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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):

4

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

9

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

10

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

11

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 =

12

)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

13

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

+++

14

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

17

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

18

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

19

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

20

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

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.