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Unformatted text preview: CO350Linear Optimization Assignment 9 Solutions 1. Consider the following optimal tableau for the LP relaxation of an integer linear program. For each constraint row, find the corresponding Gomory cutting plane inequality, then write it in a form that can be added to the tableau. z + 1 14 x 4 + 3 14 x 5 + 5 4 x 6 = 27 4 x 1 x 5 5 4 x 6 = 9 4 x 2 + 1 2 x 4 1 2 x 5 + 17 4 x 6 = 13 4 x 3 3 7 x 4 + 5 7 x 5 + 3 x 6 = 1 7 Solution: The Gomory cutting plane inequalities are 3 4 x 6 1 4 1 2 x 4 + 1 2 x 5 + 1 4 x 6 1 4 4 7 x 4 + 5 7 x 5 1 7 . To write them in a form that can be added to the tableau, we negate them and add slack variables, getting 3 4 x 6 + x 7 = 1 4 1 2 x 4 1 2 x 5 1 4 x 6 + x 8 = 1 4 4 7 x 4 5 7 x 5 + x 9 = 1 7 . 2. Solve the following integer linear program using the cutting plane algorithm, starting from the optimal basis { 1 , 2 , 5 } for its LP relaxation. max x 1 + 3 x 2 s.t. x 1 + 2 x 2 + x 3 = 2 x 1 + 2 x 2 + x 4 = 4 2 x 1 2 x 2 + x 5 = 2 x 1 , x 2 , x 3 , x 4 , x 5 , integers . Solution: The tableau for basis { 1 , 2 , 5 } is z + 5 4 x 3 + 1 4 x 4 = 7 2 x 1 1 2 x 3 + 1 2 x 4 = 1 x 2 + 1 4 x 3 + 1 4 x 4 = 3 2 3 2 x 3 1 2 x 4 + x 5 = 3 . The corresponding basic solution is not integral. The Gomory cut for the x 2row is 1 4 x 3 + 1 4 x 4 1 2 , which we can write as 1 4 x 3 1 4 x 4 + x 6 = 1 2 to insert it in the tableau. We then start the dual simplex method on the new tableau. For the first iteration, the leaving variable is x 6 , and min c j a j 6 : a j 6 < = min , , 5 / 4 1 / 4 , 1 / 4 1 / 4 , , = 1 / 4 1 / 4 , so x 4 is the entering variable. After pivoting, the new tableau is z + x 3 + x 6 = 3 x 1 x 3 + 2 x 6 = 0 x 2 + x 6 = 1...
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This note was uploaded on 05/28/2011 for the course CO 350 taught by Professor S.furino,b.guenin during the Winter '07 term at Waterloo.
 Winter '07
 S.Furino,B.Guenin

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