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# hw4 - and its corresponding big M LP min cx 1 Mx a | Ax Ix...

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AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Homework Set # 4 Due in class on Tuesday, October 12, 2010. 1). Solve using the big M method: (Note: you can check your answer using Lindo or ampl.) max z = 2 x 1 - 5 x 2 + x 3 s . t . - x 1 + x 2 + x 3 5 - x 1 - x 2 + x 3 = - 1 5 x 1 + 3 x 2 - x 3 9 x 1 , x 2 , x 3 0 2). Solve the previous problem again, this time using the 2-phase method. 3). Consider an LP problem { min cx | Ax = b, x 0 } , and the corresponding big M problem { min cx + 1 Mx a | Ax + Ix a = b, x 0 , x a 0 } . Is it possible that the optimal solution value of the big M problem is unbounded and the optimal solution of the original problem is bounded? Explain. 4). (From last year’s midterm) Consider an LP (P) min { cx | Ax = b, x
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Unformatted text preview: } , and its corresponding big M LP, { min cx + 1 Mx a | Ax + Ix a = b, x ≥ , x a ≥ } , called P(M). (a). Prove that if M 1 ≤ M 2 then z ( M 1 ) ≤ z ( M 2 ). (b). Suppose that (P) has a ±nite optimal solution and denote it z ∗ . Prove that z ( M ) ≤ z ∗ . (c). Show that there is a value M ′ such that for M ≥ M ′ , z ( M ) = z ∗ , and so we can conclude that the big-M method will produce the right solution for large enough M . (For this part, suppose as in the previous part that (P) has a ±nite optimal solution z ∗ .)...
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