sol8 - AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Homework...

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AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Homework Set # 8: Solution notes 1). (a). Dual solution is w * 1 = 4, w * 2 = 0, w * 3 = 10, objective = 440. Therefore, I would choose to increase the amount of resource 3, it has the highest shadow price. (b). Computing the update z 4 - c 4 = wa 4 - c 4 = - 2 < 0, so x 4 wants to enter the basis. After one pivot we get a new optimal tableasu: z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHS z 1 11 4 0 0 8 0 10 480 x 4 0 1 2 0 1 1 0 0 20 x 6 0 -1 1 0 0 0 1 -1 18 x 3 0 1 -2 1 0 -1 0 1 16 (c). Computing the new right hand side B - 1 b = (20 , - 2 , 36) so the solution is “optimal” but no longer feasible. We do a dual simplex pivot, with x 6 leaving the basis and x 4 entering. The new optimal tableau is: z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHS z 1 3 0 0 0 2 4 6 512 x 2 0 -1 1 0 0 0 1 -1 18 x 4 0 3 0 0 1 1 -2 2 4 x 3 0 -1 0 1 0 -1 2 -1 32 (d). Check, the current opt violates the new constraint. Adding the constraint 3 x 1 + 2 x 2 + 2 x 3 + x 4 + x 8 = 80, clean up tableau (to make the basic variables look basic in this new row) and use
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This note was uploaded on 01/31/2011 for the course AMS 540 taught by Professor Arkin,e during the Fall '08 term at SUNY Stony Brook.

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sol8 - AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Homework...

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