Unformatted text preview: z = 18 , x 1 = 2; x 2 = 2 #2, 9.8: The second contraint: x 2e 21 = 3 51 5 e 12 5 e 2 . Therefore the cut is 3 51 5 e 12 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.
 Spring '08
 Arkin,E

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