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Unformatted text preview: CO 350 Assignment 7 Winter 2010 Solutions Due: Friday March 19 at 10:25 a.m. Question # Max. marks 1 3 2 8 3 8 4 6 5 5 6 (5) Total 30 + (5) 1. (a) Study the example of cycling on Page 109 of the Course Notes. Notice that it uses the largest coefficient rule to choose the entering variable. (b) Apply the following perturbation to the first tableau: Change the righthand side of the x 5row to 0.01 and change the righthand side of the x 6row to 0.0001. Now apply the simplex method to the resulting problem, still using the largest coefficient rule to choose the entering variable. Solution: The first tableau is z 2 x 1 3 x 2 + x 3 + 12 x 4 = 2 x 1 9 x 2 + x 3 + 9 x 4 + x 5 = . 01 1 / 3 x 1 + x 2 1 / 3 x 3 2 x 4 + x 6 = . 0001 Now x 2 enters and x 6 leaves. The new tableau is: z x 1 + 6 x 4 + 3 x 6 = . 0003 x 1 2 x 3 9 x 4 + x 5 + 9 x 6 = . 0109 1 / 3 x 1 + x 2 1 / 3 x 3 2 x 4 + x 6 = . 0001 Now x 1 enters. We compute t = min( . 0001 1 3 , . 0109 1 ) = . 0003. x 2 leaves. The new tableau is z + 3 x 2 x 3 + 6 x 6 = . 0006 3 x 2 x 3 3 x 4 + x 5 + 6 x 6 = . 0106 x 1 + 3 x 2 x 3 6 x 4 + 3 x 6 = . 03 Now x 3 enters, and the algorithm terminates with unboundedness. 2. Let A =  1 3 1 1 0 0 1 1 3 0 1 0 1 2 2 0 0 1 , b = 2 5 1 c = 1 , 1 , 2 , , , 3 T define the data for an LP problem in standard equality form. Start with the basis B = { 1 , 4 , 5 } and associated basic feasible solution x * = (1 , , , 3 , 4 , 0) T and solve the problem by the revised simplex method. In every iteration, use the smallest subscript rule to choose the entering variable. Solution: 1 Iteration 1: B = { 1 , 4 , 5 } ,x * = (1 , , , 3 , 4 , 0) T . Solve A T B y = c B , i.e.  1 1 1 1 1 y 1 y 2 y 3 = 1 . So y 1 = y 2 = 0 ,y 3 = 1. Compute c 2 = c 2 A T 2 y = 1 (3 , 1 , 2)  1 = 3 > , so x 2 enters. Solve A B d = A 2 , i.e....
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This note was uploaded on 05/28/2011 for the course CO 250 taught by Professor Guenin during the Fall '10 term at Waterloo.
 Fall '10
 GUENIN

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