Unit-8 DUALITY THEOREMS

8/6/2019 Unit-8 DUALITY THEOREMS

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UNIT 8 DUALITY THEOREMS

Structure

8.1 Introduction

Objectives

8.2 Weak Duality!I

8.3 Strong Duality

8.5 Summary

8.6 AnswerslHints/Solutions- -


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Simplex Method and Duality Its dual :Minimize

D = b 1 y l + b 2 y 2 + ... + b m y mSubject to

Iallyl+ a2ly2+ ... + a m l y i 2 cia12y1+ a2~y2 ... + am2Ym 2 C 2

(Dual) ialnyl+ ... -I-mnYm2 Cny1 2 0 , ..., y m 2 0

In the matrix notations, the two forms are as follows :

Maximizez - C XI

I

I

Subject toI

Ii A X 5 BIi/ Le t K = (X : A X 5 B, X 2 0) be the set of all feasible solutions,

for the primal.

Dual!

- - , .

, MinimizeD = BY

Subject toATY 2 C"Y 20

Le t E = ( Y : ATY z CVr,Y > 0) be the set of all feasible solutions for. the dual.I - _

THEOREM 1. (Weak duality theorem).Ifprimal and dual problemsare both feasible with K andE as their respective feasible regions. The

Z 5D for every XEK. and YEE.1 PROOF: Let X,E K and YOE. Accordingly wehaveAX, < BX, 20


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!i!t SimplexMethodandDualily EXERCISE 1 - - I11

Consider the following dual pair of linear programming problems, 1Maximize

Z= 5x1+ 6x24x,Primal

Subject to

Minimize

. D = 5y,+ 2y2Dual

Subject to

Y, + 2 ~ , 252y,- y22 12Y, + 3y, 2 4

(i) Verify thatI 1x l =1, x2=1, x,=- andy, =7, y 2= 23'

., are feasible solutiohs for the two problems respectively. \

(ii) Verify the weak duality theorem.

THEOREM 2: Let primal and dual have feasible solutions X,and Yot respectively, such that

I

II

.

4 CX, =BTYo

( 6 ) 1f , I[ I , Then Xois the optimal solution for the primal and Yo is optimal for the? Ii dtkal.f,

I

PROOF:Let X be any feasible solution of the primal problem, thenaaccording to weak duality theorem1

CX5BTY,I

where Y is any arbitrary feasible solution of the dual *

Now, relation (7) is in particular true for YoDi.e. CX 5BTY, ,

=CX, (using (6 )...CX" C$,, V XEK .'Y - 7.

% - --- -


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x, and x5a .slack variables, initial solution to this primal problem isX1=o,X2=o,x4=2,X5 2 ; z = o. .'We, therefore, put the primal problem in the tableau form -..-

.

, . - I. . i P ,' 1 1 1 0 0

You may notice, all the three non-zero values ofZj-Cj an equal, i.e.thereis a tie, therefore any one of the three vectors A,! A, or A, can enter thebasis,We let4 enter the basis. To decide the departingvector, wecompute

2 2 2min {- , -)=z,which indica!es vectorA, is to be removedfrom the basis1 - 2

and is the pivot element. Next tableau is given as followsa

Cs..

00

., zj-cj I O I 0 I 0

Variablesin the basis

x4

X5

A,

2

4

1.Most negativeZj-Cjis -- , therefore column4 enters into the basisand2

2 .= min{i, 2) = ? , vectorA, k to beremoved from

2 2

is the.pivot element, .

.A,

1

p

(40

1

, . - .. . . .DualityTheorems

I l l " l - : l ~ l i ' -1'r , 8

A,

0

'2

Variablesin thebasis

x4

X2

A,

2

1

A,

1 0

0

A20 '

1

A41

0 -

A,m .1 .2

A, .

1

Solution

2

2 +

AS1

--2

1

2

Solution

1 -+*-

1


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I

Simplex Method and Next table isDuality

Since all Z, -Cj2 0,we now have the optimal solution which is givenby '

2x 1 = 0 , x2=- 2 43 , X 3 = ? and maximum Z= -.3Let us now solve the dual problem using Simplex method (i.e. two phasemethod that you learnt in Unit 6).

1 VC, -1 1 1 0 1 0 1 3 1 3 13

41- -3

2-3

1

C ,

1

1

Convert the Constraints of the dual intp equations by introducing surplusvariables y,, y4 a i d y, and constraints take form

Solution

2-3

2-3

4

IIt is obvious that initial identity matrix is not present in the coefficient Imatrix of this constraint system, therefore we i n t~o d u ceartificial variablesy,', y; and y, and in phase I, we solve

8 4

2-3

1--3

1

Variables

in the basis

*3

X 2

*2

0

1

iMaximize

Do= y:-y,'-y,''!

A,

0

2

A,

1

0

Subject to

2Y, 4~2-3 + Y; = 1Y,+ 2 ~ 2- Y, + Y = 1ZY, Y, -Y, + Y = 1y, 9 Y,, y," . . y," 20

Initial solution i s obtainedby putting non-basic vanables equal to zero i.e.,

Y1 :Yz= Y3 Y4 =y5=0 . ,I-.


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rherefore, yf - 1, y i = 1, y = 1, and =-3We put the problem in the tableau form.

c; 0 0 0 0 0 -1 - 1 - 1

Most negative Z; - C is-7, therefore column A, enters the basis and we

Ci

-1 -

-1

-1

1 1 1 1choose minimum {-, - -} =-. Hence column ofmificial vector f is to

4 2 ' 1 4

be removed from the basis and - is the pivot element. The new table is

Variables

in the basis-fYZ"u

zp-cp

3Most negative Z; -C; is--. therefore column A, enters the basis and we

2

1 / 4 314h o e m i - ,- = , min {A , } i.c. there is.tie, ir his '1 / 2 312 . 2 2 ..

A,

2

-5

A,

1 2

-7

A, A,

2 0 - 1 0 0 1 0 00 - 1 0 0

1 0 0 - 1 0

1 1

A,--------f1 0

A', A j Solution

1


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SimplexMethod and case we decide that an-tificial vecior corresponding to yi is to be removedDuality 3

from the basis and pivot element is -, again we move to the next table.2

1Most negative 2;- C; is--, is therefore column vector A, enters the basis

2

and choose minimum

That is to say {- , } =-in u.3)= 1.This implies column corresponding to artificial vector y; is to be removedand is the pivot element. Accordingly the next table is

I


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Phase 1 terminates, therefore we proceed with ptiase 11. Objective functionforphase I% s

The final table for phase 1gives initial table for phase 11 (without columnscorrespondingto artificial vectors)

As all 2 - 2 . 5O ,we have reached the optirnal and optimal for the dual isJ J

given by

1 1 4x = - , K = y , 3 O, y,=O and Max D=- -3 3

Hence for optimal solutions of primal and dual problems wc have

4Maximum Z =Minimum D = -

3 '

You may note ifthe initial Simplex tableau for the primal problem containsa unit matrix (initial basis matrix corresponding to columns ofslackvectors),the solution of the primal problem by Simplex procedure yieldsan explicitsolution to the dual.

In fact, the values ofZj- Cj in the final Simplex tableau for the primal1

problem corresponding to slack variablesx, and x, i.e.2,-C,= - andb . 3

12, - C,=- are actually the values ofdual variables y, and y, i.e.3

Duii lity Theorems


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"s I i:giE:s I A, I A, I A, 1 A, ISolution

Most negative Zj-Cjib -2. Therefore column A, enters the basis and we3. 2 2chm se rnin {- -} =-2' 3 3 '

Accordingly column A, is to be removed from the basis and pivotelement is m.Next table is

1Most negative Zj-Cjis --, therefore vectorA, enters the basis andwe

3choose

5 1 3 2 1 3 5 2 5minimum {-, -} = min {- -} =-4 /3 1/3 4 ' 1 4'

c~

0

2

which shows that vector A, is to be removed from the basis and

the pivot element. Next table isi

Variables

in the basis

X3

X1

zj-cj

At

'1

0

Solution

. -35 +

2-3

4-3

4

0 1 , l-3

--3

- 7 .

A,

l o

l o

A,

--23

1-3

2-3


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ISimplex MethodandDuality

Since all Zj- Cj 2 O , tl~ereiorecurrent solution is optimal fo r the du;ll.Optimal Solution ofdual is given by

1 5 7"1=4 X 2 = - and Max Z =-4 4As pointed ou t earlier Zj- Cj corresponding to the colutnns of slackvariables, provide us with the values of the dual vnrinbles (in this caseprimal variables),

' 1 1 7i.e. y l= ; i and yz= - and Minimum D =-2 4

Hence optimal table for the dual problem also explicitly provides with the

primal optimal solution.

3You may notice here by inspection y, = 1, y, = 0 and x,= 0, x =- are any

2

3 Ifeasible solutions of the primal and dual problems with D = 3 and Z =-

2

7i.e. Z


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Data for manufacture of two prpducts. iProduct Hours needed on Hours needed on

machine Munit machine B per unitof the product of the product

I 2 I

11 3 1Total time available 24 10(Hours/week)Let us assume that number of units manufactured is x, for the first and x, forthe second product. The mathematical formulation ofthe above problem is

MaximizeZ= 5x1 t 6x,

Subject to2x , + 3x 2 5 24x1 + x2 I10X 1 p X 2 r 0

Where Z represents the profit in rupees; first and second constraint reflectthe time resh-ictions (in hours) placed on the production of two products onmachinesA and B respectively. The last two constraints express the fact thatnegative number of units of either product cannot be produced. Optimalsolution of this primal problem is given by

xp=6, x: = 4 and maximum Z =54Dud to' the above linsar programming probrem is : .

Minimize

D = 24y, + 10y2Subject to

DualityTheorems .

Optimal solution to this dual problem is given by

Unit-8 DUALITY THEOREMS

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