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THE DUAL IN LP 

 

ASSOCIATED WITH EVERY LP PROBLEM IS ANOTHER

LP PROBLEM CALLED ITS DUAL.

  EACH MAXIMISING LP PROBLEM HAS IT

CORRESPONDING DUAL, A MINIMISING PROBLEM

  EACH MINIMISING LP PROBLEM HAS ITS

CORRESPONDING DUAL, A MAXIMISING PROBLEM.

CONSIDER THE FOLLOWING PRIMAL PROBLEM

MAXIMISE Z =2T+4C+3B

SUBJECT TO

3T+4C+2B  60

2T+1C+2B   40

1T+3C+2B  80

T,C,B   0

WHERE

T= TABLES, C=CHAIRS & B=BOOK CASE.

60=HOURS AVAILABLE IN ASSEMBLY.40= HOURS AVAILABLE IN FINISHING

80= HOURS AVAILABLE IN PACKING

Z=TOTAL PROFIT.

UNIT PROFIT ON TABLE =2, CHAIR =4 &BOOK CASE =3

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SOLUTION TO THE PRIMAL PROBLEM

CB  XB 

C j 

X j 

------

RHS

2

T

4

C

3

B

0

Sa 

0

Sf  

0

S p 

RATIO

0

0

0

Sa  60 3 4 2 1 0 0 60/4 *

Sf   40 2 1 2 0 1 0 40/1

S p  80 1 3 2 0 0 1 80/3

Z j  0 0 0 0 0 0 0

C j-z j  2 4 3 0 0 0

CB  XB 

C j.

X j 

RHS

2

T

4

C

3

B

0

SA 

0

SF 

0

SP 

RATIO

4

0

3

C 15 3/4 1 1/2 1/4 0 02

1

15 =30

SF  25 5/4 0 3/2 -1/4 1 02

3

25 =3

216 *

SP  35 -5/4 0 1/2 -3/4 0 1 2135

 =70

Z j  60 3 4 2 1 0 0

C j-Z j  -1 0 1 -1 0 0

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CB  XB 

C jX j 

RHS

2

T

4

C

3

B

0

SA 

0

SF 

0

SP 

4

3

3

C 63

2   1/3 1 0 1/3 -1/3 0

B 163

2   5/6 0 1 -1/6 2/3 0

SP  26 32   -5/3 0 0 -2/3 -1/3 1

Z j  76 32   23/6 4 3 5/6 2/3 0

C j-Z j  -11/6 0 0 -5/6 -2/3 0

THE PRIMAL SOLUTION IS:

C= 63

2  ; B=163

2 ; T=0 ; Z= 763

2  

SIGNIFICANCE OF

1)C j-Z j ROW IN OPTIMAL TABLEAU?2)SP?3)WHERE WOULD YOU ADD CAPACITY ?

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THE DUAL PROBLEM 

 

THE PRIMAL IS CONCERNED WITH MAXIMIZING

PROFIT FROM THREE PRODUCTS (PRODUCT

MIX).

  THE DUAL WOULD BE CONCERNED WITH THE

RESOURCE HOURS (EVALUATION) TO PRODUCE

PRODUCTS.

  LET SOMEBODY WANTS TO BUY ALL THE

CAPACITIES WITH THE RATES

ASSEMBLY Rs. A/HOUR

FINISHING Rs. F/HOURPACKING Rs. P/HOUR.

TOTAL RENTAL WOULD BE Y = 60A+40F+80P

RENTER WANTS TO FIND THE VALUES OF A, F

AND P TO MINIMIZE Y.

THE COMPANY OWNER HAS HIS OWN

REQUIREMENTS OF PROFIT.

(PRIMAL) Z >= (DUAL) Y2T+4C+3B   60A+40F+80P

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EACH TABLE REQUIRES 3 ASSEMBLY HOURS. ,

2 FINISHING HOURS AND 1 PACKAGING

HOURS., IT WOULD FETCH RUPEES 2 AS

PROFIT/UNIT.

3A+2F+1P > 2

SIMILARLY FOR CHAIR AND BOOK CASE

THAT IS

4A+1F+3P  4

2A+2F+2P 3

OF COURSE, THE RENTS A, F, P  0

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 THE PRIMAL PROBLEM

MAX Z=2T+4C+3B

S.T. 3T+4C+2B   60

2T+1C+2B   401T+3C+2B   80

T,C,B  0

TURNS OUT TO BE IN ITS DUAL

MIN Y =60A+40F+80PS.T. 3A+2F+1P   2

4A+1F+3P   4

2A+2F+2P   3A, F, P   0

WHAT ARE THE ADVANTAGES?

 

COMPUTATIONAL

  ECONOMICAL

  MORE INSIGHT.

  MAY HELP AVOIDING ARTIFICIAL VARIABLES

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SOLUTION TO THE DUAL PROBLEM

INITIAL TABLEAU.

CB  XB  C j 

X j

RHS

60

A

40

F

80

P

0

S1

0

S1

0

S3

M

A1

M

A2

M

A3

RATIO

M

A1  2 3 2 1 -1 0 0 1 0 0 2/3

MA2  4 4 1 3 0 -1 0 0 1 0 1

M

A3  3 2 2 2 0 0 -1 0 0 1 3/2

Z j  9M 9M 5M 6M -

M

-

M

-

M

M M M

C j-

Z j 

60-

9M

40-

5M

80-

6M

M M M 0 0 0

Carryout iterations until the final table is reached

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FINAL TABLEAU (5TH)

CB  XB  C j 

X j

RHS

60

A

40

F

80

P

0

S1

0

S1

0

S3

M

A1

M

A2

M

A3

60 A 65  

1 0 32  

0 31  

6

1  0 3

1  - 6

1

 0 S1  6

11   0 0 35   1 3

1  6

5   -1 31  

65  

40 F 32   0 1 3

1   0 31  

32   0

31 - 3

2  

Z j 3276   60 40 3

153

 

0 326

 3

216

 

0 326  

3216  

C j-Z j  0 0 3226

 

0 326  

3216   M M 326   M- 3

216  

THE OPTIMAL SOLUTION IS

A=6

5  , F=3

2  , S1= 611 ; P=0,

Z = 3276  = Y

THE OPTIMAL OBJECTIVE VALUES OF THE

PRIMAL AND DUAL PROBLEMS ARE THE SAME.

(THIS IS ALWAYS THE CASE).

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SOME IMPORTANT OBSERVATIONS

1. THE OPTIMAL VALUES OBTAINED IN THE

DUAL PROBLEM FOR A, F AND P ARE ALSO

FOUND IN THE PRIMAR’S FINAL TABLEAU.

SEE C j – Z j ROW UNDER SA, SF AND SP.

2. SIMILARLY THE OPTIMAL VALUES OBTAINED

IN THE PRIMAL FOR T, C AND B CAN BE NOTICED

IN THE FINAL TABLEAU OF DUAL IN THE C j – Z j

ROW UNDER SA, SF AND SP.

3.  THUS SOLVING THE EXISTING PRIMAL OR

DUAL GETS US THE SOLUTION TO BOTHPROBLEMS.

WHICH METHOD SHOULD BE APPLIED?

CONSIDER NUMBER OF CONSTRAINTS vis a vis NUMBER OF VARIABLES

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ECONOMIC INTERPRETATION

A PRIMAL PROBLEM:

MAXIMISE:  Z = X1+2X2-3X3+4X4 

S.T. X1+2X2+2X3-3X4  25

2X1+X2-3X3+2X4  15

X1,X2,X3,X4  0.

THIS L.P. HAS TWO CONSTRAINT EQUATIONSAND 4 VARIABLES.

DUAL PROBLEM :

MINIMISE: W=25Y1+15Y2 

S.T. Y1+2Y2   1

2Y1+Y2   2

2Y1-3Y2   -3

-3Y1+2Y2   4

Y1,Y2   0

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COMMENTS:

 

Y1 AND Y2 ARE CALLED DUAL VARIABLES.

  THE COEFFICIENTS OF OBJECTIVE FUNCTION

OF PRIMAL PROBLEM HAVE BECOME THE RHSCONSTANTS OF THE DUAL

 

THE RHS OF THE PRIMAL HAVE BECOME THE

COEFFICIENTS OF THE DUAL PROBLEM

  THE INEQUALITY SIGNS HAVE BEEN REVERSED.

  THE OBJECTIVE IS CHANGED FROM

MAXIMIZATION TO MINIMIZATION

  THE DUAL OF THE DUAL IS THE PRIMAL

PROBLEM

  EACH COLUMN IN THE PRIMAL CORRESPONDS

TO A CONSTANT (ROW) IN THE DUAL.

THUS THE NUMBER OF CONSTRAINTS OF THE DUAL

 IS EQUAL TO THE NUMBER OF PRIMAL VARIABLES.

  EACH CONSTRAINT (ROW) IN THE PRIMAL

CORRESPONDS TO A COLUMN IN THE DUAL.

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HENCE THERE IS ONE DUAL VARIABLE FOR

EVERY PRIMAL CONSTRAINT.

  IN BOTH THE PRIMAL AND DUAL PROBLEMS,

THE VARIABLES ARE NON-NEGATIVE AND THE

CONSTRAINTS ARE IN EQUALITIES.

  SUCH PROBLEMS ARE CALLED SYMMETRIC

DUAL LP

INTERPRETATION OF DUAL SOLUTIONS ASSHADOW PRICES

IN ECONOMIC SENSE OPTIMAL DUAL SOLUTION

CAN BE INTERPRETED AS THE PRICE ONE PAYS

FOR THE CONSTRAINT RESOURCES.

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SYMMETRIC DUAL LP:

A LP IS SAID TO BE IN SYMMETRIC FORM IF:

  ALL THE VARIABLES ARE RESTRICTED TO BE

 NON-NEGATIVE.

  ALL THE CONSTRAINTS ARE INEQUALITIES.

vector rowmanis y

vector columnnanis x

vector rownanisc

vector columnmanisb

matrixnmanis A

Where

Y  X 

cYAt sb AX t s

Yb Z  Min DualcX  Z  Maximal

 problemin

 problemin

)1(

)1(

)1(

)1(

)(

:

00

.....

.:Pr 

min

max

 

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GENERAL RULES FOR WRITING THE DUAL OF A

LP SYMMETRIC FORM. 

A LP IS SAID TO BE IN SYMMETRIC FORM IF:

  ALL THE VARIABLES ARE RESTRICTED TO BE

 NON-NEGATIVE.

  ALL THE CONSTRAINTS ARE INEQUALITIES.

RULES

1 

DEFINE ONE (NON NEGATIVE) DUAL VARIABLE

FOR EACH PRIMAL CONSTRAINT.

2  MAKE THE COST VECTOR OF THE PRIMAL THERHS VECTOR OF THE DUAL.

3 

MAKE THE RHS VECTOR OF THE PRIMAL THE

COST VECTOR OF THE DUAL.

4  THE TRANSPOSE OF THE COEFFICIENT MATRIX

OF THE PRIMAL BECOMES THE CONSTRAINTMATRIX OF THE DUAL.

5 

REVERSE THE DIRECTION OF THE CONSTRAINTINEQUALITIES.

6 

REVERSE THE OPTIMIZATION DIRECTION.

7  VERIFY.

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EXAMPLE:

PRIMAL:

0,

..

.

21

2211

11212111

2211

n

mnmnmm

nn

nn

 X  X  X 

b X a X a X a

b X a X a X at s

 X c X c X c Z  Max

 

DUAL:

nmnnn

m

mm

mm

cY amY aY a

cY amY aY a

cY aY aY at sY bY bY bW  Min

2211

22222212

11221111

2211

1..

.

 

0........1   m y y  

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INTERPRETATION OF DUAL SOLUTIONS AS

SHADOW PRICES

IN ECONOMIC SENSE OPTIMAL DUAL SOLUTION

CAN BE INTERPRETED AS THE PRICE ONE PAYS

FOR THE CONSTRAINT RESOURCES.

BY MAIN DUALITY THEOREM

THE OPTIMAL VALUES OF THE OBJECTIVEFUNCTIONS OF THE PRIMAL AND DUAL ARE

EQUAL.

IF X 0 , Y 0  ARE THE RESPECTIVE OPTIMAL

SOLUTIONS

Z0=CX 0 =Y 0 B=W0

THAT IS: THE OPTIMAL VALUE OF LP IS GIVEN

BY:

Z0 =Y1 0 B1+Y2 0 B2+………….+YM 0BM 

WHERE:

B1,B2………..BM REPRESENT THE LIMITED

QUANTITIES OF RESOURCES.

Y1 0 ,Y2 0…………YM 0  ARE THE OPTIMAL VALUES OF

THE DUAL VARIABLES.

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SUPPOSE

THAT LEVEL OF RESOURCES (say 1(B1)) BE

ALTERED.

FOR SMALL VALUES OF B1, SAY ( 1b , THE NET

CHANGE IN THE OPTIMAL VALUE OF LP Z0 IS

GIVEN BY Y1 0 ( 1b )

IN OTHER WORDS:

THE OPTIMAL VALUE OF THE DUAL VARIABLEFOR EACH PRIMAL CONSTRAINT GIVES THE NET

CHANGE IN THE OPTIMAL VALUE OF THE

OBJECTIVE FUNCTION FOR UNIT INCREASE INTHE RHS CONSTANTS.

THESE ARE HENCE CALLED 'SHADOW PRICES' ON

THE CONSTRAINT RESOURCES.

THESE COULD BE USED TO DETERMINE

WHETHER IT IS ECONOMICAL TO GET

ADDITIONAL RESOURCES AT PREMIUM PRICES.