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# sol4 - 3 10 M 3 4-1/3-1 1-2/3 10/3 1 2/3 1/3 4/3 This is an...

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AMS 341 Spring, 2010 Homework Set # 4 — Solution Notes #3, 4.7: Initial tableau is z x 1 x 2 x 3 s 1 s 2 RHS 1 -1 -1 0 0 0 0 0 1 1 1 1 0 1 0 1 0 2 0 1 1 We can choose x 1 as the entering variable, so s 1 wins the min ratio test and leaves: z x 1 x 2 x 3 s 1 s 2 RHS 1 0 0 1 1 0 1 0 1 1 1 1 0 1 0 0 -1 1 -1 1 0 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 x 3 s 1 s 2 RHS 1 0 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 2 0 1 1 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 are no competitors in the minimum ratio test, so the LP is unbounded. #4, 4.12: Big M method. Initial tableau is z x 1 x 2 e 1 a 1 a 2 RHS 1 5 M - 3 3 M - M 0 0 10 M 0 2 1 -1 1 0 6 0 3 2 0 0 1 4 Now enter x 1 in the basis in row 2. z x 1 x 2 e 1 a 1 a 2 RHS 1 0 (6 - M ) / 3 - M 0 (3 - 5
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Unformatted text preview: ) / 3 10 M/ 3 + 4-1/3-1 1-2/3 10/3 1 2/3 1/3 4/3 This is an optimal tableau, but a 1 is still positive, so the original problem is not feasible. #4, 4.12: Two-phase simplex method. Initial Phase I tableau is w x 1 x 2 e 1 a 1 a 2 RHS 1 5 3-1 10 2 1-1 1 6 3 2 1 4 Now enter x 1 in the basis in row 2. w x 1 x 2 e 1 a 1 a 2 RHS 1-1/3-1-5/3 10/3-1/3-1 1-2/3 10/3 1 2/3 1/3 4/3 This is an optimal tableau to phase I, but there is still an arti±cial variable a 1 which is positive, so the original problem is not feasible. Note, in this case there is no phase II....
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