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Unformatted text preview: CO 350 Linear Optimization (Spring 2009) Solution to Assignment 4 [10] 1. Note that there is no ndimensional nonnegative vector x such that d ≤ Ax ≤ b if and only if the following system has no solution: Ax ≤ b Ax ≤  d x ≥ . By Farkas’ Lemma (Theorem 4.11), this is true if and only if there exist mdimensional vectors y and z such that y,z ≥ , A T y A T z ≥ , and b T y d T z < , i.e., if and only if there are mdimensional nonnegative vectors z and y such that y T A ≥ z T A and y T b < z T d . [10] 2. (a) Using the data to form the following. z 2 x 1 2 x 2 6 x 3 9 x 4 3 x 5 x 6 = 0 2 x 1 + x 3 + 4 x 4 + 2 x 5 + x 6 = 5 x 2 + x 3 + 5 x 4 + x 5 = 4 x 3 + x 4 + x 5 + x 6 = 1 . We transform this to a tableau for B = { 1 , 2 , 6 } . z 3 x 3 + 5 x 4 + x 5 = 13 x 1 + 3 2 x 4 + 1 2 x 5 = 2 x 2 + x 3 + 5 x 4 + x 5 = 4 x 3 + x 4 + x 5 + x 6 = 1 . The corresponding basic feasible solution is [2 , 4 , , , , 1] T with objective value 13. The only candidate for entering variable is x 3 . For the leaving variable, the ratios are { , 4 , 1 } , so x 6 is the leaving variable. We pivot to get the following tableau. z + 8 x 4 + 4 x 5 + 3 x 6 = 16 x 1 + 3 2 x 4 + 1 2 x 5 = 2 x 2 + 4 x 4 x 6 = 3 x 3 + x 4 + x 5 + x 6 = 1 . The current basis is B = { 1 , 2 , 3 } with corresponding basic feasible solution [2 , 3 , 1 , , , 0] T . The objective value is 16. Since all reduced costs are nonpositive, this is an optimal solution. (b) The dual of the LP is min 5 y 1 + 4 y 2 + y 3 s.t. 2 y 1 ≥ 2 y 2 ≥ 2 y 1 + y 2 + y 3 ≥ 6 4 y 1 + 5 y 2 + y 3 ≥ 9 2 y 1 + y 2 + y 3 ≥ 3 y 1 + y 3 ≥ 1 . 1 (c) The optimal tableau from part (a) has the basis B = { 1 , 2 , 3 } . To find an optimal solution for the dual, we need to solve A T B y = c B , which is 2 0 1 0 1 1 0 0 1 T y 1 y 2 y 3 = 2 2 6 . This gives 2 y 1 = 2, so y 1 = 1; y 2 = 2; and y 1 + y 2 + y 3 = 6, so y 3 = 3. One can easily check that y = [1 , 2 , 3] T is feasible for the dual and has objective value 16, so this confirms that y is optimal for the dual. (d) First note that the optimal solution from part (a) is nondegenerate. Then we may use Theorem 6.2 and conclude that the optimal value for the new problem is 16 + εy 2 = 16 + 2 ε . [10] 3. The first tableau for Phase 1 is: w 4 x 2 7 x 3 + 5 x 4 = 3 2 x 1 + x 2 + 3 x 3 4 x 4 + x 5 = 2 2 x 1 + 3 x 2 + 4 x 3 x 4 + x 6 = 1 The second tableau for Phase 1 is: w + 8 / 3 x 1 5 / 3 x 3 + 11 / 3 x 4 + 4 / 3 x 6 = 5 / 3 8 / 3 x 1 + 5 / 3 x 3 11 / 3 x 4 + x 5 1 / 3 x 6 = 5 / 3 2 / 3 x 1 + x 2 + 4 / 3 x 3 1 / 3 x 4 + 1 / 3 x 6 = 1 / 3 The third and optimal tableau for Phase 1 is: w + 7 / 2 x 1 + 5 / 4 x 2 + 13 / 4 x 4 + 7 / 4 x 6 = 5 / 4 7 / 2 x 1 5 / 4 x 2 13 / 4 x 4 + x 5 3 / 4 x 6 = 5 / 4 1 / 2 x 1 + 3 / 4 x 2 + x 3 1 / 4 x 4 + 1 / 4 x 6 = 1 / 4 Since the optimal value of the auxiliary problem is less than 0, the original LP is infeasible....
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This note was uploaded on 04/11/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Fall '07 term at Waterloo.
 Fall '07
 S.Furino,B.Guenin

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