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INTRODUCTION TO OPERATIONS

RESEARCH

Duality Theory

AUGMENTING THE DUAL PROBLEM

Since dual problems are also a linear programming problem, it also has CP solutions.

Since dual problems have function constraints in ≥ form, must augment by subtracting surplus variables from functional constraints.

Use surplus variable: z j

– c j for each constraint j

After augmenting dual problem:

(y

1

, y

2

, ..., y m

, z

1

-c

1

, z

2

-c

2

, ..., z n

-c n

)

PRIMAL/DUAL PROBLEMS

Primal Variable

Decision Variables: x j

Slack Variables: x n+i

Associated Dual Variable

Surplus Variables: z j

-c j

Decision Variables: y i

Complementary Basic Solution Property – Each basic solution in the primal problem has a complementary basic solution in the dual problem s.t. Z = W

Complementary Slackness Property – Given the association between variables, the variables in the primal basic solution and the complementary dual basic solution satisfy the complementary slackness relationship:

Primal Dual

Basic Non-Basic

Non-Basic Basic

WYNDOR GLASS CO. EXAMPLE

Maximize Z = 3x

1

Subject to:

+ 5x

2 x

1

+ x

3

3x

1

2x

2

+ 2x

2

+ x

4

+ x

5

= 4

= 12

= 18

Minimize W = 4y

1

Subject to:

+ 12y

2

+ 18y

3 y

1

2y

2

+ 3y

3

+ 2y

3

– ( z

1

– c

1

)

– ( z

2

– c

2

)

= 3

= 5

WYNDOR GLASS CO. EXAMPLE

Optimal Solution

Basic

Variable

Z x

3 x

2 x

1

Z

1

0

0

0

Dual Solution

0

1 x

1

0

0

Coefficient of: x

2

0

0 x

0

1

3 x

4

3/2

1/3

1

0

0

0

1/2

-1/3 x

5

1

-1/3

0

1/3

Right

Side

36

2

6

2

Primal Solution

WYNDOR GLASS CO. EXAMPLE

Iteration

6

7

4

5

8

1

2

3

Basic

Solution

(0,0,4,12,18)

(4,0,0,12,6)

(6,0,-2,12,0)

(4,3,0,6,0)

(0,6,4,0,6)

(2,6,2,0,0)

(4,6,0,0,-6)

(0,9,4,-6,0)

Feasible?

Yes

Yes

No

Yes

Yes

Yes

No

No

27

30

36

42

45

0

12

18

Z=W

Basic

Solution

(0,0,0,-3,-5)

(3,0,0,0,-5)

(0,0,1,0,-3)

(-9/2,0,5/2,0,0

)

(0,5/2,0,-3,0)

(0,3/2,1,0,0)

(3,5/2,0,0,0)

(0,0,5/2,9/2,0)

Feasible?

No

No

No

No

No

Yes

Yes

Yes

COMPLEMENTARY BASIC SOLUTIONS

Complementary Optimal Basic Solutions

Property – Each optimal basic solution in the primal problem has a complementary optimal basic solution in the dual problem such that Z = W.

Feasible?

Yes

No

Satisfies Condition for Optimality?

Yes No

Optimal

Super-Optimal

Sub-Optimal

Neither Feasible nor Optimal

NONSTANDARD LP

The approach here is similar to when we dealt with non-standard formulations in the context of the simplex method.

There is one exception: we do not add artificial variables. We handle “=“ constraints by writing them as “<=“ constraints.

This is possible here because we do not require here that the RHS is non-negative.

NONSTANDARD LP

Greater than or equal to constraints:

Multiply through the inequality constraint by -1 to make it a less than constraint: x

1

( x

1

+ 2x

2

+ 2x

2

≥ 5

≥ 5 ) (-1)

-x

1

2x

2

5

NONSTANDARD LP

Equality constraints:

Convert the equality constraint to a pair of inequality constraints: x

1

+ 2x

2

= 5 and x

1 x

1

+ 2x

2

+ 2x

2

≤ 5

≥ 5 x

1

-x

1

+ 2x

2

2x

2

≤ 5

5

NONSTANDARD LP

Unrestricted variables:

Replace the variable unrestricted in sign, by the difference of two nonnegative variables: x

1

+ 2x

2

≤ 5 x

1 x

2

≥ 0, unrestricted x

1

+ 2(x

2

– x

2

) ≤ 5

Maximize Z = x

1

Subject to:

+ x

2

+ x

3 x

1 x

1

2x

2

– 3x

2

– 2x

2

– x

3

+ 4x

3

≥ 4

= 5

≤ 18 x

1

, x

2 x

3

≥ 0, unrestricted

EXAMPLE

Maximize Z = x

1

Subject to:

+ x

2

+ (x

3

– x

3

) x

1 x

1

2x

2

– 3x

2

– 2x

2

(x

3

+ 4(x

3

– x

3

) x

3

)

≥ 4

= 5

≤ 18 x

1

– x

1 x

1

– 2x

2

– 3x

2

+ 3x

2

– 2x

2

+ x

3

– x

3

+ 4x

3

– 4 x

3

4x

3

+ 4 x

3

≤ –4

≤ 5

≤ – 5

≤ 18

EXAMPLE

Primal Problem

Maximize Z = x

1

Subject to:

+ x

2

+ x

3

– x

3 x

1

– x

1 x

1

– 2x

2

– 3x

2

+ 3x

2

– 2x

2

+ x

3

– x

3

+ 4x

3

– 4 x

3

4x

3

+ 4 x

3

≤ –4

≤ 5

≤ – 5

≤ 18

Dual Problem

Minimize W = – 4y

1

+5y

2

– 5y

3

+3y

4

Subject to:

– 2y

1

–y

1 y

1

+ y

2

3y

2

– y

3

+ y

4

+ 3y

3

– 2y

4

4y

2

+ 4y

3

4y

2

– 4y

3

1

1

≥ –1

≥ 1

Dual Problem

Minimize W = – 4y

1

+5y

2

– 5y

3

+3y

4

Subject to:

– 2y

1

–y

1 y

1

+ y

2

3y

2

– y

3

+ y

4

+ 3y

3

– 2y

4

4y

2

+ 4y

3

4y

2

– 4y

3

1

1

≥ –1

≥ 1

EXAMPLE

Dual Problem

Minimize W = – 4y

1

+5y

2

+3y

3

Subject to:

– 2y

1 y

1

+ y

2

3y

2

4y

2

+ y

3

– 2y

3

1

1

= 1 y

1

, y

3 y

2

≥ 0, unrestricted

Equality constraint

(y

2

– 3(y

2

– y

3

)

– y

3

)

– 4(y

2

– y

3

)

4(y

2

– y

3

)

Unrestricted variable (y

2

– y

3

)

STREAMLINING THE CONVERSION

An equality constraint in the primal generates a dual variable that is unrestricted in sign.

An unrestricted in sign variable in the primal generates an equality constraint in the dual.

Primal Problem opt=max

Constraint i :

<= form

= form

Variable j: x j x j

>= 0 urs

Dual Problem opt=min

Variable i : y i y i

>= 0 urs

Constraint j:

>= form

= form

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