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