Duality by Example Duality by Example: II max z = 4 x 1 + x 2 + 3 x - PowerPoint PPT Presentation

Duality by Example Duality by Example: II max z = 4 x 1 + x 2 + 3 x 3 max z = 4 x 1 + x 2 + 3 x 3 s.t. x 1 + 4 x 2 ≤ 1 s.t. x 1 + 4 x 2 ≤ 1 3 x 1 − x 2 + x 3 ≤ 3 3 x 1 − x 2 + x 3 ≤ 3 x 1 , x 2 , x 3 ≥ 0 x 1 , x 2 , x 3 ≥ 0 1. Add the first inequality (multiplied by 2) to the second 1. η : maximal possible value of target function. inequality (multiplied by 3): 2. Any feasible solution ⇒ a lower bound on η . 3. In above: x 1 = 1 , x 2 = x 3 = 0 is feasible, and implies z = 4 and thus η ≥ 4 . 2 ( x 1 + 4 x 2 ) ≤ 2 ( 1 ) 4. x 1 = x 2 = 0 , x 3 = 3 is feasible = ⇒ η ≥ z = 9 . + 3 ( 3 x 1 − x 2 + x 3 ) ≤ 3 ( 3 ) . 5. How close this solution is to opt? (i.e., η ) 6. If very close to optimal – might be good enough. Maybe 2. The resulting inequality is stop? 11 x 1 + 5 x 2 + 3 x 3 ≤ 11 . (1) Duality by Example: II Duality by Example: III z = 4 x 1 + x 2 + 3 x 3 z = 4 x 1 + x 2 + 3 x 3 max max s.t. x 1 + 4 x 2 ≤ 1 s.t. x 1 + 4 x 2 ≤ 1 3 x 1 − x 2 + x 3 ≤ 3 3 x 1 − x 2 + x 3 ≤ 3 x 1 , x 2 , x 3 ≥ 0 x 1 , x 2 , x 3 ≥ 0 1. got 11 x 1 + 5 x 2 + 3 x 3 ≤ 11 . 1. For any feasible solution: z = 4 x 1 + x 2 + 3 x 3 ≤ 11 x 1 + 5 x 2 + 3 x 3 ≤ 11 , 2. inequality must hold for any feasible solution of L . 2. Opt solution is LP L is somewhere between 9 and 11 . 3. Objective: z = 4 x 1 + x 2 + 3 x 3 and x 1 , x 2 and x 3 are all non-negative. 3. Multiply first inequality by y 1 , second inequality by y 2 and add them up: 4. Inequality above has larger coefficients than objective (for y 1 ( x 1 + ) ≤ y 1 ( 1 ) corresponding variables) 4 x 2 + y 2 ( 3 x 1 - + ) ≤ y 2 ( 3 ) x 2 x 3 5. For any feasible solution: ( y 1 + 3 y 2 ) x 1 + ( 4 y 1 − y 2 ) x 2 + y 1 + 3 y 2 . y 2 x 3 ≤ z = 4 x 1 + x 2 + 3 x 3 ≤ 11 x 1 + 5 x 2 + 3 x 3 ≤ 11 ,

Duality by Example: IV Duality by Example: IV Dual LP : � L max z = 4 x 1 + x 2 + 3 x 3 Primal LP : min y 1 + 3 y 2 max z = 4 x 1 + x 2 + 3 x 3 s.t. x 1 + 4 x 2 ≤ 1 s.t. y 1 + 3 y 2 ≥ 4 s.t. x 1 + 4 x 2 ≤ 1 3 x 1 − x 2 + x 3 ≤ 3 4 y 1 − y 2 ≥ 1 3 x 1 − x 2 + x 3 ≤ 3 x 1 , x 2 , x 3 ≥ 0 y 2 ≥ 3 x 1 , x 2 , x 3 ≥ 0 1. ( y 1 + 3 y 2 ) x 1 + ( 4 y 1 − y 2 ) x 2 + y 2 x 3 ≤ y 1 + 3 y 2 . y 1 , y 2 ≥ 0 . 1. Compare to target function – require 1. Best upper bound on η (max value of z ) then solve the 4 y 1 + 3 y 2 ≤ expression bigger than LP � L . 1 ≤ 4 y 1 − y 2 target function in each 2. � L : Dual program to L . 3 ≤ y 2 , variable. 3. opt. solution of � = ⇒ z = 4 x 1 + x 2 + 3 x 3 ≤ L is an upper bound on optimal solution for L . ( y 1 + 3 y 2 ) x 1 + ( 4 y 1 − y 2 ) x 2 + y 2 x 3 ≤ y 1 + 3 y 2 . Primal program/Dual program Primal program/Dual program m n � � min b i y i max c j x j i = 1 j = 1 m � n � s.t. a ij y i ≥ c j , s.t. a ij x j ≤ b i , i = 1 j = 1 for i = 1 , . . . , m , for j = 1 , . . . , n , x j ≥ 0 , y i ≥ 0 , for j = 1 , . . . , n . for i = 1 , . . . , m . c T x y T b max min y T A ≥ c T . s. t. s. t. Ax ≤ b . x ≥ 0 . y ≥ 0 .

Primal program/Dual program Primal program / Dual program in standard form What happens when you take the dual of the dual? n � m � m max c j x j n � � max ( − b i ) y i min b i y i max c j x j j = 1 i = 1 i = 1 j = 1 n � m � m � s.t. a ij x j ≤ b i , n � s.t. ( − a ij ) y i ≤ − c j , s.t. a ij y i ≥ c j , s.t. a ij x j ≤ b i , j = 1 i = 1 for i = 1 , . . . , m , i = 1 j = 1 for j = 1 , . . . , n , for i = 1 , . . . , m , for j = 1 , . . . , n , x j ≥ 0 , y i ≥ 0 , x j ≥ 0 , y i ≥ 0 , for j = 1 , . . . , n . for i = 1 , . . . , m . for j = 1 , . . . , n . for i = 1 , . . . , m . Dual program in standard form / Dual of dual Dual of dual program / Dual of dual program program written in standard form n n � � m n � � min − c j x j min − c j x j ( − b i ) y i max max c j x j j = 1 j = 1 i = 1 j = 1 n n � � m � n � s.t. ( − a ij ) x j ≥ − b i , s.t. ( − a ij ) x j ≥ − b i , s.t. ( − a ij ) y i ≤ − c j , s.t. a ij x j ≤ b i , j = 1 j = 1 i = 1 j = 1 for i = 1 , . . . , m , for i = 1 , . . . , m , for i = 1 , . . . , m , for j = 1 , . . . , n , x j ≥ 0 , x j ≥ 0 , x j ≥ 0 , y i ≥ 0 , for j = 1 , . . . , n . for j = 1 , . . . , n . for j = 1 , . . . , n . for i = 1 , . . . , m . = ⇒ Dual of the dual LP is the primal LP !

Result Weak duality theorem Proved the following: Theorem Lemma If ( x 1 , x 2 , . . . , x n ) is feasible for the primal LP and Let L be an LP , and let L ′ be its dual. Let L ′′ be the dual to ( y 1 , y 2 , . . . , y m ) is feasible for the dual LP , then L ′ . Then L and L ′′ are the same LP . � � c j x j ≤ b i y i . j i Namely, all the feasible solutions of the dual bound all the feasible solutions of the primal. Weak duality theorem – proof The strong duality theorem Proof. Theorem (Strong duality theorem.) By substitution from the dual form, and since the two If the primal LP problem has an optimal solution � � solutions are feasible, we know that x ∗ = x ∗ 1 , . . . , x ∗ then the dual also has an optimal solution, n � �   y ∗ = � m � y ∗ 1 , . . . , y ∗ , such that � � � � � � m  y i ≤ c j x j ≤ y i a ij x j ≤ a ij x j b i y i . � � i = 1 j j i j i c j x ∗ b i y ∗ j = i . j i 1. y being dual feasible implies c T ≤ y T A 2. x being primal feasible implies Ax ≤ b 3. ⇒ c T x ≤ ( y T A ) x ≤ y T ( Ax ) ≤ y T b