 = 13.75 hours respectively i.e, optimal
number of working hours of biophysicists and biochemist per day together with associated
total cost is total cost = $434.37 per day
Similarly, optimal values of dual variables yl, y2, and y3 (y1 = 2.07, y2 = 1.63, y3 =0 from
Table 4.4) could be obtained from the optimal primal solution given by Table 4.3 either under
the slack variables s1, s2, and s3.
EXAMPLE 4.6
A company makes three products X, Y and Z out of three raw materials A, B and C. The
number of units of raw materials required to produce one unit of the product is as given in the
Table4.5
TABLE 4.5
Raw materials
Products
X Y Z
A 1 2 1
B 2 1 4
C 2 5 1
The unit profit contribution of the products X, Y and Z is Rs. 40, 25 and 50 respectively. The
number of units of raw materials available are 36, 60 and 45 respectively.
(i) Determine the product mix that will maximize the total profit.
(ii) From the final simplex table, write the solution to the dual and give the economic
interpretation.
Solution
(i) Let the company produce x1 x2 and x3 units of the products X, Y and Z respectively. Then
the problem can be expressed mathematically as
321 x50x25x40zMax ++= subject to
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0x,x,x
45xx5x2
60x4xx2
36xx2x
321
321
321
321
++++++
Adding slack variables 321 sands,s , we get
321321 s0s0s0x50x25x40zMax +++++= subject to
0s,s,s,x,x,x
45sxx5x2
60sxxx2
36sxx2x
321321
3321
2321
1321
=+++
=+++=+++
Initial basic feasible solution is 45s,60s,36s,0x,0x,0x 321321 ====== and z = 0. This solution and further improved solutions are given in the following tables
TABLE 4.6 Initial Table (Example 4.6) 1x 2x x3 s1 s2 s3 b B.V.
1 2 1 1 0 0 36 A1 2 1 4 0 1 0 60 s1 2 5 1 0 0 1 45 s2
-40 -25 -50 0 0 0 20 r
TABLE 4. 7 First iteration of simplex tableau 1x 2x x3 s1 s2 s3 b B.V.
1 2 1 1 0 0 36 A1 2 1 0 1 0 60 s1 2 5 1 0 0 1 45 s2
-40 -25 -50 0 0 0 20 r
4
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TABLE 4. 8 Second iteration of simplex tableau 1x 2x x3 s1 s2 s3 b B.V.
0.5 1.75 0 1 0 0 21 s1 0.5 0.25 1 0 0.25 0 15 x3
4.75 0 0 -0.25 1 30 s2 -15 -12.5 0 0 12.5 0 750 z
TABLE 4. 9 Optimal simplex tableau of primal problem 1x 2x x3 s1 s2 s3 b B.V.
0 0.17 0 1 -0.17 -0.33 11 s1 0 -1.33 1 0 0.33 -0.33 5 x3 1 3.17 0 0 0.33 -0.67 20 x1
0 35 0 0 10 10 1050 z
Optimal solution to the given (primal) problem is x1 = 20 units, x2 = 0, x3 = 5units;
Z = Rs. 1,050.
(ii) The dual problem is
321 y45y60y36zMinimize ++= subject to
0y,y,y
50yy5y2
25y5yy2
40y2y2y
321
321
321
321
++++++
From Table 4.9, .1050zand10y,10y,0y 321 ==== . Economic Interpretation
Suppose the manager of the company wants to sell the three raw materials A, B and C instead
of using them for making products X, Y and Z and, then, by selling the products earn a profit
of Rs. 1,050. Suppose the selling prices were Rs. yl, Rs. y2 and Rs.y3 per unit of raw materials
A, B and C respectively. Then the cost to the purchaser of all the three raw materials will be
).y45y60y36(.Rs 321 ++ Of course, the purchaser will like to set the selling prices of A, B
1.5
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and C so that the total cost is minimum. So the objective function will be to minimize
).y45y60y36( 321 ++ Table 4.9 indicates that the marginal values of raw materials, A, B and C are Rs. 0, Rs. 10 and
Rs. 10 per unit respectively. Thus if the manager sells the raw materials A, B and C at price
Rs. 0, Rs. 10 and Rs. 10 per unit respectively, he will get the same contribution of Rs. 1,050
which he is going to get if he uses these resources for producing products X, Y and Z and
then sells them.
EXERCISES
1 Convert the following linear programming into standard form
321 xx4x3zMaximize ++= subject to
0x,x,x5xxx30xx2x610x2x3x
321
321
321
321
=++++++
(b) Write (i) the dual of the linear program in (a), and (ii) the dual of the standard form
of the linear program in (a). Show that these two dual problems are equivalent.
2 (a) Find the dual of the following problem:
4321 x12x10x6x4zMaximize +++= subject to
0x,x,x,x15x3x5xx5x4x2x3x
4321
4321
4321
++++++
(b) Graph the dual, and, using DR6 (Complementary Slackness Theorem), find the solution
of the primal. Check your answer using DR5 (Duality Theorem). Show that your solution to
the primal is an extreme point to that problem.
3 For each of the following problems, write out the dual (not the standard form dual).
(a) 321 xx2x4zMaximize ++= subject to
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0x,x,x2x2xx2
10x2x2x
321
321
321
=+
++
(b) 321 x3x2x4zMaximize ++= subject to
0x,x,x12x3xx28xxx
321
321
321
+++
(c) 321 x4x9x2zMinimize += subject to
signintedunrerstricx,0x,x4x4x6x
12x4x3x2
321
321
321
+
++
(d) 21 x2x3zMaximize = subject to
0x,x4x1xx5xx
21
2
21
21
=+
4 For each of the problems in exercise 3, find:
(a) The standard form primal.
(b) The standard form dual.
find the optimal primal solution. Use the quickest check you know to verify the optimality of
the primal solution.
4.7 Give the interpretations of the optimal values of w2 and w, of the power
generating problem, summarized in Table 3-3.
4.8 For the problem in exercise 4.3(b), after making the changes necessary to solve
it by the big M method, we have the following optimal tableau.
Ci 4 2 3 0 M
Ci Basis Solution X1 x2 x3 X4 x5
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21
3
3 x3 18 'T 0 3'
4 14 i 0 X1 T 1 5 5 5
Zi Ci 22 0 3 0 2 1+M
(a) Using this tableau, write down (i) the optimal values of the variables and z, and (ii) the
optimal values of the dual variables.
(b) Interpret wl, the first dual variable.
4.9 Solve the following by the dual simplex method: minimize z = 3x1 + 2x2
subject to X, + x2 10
2x1 + x2 14
X1, X2 0 4.10 Solve the following by the dual simplex method:
minimize z = 2x, + x2
subject to X1 + X2 15
X1 - X2 1
X1, X2 0 4.11 Solve the following by the dual simplex method:
maximize z = xi X2 2x3 subject to 2x, + x2 - x3 8
X1 3x, 3
X11 X21 x3 --0
4.12 For the problem in exercise 4. 10, use the dual simplex method to show that there is no
feasible solution if a third constraint, 2x1 + x2 = 7, is added.
Example 5 Find the maximum of 21 x8x6zMax += subject to
-
22
10x2x20x2x5
21
21
++
0x,x 21 by solving its dual problem
Solution The dual of the given primal problem is given below. as there are two constraints in
the primal, we have two dual variables namely , 21 y,y .
21 y10y20zMin += subject to
0y,y8yy26yy5
21
21
21
++
we solve the dual problem using the big M method. since this method involves artificial
variables, the problem is reformulated and we have
112121 MAMAS0S0y10y20zMax ++= Subject to
0A,A,S,S,y,y
8ASyy26ASyy5
212121
2221
1121
=++=++
i.e., 2121 MSMSy)10M3(y)20M7(zMax +=
Initial Simplex Tableau
basic y1 y2 S1 S2 A1 A2 rhs
Z
A1 A2
20-7M 10-3M -M -M 0 0
5 1 -1 0 -1 0
2 2 0 -1 0 1
-14M
6
8
Optimum tableau
basic y1 y2 S1 S2 A1 A2 rhs
Z
A1 A2
0 0 5/2 15/4 .. ..
1 0 -1/4 1/8 . ..
0 1 -5/8 0 0
-45
1/2
7/2
-
23
Since the problem is not in the canonical form, we interchange the inequality of second
constraint.
Min z/ =-45 and therefore Max z = 45.
Example 6
Apply the principal of duality to solve the LPP.
21 x2x3zMax += subject to
3x10x2x7x2x1xx
2
21
21
21
+++
0x,x 21 by solving its dual problem Solution The dual of the given primal problem is given below. As there are 4 constraints in
the primal problem we have four variables 4321 y,y,y,y in its dual.
21 x2x3zMax += |S
Subject to
3x10x2x7x2x
1xx
2
21
21
21
++
0x,x 21 Dual
4321 y3y10y7yzMin +++= Subject to
0y,y,y,y2yy2yy3y0yyy
4321
4321
4321
++++++
Apply Big M method to get the solution of the dual problem , as it involves artificial
variables, the problem is reformulated and we have
21214321 MAMAS0S0y3y10y7yzMax ++=
-
24
Subject to
0A,A,S,S,y,y,y,y2ASyy2yy
3ASy0yyy
21214321
224321
114321
=++++=++++
The optimal solution of the dual problem is
3y,0yyyand,21zMin 3431 ===== Therefore optimal solution of the primal problem is
0y,7xand,21zMax 21 ===
Important Results in Duality
1. the dual of the dual is primal.
2. if one is a maximization problem then the other is a minimization one.
3. the necessary and sufficient condition for any LPP and its dual to have an optimal solution
is that both must have feasible solution.
4. fundamental duality theorem states if either the primal or dual problem has a finite optimal
solution, then the other problem also has a finite
optimal solution and also the optimal values of the objective function in both the problems are
the same ie max z=min z'. the solution of the other problem can be read from the z. objective
row below the columns of slack, surplus variables.
5. existence theorem states that, if either problem has an unbounded solution then the other
problem has no feasible solution. 6. complementary slackness theorem: according to which
(i) if a primal variable is positive, then the corresponding dual constraint is an equation at the
optimum and vice versa.
(ii) if a primal constraint is a strict inequality then the corresponding dual variable is zero at
the optimum and vice versa.