# sol9 - z = 18 x 1 = 2 x 2 = 2#2 9.8 The second contraint x...

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AMS 341 Spring, 2010 Homework Set # 9 — Solution Notes #6, 9.3: Subproblem 1 (LP relaxation of the IP): z = 18 . 7 , x 1 = 1 . 8; x 2 = 2 . 3 Subproblem 2 (Subproblem 1 + x 1 1): z = 16 . 5 , x 1 = 1; x 2 = 2 . 5 Subproblem 3 (Subproblem 1 + x 1 2): z = 18 , x 1 = 2; x 2 = 2 Subproblem 3 gives us a candidate solution. Using it, we can fathom subproblem 2, which had a z worse than the candidate z = 18. We stop, and have the optimal solution:
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Unformatted text preview: z = 18 , x 1 = 2; x 2 = 2 #2, 9.8: The second contraint: x 2-e 2-1 = 3 5-1 5 e 1-2 5 e 2 . Therefore the cut is 3 5-1 5 e 1-2 5 e 2 ≤ 0. We can use the dual simplex algorithm (page 316) or LINDO, we get the optimal solution: z = 20 , x 1 = 2 , x 2 = 1, which is integer, and so we are done....
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## This note was uploaded on 08/08/2010 for the course AMS 341 taught by Professor Arkin,e during the Spring '08 term at SUNY Stony Brook.

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