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Unformatted text preview: ORIE 3300/5300 Assignment 12 SOLUTION Fall 2008 Problem 1 (a) Since the variables are binary, on the left branch we place the restriction x = 0 and on the right branch x = 1. (b) The branchandbound tree is here. The optimal solution is x = (1 , 1 , 0) with value z = 16. (1) Original problem z=21.6 x=(F,1,1) x 1 =1 (2) z=20 x=(1,F,F) x 2 =1 (3) z=19.3 x=(1,1,F) x 3 =1 (4) infeasible x 3 =0 (5) z=16 x=(1,1,0) current incumbent = 16 x 2 =0 (6) z=18.5 x=(1,0,F) x 3 =1 (7) infeasible x 3 =0 (8) z=11 worse than incumbent x 1 =0 (9) z=15 worse than incumbent 1 (c) The new branchandbound tree is here. Note that in the first step, the only variable with fractional value is x 1 and so we will branch on x 1 again (and not on x 3 as few of you did). However, in the second step, we have a choice between x 2 and x 3 and in this case, we will choose x 3 because it has higher index. The optimal solution is again x = (1 , 1 , 0) with value z = 16. This time the tree is much smaller. (1) Original problem z=21.6 x=(F,1,1) x 1 =1 (3) z=20 x=(1,F,F) x 3 =1 (5) infeasible x 3 =0 (4) z=16 x=(1,1,0) new incumbent x 1 =0 (2) z=15 x=(0,1,1) current incumbent =15 Comments: As I already mentioned, branching on nonfractional variable was the most common mistake. Also, many of you started your tree in x 1 = 1 node, and thus did not investigate the whole x 1 = 0 half of the tree. If the optimal solution were there, you wouldnt find it. Another mistake was branching on more than one variable at the same time  if you do this, you are investigating...
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 Fall '08
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