# sol4 - are no competitors in the minimum ratio test so the...

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AMS 341 Spring, 2011 Homework Set # 4 — Solution Notes #5, 4.7: Initial tableau is z x 1 x 2 s 1 s 2 RHS 1 -2 -2 0 0 0 0 1 1 1 0 6 0 2 1 0 1 13 We can choose x 1 as the entering variable, so s 1 wins the min ratio test and leaves: z x 1 x 2 s 1 s 2 RHS 1 0 0 2 0 12 0 1 1 1 0 6 0 0 -1 -2 1 1 This tableau is optimal. However, we can enter x 2 without changing the value of the objective, and so get another optimal solution: z x 1 x 2 s 1 s 2 RHS 1 0 0 2 0 12 0 1 1 1 0 6 0 1 0 -1 1 7 So, we found 2 BFS that are optimal. #3, 4.8: From the given tableau, we know that z = 2 x 2 ,x 1 = 0 ,x 3 = 3 + x 2 ,x 4 = 4 are feasible solutions of the LP. Therefore, z is unbounded by letting x 2 be large. In other words, x 2 can be chosen as an entering variable, and there
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Unformatted text preview: are no competitors in the minimum ratio test, so the LP is unbounded. #9, Ch4 Review: Big M method: z x 1 x 2 e 1 e 2 a 1 a 2 RHS 1 3-1-M-M-M 5 M 1-2-1 1 2-1 1-1 1 3 1 5-M 3-M-M-3-6 + 5 M 1-2-1 1 2-1-1-1 1 1 5 Two-phase method: w x 1 x 2 e 1 e 2 a 1 a 2 RHS 1-1-1-1 5 1-2-1 1 2-1 1-1 1 3 From both methods we have that the LP is infeasible since arti±cial variables can’t be reduced to 0....
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