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homework-8-solutions

homework-8-solutions - MthSc810 Mathematical Programming...

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MthSc810 – Mathematical Programming Fall 2011, solutions to Homework #8. Problem 1. Write the dual of the following problems: 1. The Klee-Minty example for general n ; 2. The bank loan model seen in lecture 2, slide 21; 3. The production planning model (same lecture, slide 11). Solutions. 1. Primal: min - n j =1 10 n j x j s.t. x i + 2 i 1 j =1 10 i j x j 100 i 1 i = 1 , 2 . . . , n x i 0 i = 1 , 2 . . . , n The dual has n constraints and n variables, just like the primal: max n i =1 100 i 1 u j s.t. u i + 2 n j = i +1 10 j i u j ≤ - 10 n i i = 1 , 2 . . . , n u i 0 i = 1 , 2 . . . , n. 2. Primal: min y 1 + y 2 s.t. y 1 0 . 05 x 1 y 1 5 + 0 . 08( x 1 - 100) y 2 0 . 03 x 2 y 2 4 . 2 + 0 . 12( x 2 - 140) x 1 + x 2 = 300 x 1 0 x 2 0 . Let’s rewrite it so that variables are on the left-hand side, and constants on the rhs (and write the dual variables’ names next to each primal constraint): min y 1 + y 2 s.t. - . 05 x 1 + y 1 0 ( u 1 ) - . 08 x 1 + y 1 - 3 ( u 2 ) - . 03 x 2 + y 2 0 ( u 3 ) - . 12 x 2 + y 2 ≥ - 12 . 6 ( u 4 ) x 1 + x 2 = 300 ( u 5 ) x 1 , x 2 0 . Dual: max - 3 u 2 - 12 . 6 u 4 + 300 u 5 s.t. - . 05 u 1 - . 08 u 2 + u 5 0 - . 03 u 3 - . 12 u 4 + u 5 0 u 1 + u 2 = 1 u 3 + u 4 = 1 u 1 , u 2 , u 3 , u 4 0 . 3. Primal: min 12 i =1 c i x i x i + y i 1 = d i + y i i = 1 , 2 . . . , 12 0 x i P i = 1 , 2 . . . , 12 y i 0 i = 1 , 2 . . . , 12 y 0 = y 12 = 0 Rewrite it to ease writing the dual. Because the two non-trivial constraints are written for various values of i , the corresponding dual will be indexed
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similarly. However, the two trivial constraints y 0 = 0 and y 12 = 0 will be associated with dual variables rather than used for simplifying the problem (which would be an equally good solution). min 12 i =1 c i x i x i + y i 1 - y i = d i i = 1 , 2 . . . , 12 ( u i ) x i P i = 1 , 2 . . . , 12 ( v i ) y 0 = 0 ( λ ) y 12 = 0 ( μ ) y i 0 i = 1 , 2 . . . , 12 x i 0 i = 1 , 2 . . . , 12 Dual: max 12 i =1 d i u i + P 12 i =1 v i u i + v i c i i = 1 , 2 . . . , 12 u 1 + λ 0 - u i + u i +1 0 i = 1 , 2 . . . , 11 - u 12 + μ 0 v i 0 i = 1 , 2 . . . , 12 Problem 2.
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