Eastern MediterraneanDepartment Of

Industrial Engineering

Duality And Sensitivity Analysis

presented by: Taha Ben Omar Supervisor: Prof. Dr. Sahand Daneshvar

Introduction for every program we solve , there is another associated linear program which we happen to be simultaneously solving. The new linear program satisfies some very important properties. It may be used to obtain the solution to the original program. Its variables provides extremely useful information about the optimal solution to the original linear program.

 Formulation of the Dual problem

Canonical form of duality•

•P: minimize cx

Subject to Ax ≥ b

X ≥ 0 

D: Maximize wb Subject to wA ≤ c

W ≥ 0  

 

Example  

P: Minimize 6x1 + 8x2

Subject to 3x1 + x2 ≥ 4

5x1 + 2x2 ≥ 7 x1 , x2 ≥ 0

 D: Maximize 4w1 + 7w2

Subject to 3w1 + 5w2 ≤ 6 W1 + 2w2 ≤ 8 W1 , w2 ≥ 0

Standard form of duality

 P: Minimize cx

Subject to Ax = b X ≥ 0

 D: Maximize wb Subject to wA ≤ c

W unrestricted

Example

P: Minimize 6x1 + 8x2 Subject to 3x1 + x2 – x3 = 4

5x1 +2x2 - x4 = 7

x1 , x2 , x3 , x4 ≥ 0

D: Maximize 4w1 +7w2

Subject to 3w1 + 5w2 ≤ 6

w1 + 2w2 ≤ 8 - w1 ≤ 0

- w2 ≤ 0

w1 , w2 unrestricted

Given one of the definitions canonical or standard , it is easy to demonstrate that the other definition is valid. For example suppose that we accept the standard form as a definition and wish to demonstrate that the canonical form is correct . Bu adding slack variables to the canonical form of a linear program , we may apply the standard form of

duality to obtain the dual problem .

P: Max cx D: Max wb

 Subject to

Subject to Ax –Ix = b wA

≤ c x x ≥ 0 w

unrestricted

since -wI ≤ 0 is the same w ≥ 0

Dual of the Dual Since the dual linear program is itself a linear program , we may wonder

what its dual might be. Consider the dual in canonical form :

Maximize wb Subject to wA ≤ c

W ≤ 0

We may rewrite this problem in a different form :  

Minimize (-bt)wt

Subject to (-At)wt ≥ (-ct)Wt ≥ 0

The dual liner program for this linear program is given by ( letting x play the role of the row vector of dual variables)- :

Maximize xt (-ct)

Subject to xt (-At) ≤ (-bt)Xt ≥ 0

But this is the same as :

 Minimize cx Subject to Ax ≥ b

X ≥ 0

Which is precisely the primal problem. Thus we have the following lemma which is known as the involuntary property of Duality .

Lemma  

The dual of the dual is the primal .This lemma indicated that the definitions may

be applied in reverse. The terms " primal" and "dual" are relative to the frame of

reference we choose . 

 

Mixed forms of Duality

Consider the following linear program . 

P: Minimize c1x1 + c2x2 + c3x3 Subject to A11x1 + A12x2 + A13x3 ≤ b1

A21x1 + A22x2 + A23x3 ≤ b2 A31x1 + A32x2 + A3x33 = b3

X1 ≥ 0 , x2 ≤ 0 , x3 unrestricted

Converting this problem to conical form by multiplying the second set of inequalities by -1 , write the equality constraint set equivalently as two inequalities , and substituting x2 = -x'2 , x3 = x'3 –

x3 ''

 Minimize c1x1 – c2x2 + c3x3 –c3x3

Subject to A11x1 – A12x2' + A13x3' – A13x3'' ≥ b1

-A21x1 + A22x2' – A23x3' + A23x3'' ≥ -b2 A31x1 – A32x2' + A33x3' – A33x3'' ≥ b3

A31x1 + A32x2' - A33x3' + A33x3'' ≥ -b3-X1 >= 0 , X'2 >= 0 , X'3 ≥ 0 , X3'' ≥ 0

Denoting the dual variable associated with the four constraints sets as w1 , w2'‘ , w3'‘ and w3'‘respectively , we obtain the dual to this problem as follows.

 

Minimize w1b1 –w'2b2 + w'3b3 – w''3b3 Subject to

w1A11 – w'21A1 + w'31A1 – w''31A1 ≤ c

- w1A12 + w'2A22 – w'3A22 + w''3A32 ≤ c2

w1A13 – w'2A23 + w'3A33 – w''3A33 ≤ c3

w1A13 + w'2A23 - w'3A33 + w''3A33 ≤ c3-w'1 >= 0 , w'2 >= 0 , w'3 >= 0 , w''3 ≥ 0

MINIMIZATIONPROBLEM

≤0 ≥0=

MAXIMIZATION PROBLEM

≥0 ≤0

Unrestricted

≥0 ≤0

Unrestricted

≤0≥ 0

=

Finally , using w2 = -w2' and w3 = w3' – w3'' , the forgoing problem may be equivalently started as

follows :

 

D: Maximize w1b1 +w2b2 + w3b3 Subject to

w1A11 + w2A21 + w3A31 ≤ c

w1A12 + w2A22 + w3A32 ≥ c

w1A13 + w2A23 + w3A33 = c

w ≥ 0 , w2 ≤ 0 , w3 unrestricted

Example Consider the following linear program

Maximize 8x1 + 3x2 + 2x3

Subject to x1 – 6x2 + x3 ≥ 2 5x1 +7x2 -2x3 = -4

x1 ≤ 0 , x2 ≥ 0 , x3 unrestricted

 Applying the results of the table, we can immediately

write down the dual.

Minimize 2w1 – 4w2 Subject to w1 +5w2 <= 8

- 6w1 + 7w2 ≥ 3

w1 – 2w2 = -2 w1 <= 0 , w2

unrestricted