MIT15_053S13_rec07sol

Since this solution is not better than the incumbent

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Unformatted text preview: solution to this relaxation is x5 = 1/2 with an objective value of 14. Since this solution is not better than the incumbent, node 13 gets fathomed. Problem 2 Consider the knapsack problem with the following decision variables for i = 1 to 4: � 1 if item i is selected; xi = 0 otherwise. The knapsack problem is formulated as follows: 2 max 19x1 + 23x2 + 30x3 + 40x4 s.t. 6x1 + 8x2 + 10x3 + 13x4 ≤ 25, (1) xi ∈ {0, 1}, for i = 1, . . . , 4. Apply the branch-and-bound algorithm to solve the problem (notice that the the LP relax­ ation of a knapsack problem can be easily solved by selecting the items. Solution. Let’s first review the branch and bound algorithm: At each iteration, we select an active node j and make it inactive. Suppose that LP(j ) represents the integer program corresponding to node j , in which the binary variables are relaxed to be fractional. Let x(j ) and ZLP (j ) be the optimal solution and the optimal value of LP. There are three cases: Case: 1 If ZLP (j ) ≤ Z ∗ , the node j gets pruned; Case: 2 If ZLP (j ) > Z ∗ , and x(j ) is integral, then update the incumbent solution by setting Z ∗ := ZLP (j ) and x∗ := x(j ); In addition, node j gets pruned, Case: 3 If ZLP (j ) > Z ∗ and x(j ) is not integr...
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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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