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# sol4 - AMS 540 MGT 540(Fall 2010 Estie Arkin Homework Set 4...

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AMS 540 / MGT 540 (Fall, 2010) Estie Arkin Homework Set # 4: Solution notes 1). Solve using the big M method: Set up a standard form, with a 1 and a 2 the artificial variables: max z = 2 x 1 - 5 x 2 + x 3 - Ma 1 - Ma 2 s . t . - x 1 + x 2 + x 3 - x 4 + a 1 = 5 x 1 + x 2 - x 3 + a 2 = 1 5 x 1 + 3 x 2 - x 3 + x 5 = 9 x 1 ,x 2 ,x 3 ,x 4 ,x 5 ,a 1 ,a 2 0 after cleaning up the objective row, the first tableau is: z x 1 x 2 x 3 x 4 x 5 a 1 a 2 RHS z 1 -2 5-2M -1 M 0 0 0 -6M a 1 0 -1 1 1 -1 0 1 0 5 a 2 0 1 1 -1 0 0 0 1 1 x 5 0 5 3 -1 0 1 0 0 9 pivot x 2 in and a 2 out and get second tableau: z x 1 x 2 x 3 x 4 x 5 a 1 a 2 RHS z 1 2M-7 0 4-2M M 0 0 2M-5 -5-4M a 1 0 -2 0 2 -1 0 1 -1 4 x 2 0 1 1 -1 0 0 0 1 1 x 5 0 2 0 2 0 1 0 -3 6 pivot x 3 in and a 1 out get third tableau: z x 1 x 2 x 3 x 4 x 5 a 1 a 2 RHS z 1 -3 0 0 2 0 M-2 M-3 -13 x 3 0 -1 0 1 -0.5 0 0.5 -0.5 2 x 2 0 0 1 0 -0.5 0 0.5 0.5 3 x 5 0 4 0 0 1 1 -1 -2 2 At this point we have a feasible solution to the original problem, we could remove the artificial columns, if we want. Pivot x 1 in and x 5 out to get the last (optimal) tableau: z x 1 x 2 x 3 x 4 x 5 a 1 a 2 RHS z 1 0 0 0 2.75 0.75 M-2.75 M-4.5 -11.5 x 3 0 0 0 1 -0.25 0.25 0.25 -1 2.5 x 2 0 0 1 0 -0.5 0 0.5 0.5 3 x 1 0 1 0 0 0.25 0.25 -0.25 -0.5 0.5

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Note that M appears throughout only in the objective row, and at the final tableau only in the coefficients of the artificial variables. The optimal solution is x 1 = 0 . 5, x 2 = 3, x 3 = 2 . 5 and z = - - 11 . 5.
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