MIT15_053S13_rec07sol

0769 therefore we rst select item 4 since it has the

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Unformatted text preview: r weight ratio of each item for problem 2 item Proﬁt(P) Weight(W) P/W 1 19 6 3.16667 2 23 8 2.875 3 30 10 3 4 40 13 3.0769 Therefore, we ﬁrst select item 4 since it has the most proﬁt per weight ratio. The knapsack still have a capacity of 12, and we then select item 1. After selecting item 1 and 4, the knapsack still has a capacity of 6. So we can select 0.6 fraction of item 3. This gives an optimal solution for LP(1) as x(1) = (1, 0, 0.6, 1) with value ZLP (1) = 75.75. At this node, Case 3 occurs, so we make the children of node 1 (i.e., nodes 2 and 3) as active. 3 We next choose node 2 and make it inactive. We then solve LP(2). Notice that at node 2, the value of the ﬁrs binary variable is set to be zero, so LP(2) is as follows: max 23x2 + 30x3 + 40x4 s.t. 8x2 + 10x3 + 13x4 ≤ 25, (LP(2)) 0 ≤ xi ≤ 1, for i = 2, . . . , 4. The optimal solution of LP(2) is x(2) = (0, 2/8, 1, 1) with value ZLP (2) = 75. Notice that Z ∗ &lt; ZLP (2) and x(2) is not integral, so we make the children of node 2 (nodes...
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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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