hw12_Solutions - ORIE 3300/5300 Assignment 12 SOLUTION Fall...

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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 branch-and-bound 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 branch-and-bound 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 non-fractional 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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hw12_Solutions - ORIE 3300/5300 Assignment 12 SOLUTION Fall...

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