Assignment3_solutions - x = [0 , 3 , , 1] T is feasible to...

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ORIE 320, Homework 3, Question 3 Consider the following tableau for a linear program in standard equality form. z + 1 3 x 1 + 1 3 x 3 = 8 1 3 x 1 - 2 3 x 3 + x 4 = 1 1 3 x 1 + x 2 - 1 6 x 3 = 3 ( * ) (a) The corresponding basis is B = [2 , 4]. (b) The optimal solution for the original problem is x = [0 , 3 , 0 , 1] T with objective value 8. Note that this is a basic feasible solution corresponding to the current basis B . Part (c) explains why this solution is feasible and (d) explains why this solution is indeed optimal. (c) Recall that the above tableau was obtained from the original linear program (in a standard equality form) z - c T x = 0 Ax = b after a finite number of elementary row operations. Since elementary row operations preserve equivalance, the new system ( * ) is equivalent with the original one. This means that any feasible solution to ( * ) is also feasible to the original problem. Hence, solution
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Unformatted text preview: x = [0 , 3 , , 1] T is feasible to the original linear program. (d) In part (c), we showed that any feasible solution to the original (maximization) problem has to be feasible also to the system ( * ). In particular, it has to solve its rst equation z + 1 3 x 1 + 1 3 x 3 = 8 which implies (by non-negativity of x i s) z = 8-1 3 x 1-1 3 x 3 8 . This shows that for any feasible solution to the original problem, the objective value z cannot be larger than 8. On the other hand, as argued in (c), the solution x = [0 , 3 , , 1] is feasible to the original program and attains objective value equal to 8. Hence, it must be optimal to the original problem. 1...
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This note was uploaded on 09/29/2011 for the course ORIE 5300 taught by Professor Todd during the Fall '08 term at Cornell University (Engineering School).

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Assignment3_solutions - x = [0 , 3 , , 1] T is feasible to...

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