chap6_handout

# chap6_handout - Chapter 6 Duality and Sensitivity Analysis...

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I Chapter 6: Duality and Sensitivity Analysis 1

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Introduction Introduction I Consider the following LP: maximize Z = 30 x 1 + 20 x 2 subject to x 1 + 2 x 2 6 (1) 2 x 1 + x 2 8 (2) x 2 2 (3) - x 1 + x 2 1 (4) x 1 , x 2 0 I Here is a feasible solution: ( x 1 , x 2 ) = (2 , 1) with Z = 80. I Is it optimal? If not, How far from optimal is it? I We could answer by solving the LP using simplex. Is there another way? 2
Introduction Lower and Upper Bounds I Since (2 , 1) is a feasible solution with Z = 80, 80 is a lower bound on the optimal objective value. I Why? I ( 10 3 , 4 3 ) is also a feasible solution, with Z = 380 3 = 126 . 67, so 126.67 is also a LB. I What if I told you that 160 is an upper bound on the optimal objective value? I Then ( 10 3 , 4 3 ) is no more than 160 - 126 . 67 = 33 . 33 units away from optimal. I If I could prove that 126.67 is also an upper bound, then ( 10 3 , 4 3 ) must be optimal. I Why? I How can we find such an upper bound? 3

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Introduction An Upper Bound max Z = 30 x 1 + 20 x 2 s.t. x 1 + 2 x 2 6 (1) 2 x 1 + x 2 8 (2) x 2 2 (3) - x 1 + x 2 1 (4) I If we multiply (2) by 20, we get 40 x 1 + 20 x 2 160 . I So any feasible solution satisfies 40 x 1 + 20 x 2 160. I The objective function is 30 x 1 + 20 x 2 . I So any feasible solution must satisfy 30 x 1 + 20 x 2 160. I Therefore 160 is an upper bound on the optimal objective value. 4
Introduction Another Upper Bound max Z = 30 x 1 + 20 x 2 s.t. x 1 + 2 x 2 6 (1) 2 x 1 + x 2 8 (2) x 2 2 (3) - x 1 + x 2 1 (4) I Multiply (1) and (2) by 10 and add them: 30 x 1 + 30 x 2 140 . I Since 30 x 1 + 20 x 2 30 x 1 + 30 x 2 140 , 140 is another (even better) upper bound. I How can we find the best (i.e., smallest) upper bound? 5

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Introduction Finding the Best Upper Bound max Z = 30 x 1 + 20 x 2 s.t. x 1 + 2 x 2 6 (1) 2 x 1 + x 2 8 (2) x 2 2 (3) - x 1 + x 2 1 (4) I We want multipliers for each constraint. I Call them y 1 , y 2 , y 3 , y 4 . I The multipliers must be chosen such that: I multipliers times x 1 ’s constraint coefficients 30. I multipliers times x 2 ’s constraint coefficients 20. I We also want the sum of the multipliers times the RHSs to be as small as possible. I Also, the multipliers have to be nonnegative. 6
Introduction Finding the Best UB, cont’d max Z = 30 x 1 + 20 x 2 s.t. x 1 + 2 x 2 6 (1) 2 x 1 + x 2 8 (2) x 2 2 (3) - x 1 + x 2 1 (4) I In other words, we want to minimize W = 6 y 1 + 8 y 2 + 2 y 3 + y 4 subject to y 1 + 2 y 2 - y 4 30 2 y 1 + y 2 + y 3 + y 4 20 y 1 , y 2 , y 3 , y 4 0 7

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Introduction The Dual Problem I This new problem is called the dual problem . I The original problem is called the primal problem . I The optimal objective function value for the dual problem gives an upper bound on the optimal objective function value for the primal problem: Z * W * . I How close is the UB? I We’ll answer this shortly. 8
The Essence of Duality Theory (Section 6.1) Formulating the Dual Formulating the Dual I In standard form, the primal and dual problems look like this: Primal: (P) maximize Z = n P j =1 c j x j subject to n P j =1 a ij x j b i i = 1 , . . . , m x j 0 j = 1 , . . . , n Dual: (D) minimize W = m P i =1 b i y i subject to m P i =1 a ij y i c j j = 1 , . . . , n y i 0 i = 1 , . . . , m 9

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The Essence of Duality Theory (Section 6.1)
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