MIT15_053S13_rec07sol

# Figure 2 gives an overview of the branch and bound

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Unformatted text preview: 4 and 5) as active. The above procedure is repeated until there is no active node, at which point the incumbent solution will be optimal. Figure 2 gives an overview of the branch and bound tree for solving Problem (1). As shown in the tree, the nodes 15,17,19,21 get pruned since the corresponding linear program at these nodes are all infeasible. In addition, we get x(4) = (0, 0, 11) with value ZLP (4) = 70 at node 4, so node gets pruned and the incumbent solution is updated. Later on, we get an improved incumbent solution at node 20 since we have x(20) = (1, 1, 1, 0) with value ZLP (20) = 73. This is the optimal solution. IP(1) 1  x1=0 x1=1 ZLP(2)=75.57 2 x2=0 x2=1 ZLP(4)=70 x2=1 x2=0 ZLP(4)=75 5 4 ZLP(3)=75 3 ZLP(6)=77 6 x3=0 x3=1 x3=0 ZLP(9) ZLP(8)=63 8 9 =74.53 x4=0 14 ZLP(14)=53 ZLP(7)=75.846 x4=1 10 x3=1 x3=0 ZLP(12) =75.846 ZLP(11) 11 =75.846 ZLP(10)=59 x4=0 x4=1 x4=0 15 16 InFeas ZLP(16)=49 InFeas 17 18 x3=1 ZLP(13) =75.846 12 x4=1 x4=0 19 20 13 x4=1 21 ZLP(18) InFeas ZLP(20) InFeas =42 =73  Figure 2: Branch and Bound Tree for Problem 1. 4 7 MIT OpenCourseWare http://ocw.mit.edu 15.053 Optimization Methods in Management Science Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms....
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## This note was uploaded on 03/18/2014 for the course MGMT 15.053 taught by Professor Jamesorli during the Spring '07 term at MIT.

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