7.1 lecture_2a.pdf

8/9/2019 7.1 Lecture_2a.pdf

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Lecture 2a:

Duality in Linear Programming

Jeff Chak-Fu WONG

Department of Mathematics

Chinese University of Hong [email protected]

MAT581SS

Mathematics for Logistics

Produced by Jeff Chak-Fu WONG 1


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TABLE OF CONTENTS

1. The Dual Problem: Motivation Through an Example

2. Construction of the Dual

3. Duality Theorem

4. A Convenient Way for Reading the Solution of the Dual

BLE OFCONTENTS 2


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INTRODUCTION

TRODUCTION 3


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By now we must be comfortable in using the simplex method forsolving linear programming problems.

One important aspect of the simplex method is that it not only

solves the given LPP (called theprimal) but also solves another

closely related LPP (called thedual).

The other LPP, namely the dual, remains hidden in the

implementation of the simplex method but its solution is readily

available in the optimal simplex tableau of the primal problem

itself.

In the study of duality in linear programming we give a well

defined procedure to construct the hidden LPP (namely the

dual) and establish those results which bring out meaningfulrelationships between the given problem (primal) and its dual.

These results, besides being of theoretical and computational

importance, give interesting and useful economic

interpretations.

TRODUCTION 4


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THE DUAL PROBLEM: MOTIVATION THROUGH AN EXAMPLE

HEDUALPROBLEM: MOTIVATIONTHROUGH ANEXAMPLE 5


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Let us consider the LPP

max 4x1+ 3x2

subject to

x1+x2 8

2x1+x2 10

x1, x2 0 (1)

This problem has already been solved earlier by the simplex method

to get its optimal solution as(x1 = 2, x

2 = 6)and its optimal value as

26.



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Also the initial and the final tableaus are:

cj : (4 3 0 0)

y1 y2 y3 y4

CB XB x1 x2 x3 x4 0 x3 8 1 1 1 0

0 x4 10 2 1 0 1

zj cj z(XB) = 0 -4 -3 0 0

cj : (4 3 0 0)

y1 y2 y3 y4

CB XB x1 x2 x3 x4

3 x2 6 0 1 2 -1

4 x1 2 1 0 -1 0

zj cj z(XB) = 26 0 0 2 1



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Given the LPP (1) let us introduce another LPP as follows

min 8w1+ 10w2

subject to

w1+ 2w2 4

w1+w2 3

w1, w2 0. (2)



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In this construction (which is purely adhoc at present) the

problem is taken in the min form as the given problem (1) is in

the max form.

The roles ofcj (c1 =4, c2 =3)andbi (b1 =8, b2 =10)have been

interchanged and in the constraints of problem (2), sign has

been taken as in problem (1) the constraints are of type.

Problem 1 Problem 2

max 4x1+3x2 min 8w1+10w2

subject to subject to

x1+x28 w1+ 2w242x1+x210 w1+w23

with x1, x2 0 with w1, w2 0



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max min4 3 8 10

10

8 4

3

Figure 1:



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max min4 3 8 10

10

8 4

3

1 1

2

1 2

11 1

=>


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Problem (2) has only two decision variables

(w1andw2), we can solve it graphically (see Fig. 3) to get its optimal

solution as(w1 = 2, w

2 = 1)and its optimal value as26.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

P

w

w2

10

w + w = 3

w + 2 w = 42

21

1

Figure 3:



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If we now look at the last tableau of problem (1) as given here,

we notice two things.

Firstly, both problems (1) and (2) have the same optimal

value, namely, 26.

Secondly, the optimal solution of problem (2) is really present

in the last row of the optimal simplex tableau of problem (1).

cj : (4 3 0 0)

y1 y2 y3 y4

CB XB x1 x2 x3 x4 Min. Ratio

3 x2 6 0 1 2 -14 x1 2 1 0 -1 0

zj cj z(XB) = 26 0 0 2 1

But, it is just a coincidence or is there something deeper in it?



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If we now look at the last tableau of problem (1) as given here,

we notice two things.

Firstly, both problems (1) and (2) have the same optimal

value, namely, 26.

Secondly, the optimal solution of problem (2) is really present

in the last row of the optimal simplex tableau of problem (1).

cj : (4 3 0 0)

y1 y2 y3 y4


3 x2 6 0 1 2 -14 x1 2 1 0 -1 0

zj cj z(XB) = 26 0 0 2 1

But, it is just a coincidence or is there something deeper in it?



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The duality theory of linear programming asserts that this is

always going to happen, i.e.

if the given LPP (primal) has an optimal solution then its dual will certainly have an optimal solution and

furtherthe optimal values of the two problemswill be equal.



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Let us proceed

to the construction of the dual for a general LPP and then

to establish basic duality relationship between this pair of LPPs.



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CONSTRUCTION OF THE DUAL

ONSTRUCTION OF THEDUAL 17


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Let the given LPP (calledprimal) be

Maximise CTX

subject to

AX bX 0, (3)

whereX Rn, C Rn, b Rm andAis an(m n)real matrix.

Now, let us construct another associated problem, called thedualof the primal problem (3), as follows

Minimise bTw

subject toATw C

w 0, (4)

wherew Rm

.



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The relationship between the primal and dual problems can be best

described by

Dual\Primal x1 x2 xn min bTw

w1 a11 a12 a1n b1

w2 a21 a22 a2n b2

......

.... . .

... ...

wm am1 am2 amn bm

maxCTX c1 c2 cn



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Primal Dual

max CTX min bTw


AX b ATw C

with X 0 with w 0

The LPPs (3) - (4) together are called theprimal-dual pair.

The components of vectorX(respectively vectorw) are

called theprimal variables(respectivelydual variables).

The constraintsAX b(respectivelyATw C)are called

theprimal constraints(respectivelydual constraints).



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w1 a11 a12 a1n b1

w2 a21 a22 a2n b2...

......

. . ....

......

wm am1 am2 amn bm

maxCTX c1 c2 cn

Here it may be noted that

the number of dual constraints equals the number of primal

variables and

the number of dual variables equals the number of primal

constraints.



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w1 a11 a12 a1n b1

w2 a21 a22 a2n b2...

......

. . ....

......

wm am1 am2 amn bm

maxCTX c1 c2 cn

Here it may be noted that

the number of dual constraints equals the number of primal

variables and

the number of dual variables equals the number of primal

constraints.



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Primal Dual

Problem (4) Problem (3)

max CTX min bTw


AX b ATw C

with X 0 with w 0

Further, if we write problem (4) in the max form (i.e. in the form

of (3)) and write its dual we get problem (3).

Thus if we take the dual of dual we get back primal, i.e.

dual (dual) = primal.

This property of LPP is called thesymmetric duality.



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As the above construction of the primal-dual pair (3) - (4) is

symmetric we call either of these two problems as primal and

the other one as its dual, i.e. what we are calling as primal, the

same may also be called as dual and vice-versa.

However, in our presentation we shall continue calling problem

(3) as the primal and problem (4) as its dual.

Primal Dual

max CTX min bTw


AX b ATw C

with X 0 with w 0

With this understanding, we already have an example of a

primal-dual pair, namely LPPs (1) and (2).



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Next let us consider the situation when the constraints of the given

LPP (we shall continue calling it primal) are of type, i.e. the given

LPP has the following form

Maximise CTX

subject to

AX b

X 0. (5)

Now problem (5) can be rewritten in the form

Maximise CTX

subject to

(A)X (b)

X 0,



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Primal Dual

max CTX min (b)TU


(A)X (b) (A)TU C

with X 0 with U 0

and then using the construction (3) - (4) we can write its dual as

Minimise (b)TU

subject to

(A)TU C

U 0.



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and then using the construction (3) - (4) we can write its dual as

Minimise (b)TU

subject to

(A)TU C

U0.

Now if we writew = U, then the above problem can be written as

Minimise bTw

subject to

ATw C

w0. (6)



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Problem (4) Problem (6)min bTw min bTw


ATw C ATw C

with w 0 with w 0

If we now compare problems (4) and (6), then we observe that

(6) is of the same form as (4) except here w 0rather than

w 0.

This is because the primal constraints are of type.


max CTX min bTw


AX b ATw C

with X 0 with w 0



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We next consider the case when the constraints of the primal are of

= type, i.e. the primal problem is

Maximise CTX

subject to

AX=b

X 0. (7)

Problem (7) can be rewritten as

Maximise CTX

subject toAX b

(A)X b

X 0. (8)



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Primal Dualmax CTX min bTU bTV


AX b ATU ATV C

(A)X b

with X 0 with U 0

V 0

Now using the construction (3) - (4), we get the dual of above

problem as

Minimise bTU bTV

subject to

ATU ATV C

U 0

V 0.



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Now using the construction (3) - (4), we get the dual of above

problem as

Minimise bTU bTV

subject to

ATU ATV C

U 0

V 0. (9)

If we now writew = (U V)then the above problem can be

written as

Minimise bTw

subject to

ATw C

wunrestricted in sign. (10)



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Problem (4) Problem (10)min bTw min bTw


ATw C ATw C

with w 0 with wunrestricted in sign

A comparison of (4) and (10) tells that the only difference here is

thatwis unrestricted in sign.

Again this has happened because the primal constraints are of

= type.


max CT

X min bT

w


AX=b ATw C

with X 0 with wunrestricted in sign



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Just as the sign of dual variables is determined by the sign of

primal constraints,because of symmetric dual nature, we

expect that the sign of primal variables will determine the signof the dual constraints.

Using the construction (3) - (4), it is not difficult to prove that

Ifxj 0, then thejth dual constraint will be of type.

Ifxj 0,

then thejth dual constraint will be of type.

Ifxjunrestricted in sign,

then thejth dual constraint will be of = type.



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The above construction of the dual is valid for any general LPP

(primal) which is given in the max form and we summarize this

construction in the form of a table calledrule table for the

construction of the dual.

Rule Table for the Construction of the Dual

Primal (max) Dual (min)

ith constraint is type wi 0


ith constraint is=type wiunrestricted in sign

xj 0 jth constraint is type


xjunrestricted in sign jth constraint is=type

We read this table from left to right as the given LPP (primal) is in

the max form.



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But what happens if the given LPP (primal) is in the min form?

Rule Table for the Construction of the Dual

Primal (max) Dual (min)ith constraint is type wi 0


ith constraint is=type wiunrestricted in sign



xjunrestricted in sign jth constraint is=type

The answer is obvious, we read the rule table from right to leftand our construction will be correct because of the symmetric

nature of the LPP dual.



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Now we conclude that

Primal (max) Dual (min)

Maximisation Minimisation

Coefficients of objective function Right-hand sides of constraints

Coefficients ofith constraint Coefficients ofith variables, one in each constraint



ith constraint is =type wiunrestricted in sign



xjunrestricted in sign jth constraint is =type

Number of variables Number of constraints

We now illustrate the construction of the dual for some examples.



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Example 1 Write the dual of the following LPP

max z= 2x1 x2

subject to

x1+x2 10x1 2x2 = 8

x1+ 3x2 9 (11)

x1 0, x2unrestricted in sign.

Solution

We observe that

X=

x1

x2

, C=

2

1

,b=

10

8

9

,



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A=

1 1

1 2

1 3

andAT =

1 1 1

1 2 3

.

As there are three primal constraints, there will be three dualvariables, i.e.,

w=

w1

w2

w3

.

Also

bTw= 10w1 8w2+ 9w3

and

ATw=

w1+w2+w3

w1 2w2+ 3w3

.



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Now the given LPP is in the max form so we read the rule table

from left to right.

This gives the dual as

min w= 10w1 8w2+ 9w3

subject to

w1+w2+w3 2

w1 2w2+ 3w3 = 1

w1 0, w3 0, w2 unrestricted in sign.

In the dual, the first constraint is of type asx1 0.

The second constraint is of = type asx2unrestricted in sign.



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Similarly

w1 0as the first primal constraint is of type,

w2unrestricted in sign as the second primal constraint is of =

type and

w3 0as the third primal constraint is of type.



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Example 2 Write the dual of the following LPP

min z= 2x1 x2+x3

subject to

2x1+x2 x3 8

x1+x3 1

x1+ 2x2+ 3x3 = 9 (12)

x1 0, x2 0, x3unrestricted in sign.



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Solution

As the given (primal) LPP is in the min form we read the rule table

from right to left and get its dual as

max w= 8w1+w2+ 9w3

subject to

2w1 w2+w3 2

w1+ 2w3 1

w1+w2+ 3w3 = 1

w1 0, w2 0, w3 unrestricted in sign.



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Remark 1

Here it may be remarked that it is not essential to use the rule table.

In fact we can always write problem (11) in the form of problem (3)

and use the construction (4) to get its dual by the first principle.

Similarly the dual of problem (12) can also be obtained by the first

principle.

We use the rule table for the convenience only and there is absolutely

no compulsion to use it.



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DUALITYTHEOREMS

UALITYTHEOREMS 45


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Let us consider the primal-dual pair as

max CTX

subject to

AX b

X 0 (13)

and

min bTw

subject to

ATw C

w 0 (14)

UALITYTHEOREMS 46


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In this part, we present various duality results relating problems

(13) and (14).

In particular we prove four main theorems, namely

theweak duality theorem;

thestrong duality theorem;

theexistence theorem;

thecomplementary slackness theorem.

In the following we shall write primal and dual problems only

and it will be understood that we are referring to problems (13)and (14) respectively.

UALITYTHEOREMS 47


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WEAK DUALITYTHEOREM

EAKDUALITYTHEOREM 48


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Theorem 1 Let Xbe feasible for theprimaland wbe feasible for thedual.

Then

CTX bTw.

Proof

SinceXis feasible for theprimal, we have

AX b, X 0. (15)

Similarly the feasibility ofwfor thedualgives

ATw C, w 0. (16)

NowAX bimpliesXTAT bT ((AX)T =XTAT), which on

multiplication byw( 0)on the right gives

XTATw bTw. (17)



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SimilarlyATw CimplieswTA CT

((ATw)T =wT

ATT

=wTA), which on multiplication by

X( 0)on the right gives

wTAX CTX. (18)

ButXTATwis a scalar and hence

XTATw= (XTATw)T =wTAX.

Therefore (17) and (18)

XTATw bTw

wTAX CTX

give CTX bTw.



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Corollary 1 Let Xbe feasible for theprimaland wbe feasible for the

dual. Also let CTX=bTw. Then X is optimal to theprimaland wis

optimal to thedual.

Proof

LetXbe an arbitrary feasible point of theprimal.

The feasibility of wto thedualtogether with the weak duality

theorem (Theorem 1) givesCTX bTw.

But asbTw=CTXis given, this gives

C

T

X C

T X,

i.e. Xis optimal for theprimal.

Similarly we can show that wis optimal to thedual.



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STRONG DUALITYTHEOREM

TRONGDUALITYTHEOREM 53


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Theorem 2

(i) Let Xbe an optimal solution of theprimal. Then there exists a w

which is optimal to thedual. Also CTX=bTw.

(ii) Let w be an optimal solution of thedual. Then there exists a X

which is optimal to theprimal. Also bT

w

=CT

X

.

Here

part(i)is called thedirect duality theoremand

part(ii)is called theconverse duality theorem.

Asdual (dual) = primal, it is enough to prove part(i)only.



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Proof

Without any loss of generality we can take the primal-dual pair

as

max CTX

subject to

AX=b

X 0,

and

min bTw

subject to

ATw C

wunrestricted in sign. (19)

Also, again without any loss of generality, we can assume thatXwhich is optimal to problem (19) has been obtained by



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solving (19) by the simplex method. Thus Xis an optimal basic

feasible solution of (19), i.e. for some basis matrixB,

X= (XB =B1b,XR =0).

But Xis optimal to problem (19) hence we have

(zj cj) 0 (j= 1, 2, , n)

i.e.

(C

T

Byj cj) 0 (j= 1, 2, , n)i.e.

(CTBB1)aj cj (j= 1, 2, , n). (20)

Now if wedefinea vector w Rm given by wT =CTBB1

then (20) givesATw C.

This implies that wis feasible to the dual problem (19) as in

(19) wis unrestricted in sign.



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Thus given an optimal solution Xof the primal (19) we have

been able to construct a vector w, given by wT =CTBB1,

such that wis feasible to the dual (19).

Now

if we prove thatCTX=bTw

then because of the weak duality theorem wwill be

optimal to the dual (19).

But this is true as

CTX=CTBXB

= CTB(B1b) useXB =B

1b

= (CTBB1)b use the associative law for MM

= wTb use wT =CTBB1

= bTw.

Hence the result.



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Example 3 Write the dual of the following LPP (primal)

max z= 4x1+ 3x2

subject to

x1+x2 8

2x1+x2 10 (21)

x1, x2 0,

and useTheorem 2to find the solution of the dual by solving the primal.



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Solution

We have already seen that the dual of problem (21) is

min z= 8w1+ 10w2

subject to

w1+ 2w2 4

w1+w2 3 (22)

w1, w2 0.

We are required to determine an optimal solution of (22) by

solving (21).

From Page 8, we know that the initial and final simplex tableaus

for problem (21) are



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cj : (4 3 0 0)y1 y2 y3 y4


0 x3 8 1 1 1 0

0 x4 10 2 1 0 1

zj cj z(XB) = 0 -4 -3 0 0

cj : (4 3 0 0)

y1 y2 y3 y4


3 x2 6 0 1 2 -1

4 x1 2 1 0 -1 0

zj cj z(XB) = 26 0 0 2 1

Thus X= (x1 = 2,x2 = 6,x3 = 0,x4 = 0) and we have to find w,

i.e. an optimal solution of problem (22).



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cj : (4 3 0 0)

y1 y2 y3 y4


3 x2 6 0 1 2 -1

4 x1 2 1 0 -1 0

zj cj z(XB) = 26 0 0 2 1

Now from the last tableau of the primal (21) we get

CB =

c2

c1

=

3

4

, (and NOT

3

4

because in the tableau

the basic variable are x2 = 6and x1 = 2), and

B1 =

2 1

1 1

, whereB=

1 1

1 2

.



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Hence

w = w1

w2

=CTBB1

= 3 4

2 1

1 1

=

2

1

,

i.e. w1 = 2and w2 = 1is optimal to the dual (22).

Further by the duality theorem, optimal value of the dual equals

the optimal value of the primal which is26.



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cj : (4 3 0 0)

y1 y2 y3 y4


3 x2 6 0 1 2 -14 x1 2 1 0 -1 0

zj cj z(XB) = 26 0 0 2 1

Remark 3

The components of the vector w, namely(2, 1)are also available in

the last row of the optimal simplex tableau of problem (21).

So the obvious question is - can we just read the optimal solution of

the dual directly from the last tableau of the primal so that there is no

need of identifyingB1 and computing CTBB1?



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The answer is in the affirmative and that we discuss in the following

section (see Page 84).



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Now to prove the existence theorem and the complementary

slackness theorem, we again consider the primal-dual pair as statedat (13) and (14), where


max CT

X min bT

wsubject to subject to

AX b ATw C

with X 0 with w 0



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EXISTENCE THEOREM

XISTENCETHEOREM 67


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Theorem 3

(i) If primal and dual both have feasible solutions

then both have optimal solutions.

(ii) If primal (dual) has unbounded solution then the dual (primal) has no feasible solution.

(iii) If primal (dual) has no feasible solution but the dual (primal) has

feasible solution

then the dual (primal) has unbounded solution.

XISTENCETHEOREM 68


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(i) If primal and dual both have feasible solutions

then both have optimal solutions.

Proof (i)

LetXandwbe feasible solutions of the primal and the dual

respectively. Then by the weak duality theorem (cf. Theorem 1),

CTX bTw.

SincebTwis finite and is an upper bound (not necessarily the

least upper bound) on the primal objective function value, we

infer that the primal has an optimal solution.

Similar arguments hold for the dual.

XISTENCETHEOREM 69


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(ii) If primal (dual) has unbounded solution

then the dual (primal) has no feasible solution.

Proof (ii)

If possible let the dual has a feasible solution.

Then as primal and dual both become feasible, by part (i), both

have optimal solution.

But this contradicts that primal has unbounded solution.

Therefore the dual should be infeasible.

XISTENCETHEOREM 70


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(iii) If primal (dual) has no feasible solution but the dual (primal)

has feasible solution

then the dual (primal) has unbounded solution.

Proof (iii)

If possible let the dual has bounded (finite) optimal solution.

Then by the strong duality theorem (cf. Theorem 2) the primal

should also have an optimal solution and hence certainly be

feasible.

But this contradicts the hypothesis that the primal is infeasible.

Therefore the dual should have unbounded solution.

XISTENCETHEOREM 71


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The statement of the existence theorem can also be understood by

the following

(Dual|)feasible infeasible

both have Primal has

feasible optimal solution unbounded solution(Primal) This case is

Dual has possible, i.e. primal

infeasible unbounded and dual both

solution could be infeasible

Cf. Our motivating Example

XISTENCETHEOREM 72


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the following

(Dual|)

feasible infeasible


feasible optimal solution unbounded solution

(Primal) This case is


infeasible unbounded and dual bothsolution could be infeasible

XISTENCETHEOREM 73

For this let the primal problem be


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max z= 2x1+x2

subject to

3x1 2x2 6

x1 x2 1

x1, x2 0.

Then its dual is

min z = 6w1+w2

subject to

3w1+w2 2

2w1 w2 1

w1, w2 0.

XISTENCETHEOREM 74

x2

2w


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3 x - 2 x = 6

x - 2 x = 1

x

1

2

3

21 3

1

1

1

2

2

(a) Primal Problem

2

21 31

- 2 w - 2 w = 1

3 w + w = 221

1

w

2

(b) Dual Problem

Figure 4:

The set of feasible solution is shown in Fig. 4a. Evidently, this problem has

an unbounded optimal solution. For setting x1 = 0allowsx2to be as

large as we please. In this case z =x2is also as large as we please.

The constraints are shown in Fig. 4b. There is no feasible solutions to this

problem, since the second constraint can never hold for non-negative

values ofw1and w2.

XISTENCETHEOREM 75


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the following

(Dual|)

feasible infeasible





infeasible unbounded and dual bothsolution could be infeasible

XISTENCETHEOREM 76


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the following

(Dual|)

feasible infeasible





infeasible unbounded and dual both

solution could be infeasible

We now give an example to illustrate that both primal and dual

could be infeasible.

XISTENCETHEOREM 77



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max 3x1+ 4x2

subject to

x1 x2 1

x1+x2 0

x1, x2 0.

Then its dual is

min w1

subject to

w1 w2 3

w1+w2 4

w1, w2 0.

The feasible regions of the primal and dual problems are given in

XISTENCETHEOREM 78


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Fig. 5a and Fig. 5b respectively.

- x + x = 0

x - x = - 1

x

x

2

1

1 2

21

(a) Primal Problem

2

- w + w = 4

w - w = 3

w1

w

2

21

1

(b) Dual Problem

Figure 5:

As we see here both feasible regions are empty, i.e. the primal and

dual, both problems are infeasible.

XISTENCETHEOREM 79


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COMPLEMENTARYSLACKNESS THEOREM

OMPLEMENTARYSLACKNESSTHEOREM 80


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Theorem 4

Let X and wbe feasible solutions to the primal-dual pair (13) - (14).

Then X and w are optimal to the respective problems (13) and (14) if and only

if

wT(AX b) = 0

and

XT(C ATw) = 0.

The above conditions are called thecomplementaryslackness conditions or in

short thec.s.conditions.


max CT

X min bT

wsubject to subject to

AX b ATw C

with X 0 with w 0


Proof


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We shall prove only the necessary part; the proof of sufficient

part is similar.

Thus assuming that Xand ware optimal to the primal and

the dual respectively, we have to show that thec.s.conditions hold.

Let = wT(AX b)and= XT(C ATw).

Then by the feasibility of Xto the primal and the feasibility ofwto the dual, we have 0and 0.

Also

+ = wT

(AX b) +

X

T

(C A

T

w)= wTAX wTb+ XTC XTATw

= wTAX bTw + CTX (wTAX)T. (23)



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+ = wTAX bTw + CTX(wTAX)T

= bTw + CTX

= 0.

But wTAXis a scalar and therefore wTAX= (wTAX)T.

Further, as Xand ware optimal to the primal and dual

respectively, bythe strong duality theoremwe have

CTX=bTw.

Hence from (23) we get that

+= 0.

But then 0, 0, += 0imply that = 0and= 0, which

prove that thec.s. conditions hold.



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A CONVENIENT WAY FOR READING THE SOLUTION OF THE DUAL

CONVENIENTWAY FORREADING THESOLUTION OF THEDUAL 84

Consider the following primal LPP


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g p

max c1x1+c2x2+ +cnxn

subject to

a11x1+ +a1nxn b1

..

.

ar1x1+ +arnxn br

ar+1,1x1+ +ar+1,nxn br+1

... (24)

ar+s,1x1+ +ar+s,nxn br+s

ar+s+1,1x1+ +ar+s+1,nxn =br+s+1

.

..

ar+s+k,1x1+ +ar+s+k,nxn =br+s+k

x1 0, , xn 0.

Herer +s+k =m

and, without any loss of generality, we can assume that



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the firstrconstraints are with sign,

the nextsconstraints are with sign and

the nextkconstraints are with = sign.

Alsob 0. The dual of problem (24) is

min bTw

subject to

ATw C

w1, w2, , wr 0 (25)

wr+1, wr+2, , wr+s 0

wr+s+1, wr+s+2, , wr+s+k unrestricted in sign,

whereA= [aij ]is the matrix of order (m n).


F Th 2 k th t


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From Theorem 2 we know that

if Xis optimal to the primal (24)

then wT =CTBB1 is optimal to (25),

whereCBandB1 are read from the last (optimal) simplex

tableau of the primal (24).

Here we may see that as the constraints in (24) are mixed,

different components ofwin (25) will be constrained to have

different signs.

The main point which we wish to bring out here is that

to find wwe do not need to evaluate CTBB

1 but rather read certainspecific elements in the last row of the optimal simplex tableau of the

primal.



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For this, let us assume that problem (24) has been solved by using the Big M

method (we may make appropriate changes in the arguments if (24) is

solved by the two phase method).

Let us try to compute the number of columns which the simplex tableau will

have. Obviously there will be

ncolumns for the variablesx1, x2, , xn;

rcolumns for therslack variablesxn+1, xn+2, , xn+r;

scolumns for thessurplus variablesxn+r+1, xn+r+2, , xn+r+s;

scolumns for thesartificial variablesxn+r+s+1, xn+r+s+2, , xn+r+2s

which have been added to these constraints; and

kcolumns for thekartificial variablesxn+r+2s+1, xn+r+2s+2, , xn+r+2s+kwhich have been added to k

equal to constraints.


Therefore the (initial) simplex tableau for the problem (24) will look like


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XB n r s s k

xB1 artificial artificial

xB2 columns slack surplus columns for columns for

..

. ofA columns columns type =type

xBm constraints constraints

z(XB) zj cj

and at the optimality, i.e. in the last tableau,(zj cj) 0,j =

1, 2, , n,

n+ 1, , n+r,

n+r+ 1, , n+r+s,

n+r+s+ 1, , n+r+ 2s,

n+r+ 2s+ 1, , n+r+ 2s+k.




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XB n r s s k



..



z(XB) zj cj


1, 2, , n,

n+ 1, , n+r,

n+r+ 1, , n+r+s,

n+r+s+ 1, , n+r+ 2s,

n+r+ 2s+ 1, , n+r+ 2s+k.




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XB n r s s k



..



z(XB) zj cj


1, 2, , n,

n+ 1, , n+r,

n+r+ 1, , n+r+s,

n+r+s+ 1, , n+r+ 2s,

n+r+ 2s+ 1, , n+r+ 2s+k.




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XB n r s s k



..



z(XB) zj cj


1, 2, , n,

n+ 1, , n+r,

n+r+ 1, , n+r+s,

n+r+s+ 1, , n+r+ 2s,

n+r+ 2s+ 1, , n+r+ 2s+k.




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XB n r s s k



..



z(XB) zj cj


1, 2, , n,

n+ 1, , n+r,

n+r+ 1, , n+r+s,

n+r+s+ 1, , n+r+ 2s,

n+r+ 2s+ 1, , n+r+ 2s+k.


N l t t i th fi t t f th ti l d l


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Now let us compute w1, i.e. the first component of the optimal dual

vector w.

As wT =CTBB1, we have

w1 = First component ofCTBB

1

= CTBB1e1, e1 = [1 0 0 0]

T

= CTB(B1e1)

= zn+1

= (zn+1 cn+1) ( 0).

This is becauseB1e1is they-column for the first slack variablexn+1,

i.e. yn+1, cn+1 = 0and(zj cj) 0for allj; so in particular for

j=n+ 1.



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XB n r s s k



... ofA columns columns type =type


z(XB) zj cj

Similarly w2 = (zn+2 cn+2) 0, , wr = (zn+r cn+r) 0.So the firstrcomponents of the optimal dual vector ware

non-negative as desired.



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Next let us compute wr+1, i.e. (r+ 1)th component of the vector w.

We have

wr+1 = (r+ 1)th component ofCTBB

1

= CTBB1er+1 = C

TBB

1(er+1)

= CTB(B1(er+1))

= zn+r+1= (zn+r+1 cn+r+1) ( 0).

This is becauseB1(er+1)is they-column for the first surplus

variable, i.e. yn+r+1, cn+r+1 = 0and (zn+r+1 cn+r+1) 0becausezn+r+1 cn+r+1 0.



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XB n r s s k



... ofA columns columns type =type


z(XB) zj cj

Similarly

wr+2 = (zn+r+2 cn+r+2) 0, , wr+s = (zn+r+s cn+r+s) 0.

So the nextscomponents of the optimal dual vector ware

non-positive as desired.



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Next let us compute wr+s+1.

We have

wr+s+1 = CTBB

1er+s+1

= CTB(B1er+s+1)

= zn+r+2s+1

This is becauseB1er+s+1is they-column for the first artificial

variable which has been added to = type constraints, i.e.

yn+r+2s+1andzn+r+2s+1can have any sign.

Similarly wr+s+2 =zn+r+2s+2, , wr+s+k =zn+r+2s+kwhich areunrestricted as desired.


In the optimal dual solution w,


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p ,

(i) those components ofwwhich are constrained to be 0, are

obtained as the values of(zj cj)forrslack columns in the

optimal simplex tableau.

(ii) those components ofwwhich are constrained to be 0, are

obtained as the values of (zj cj)forssurplus columns in the

optimal simplex tableau.

(iii) those components ofwwhich are unrestricted in sign, are

obtained as the values ofzjforkartificial columns (for the =

constraints) in the optimal simplex tableau.

Therefore if the optimal simplex tableau of the primal is given then

the optimal solution wof the dualcan just be read from the last row

as specified aboverather than evaluatingCTBB1.



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Example 4 Write the dual of the following (primal) LPP and solve the

dual by solving the primal

max 2x1 x2

subject to

3x1+x2 = 3

4x1+ 3x2 6

x1+ 2x2 3

x1, x2 0.


S l ti U i th l t bl t th d l


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Solution Using the rule table, we get the dual as

min 3w1+ 6w2+ 3w3

subject to3w1+ 4w2+w3 2

w1+ 3w2+ 2w3 1

w1 R, w2 0, w3 0.

Now we wish to obtain an optimal solution of the above problem

(dual) from the optimal solution of the given LPP (primal).

For this let us solve the given LPP by the BigMmethod, and get theinitial and final simplex tableau as


cj 2 1 0 M M 0

CB XB y1 y2 y3 ya1 ya2 y4


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M xa1 = 3 3 1 0 1 0 0

M xa2 = 6 4 3 -1 0 1 0

0 x4 = 3 1 2 0 0 0 1

9M (2 7M) (1 4M) M 0 0 0

and

XB y1 y2 y3 ya1 ya2 y4

x1 = 3/5 1 0 1/5 3/5 -1/5 0

x2 = 6/5 0 1 -3/5 -4/5 3/5 0

x4 = 0 1 0 1 1 -1 1

12/5 0 1/5 1/5 (M 2/5) (M 1/5) 0

respectively.

Therefore an optimal solution of the primal is x1 = 3/5,x2 = 6/5and the

optimal value is 12/5.




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-2 x1 = 3/5 1 0 1/5 3/5 -1/5 0

-1 x2 = 6/5 0 1 -3/5 -4/5 3/5 0

0 x4 = 0 1 0 1 1 -1 1

12/5 0 1/5 1/5 (M 2/5) (M 1/5) 0

Let us compute the optimal solution of the dual:

wT = 2 1 0

3 1 0

4 3 0

1 2 1

1

= 2/5 1/5 0 .


Now we wish to get an optimal solution of the dual by reading certain


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specific elements in the last row of the above tableau.


xa1 = 3 3 1 0 1 0 0xa2 = 6 4 3 -1 0 1 0

x4 = 3 1 2 0 0 0 1

9M (2 7M) (1 4M) M 0 0 0

and


x1 = 3/5 1 0 1/5 3/5 -1/5 0

x2 = 6/5 0 1 -3/5 -4/5 3/5 0x4 = 0 1 0 1 1 -1 1

12/5 0 1/5 1/5 (M 2/5) (M 1/5) 0



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For this we note thatw1is unrestricted in sign and hence the value of w1in

the optimal dual solution wwill be the value ofzjfor that artificial variable

which corresponds to = constraint.

Thus

w1 = value ofzj for theya1 column

= (zj cj) + cj for theya1 column

= (M 2/5) + (M)

= 2/5.



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Also

w2 = value of (zj cj) for the surplus column y3

= 1/5.

w3 = value of (zj cj) for the slack columny4

= 0.



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xa1 = 3 3 1 0 1 0 0

xa2 = 6 4 3 -1 0 1 0x4 = 3 1 2 0 0 0 1

9M (2 7M) (1 4M) M 0 0 0

and


x1 = 3/5 1 0 1/5 3/5 -1/5 0

x2 = 6/5 0 1 -3/5 -4/5 3/5 0

x4 = 0 1 0 1 1 -1 112/5 0 1/5 1/5 (M 2/5) (M 1/5) 0



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xa1 = 3 3 1 0 1 0 0

xa2 = 6 4 3 -1 0 1 0

x4 = 3 1 2 0 0 0 1

9M (2 7M) (1 4M) M 0 0 0

and


x1 = 3/5 1 0 1/5 3/5 -1/5 0

x2 = 6/5 0 1 -3/5 -4/5 3/5 0

x4 = 0 1 0 1 1 -1 112/5 0 1/5 1/5 (M 2/5) (M 1/5) 0



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The optimal value of the dual is same as the optimal value of the primal, i.e.

z = 3 w1+ 6 w2+ 3 w3

= 3(2/5) + 6(1/5) + 3(0)

= 12/5.



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Remark 4 Let us recollect the revised simplex method and note that the solution of

the dual problem, namely wT = CTBB1 is automatically available in the first row

ofB1R

(see Page 59, Lecture note 1g) as stored in the revised simplex tableau in a

tableau form.


7.1 lecture_2a.pdf

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