MIT15_053S13_rec07sol

We rst need an initial incumbent solution we set x 0 0

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Unformatted text preview: al, then make the childeren of node j as active; We now apply this algorithm to solve Problem (1)). We first need an initial incumbent solution. We set x∗ = (0, 0, 0, 0) with value z ∗ = 0 as the initial incumbent solution and then start with node 1, which represents problem IP(1) (notice that IP(1) is the original knapsack problem, that is, Problem (1)). Remember that this node is considered as an active node. We select node 1 and then solve LP(1). Notice that LP(1) is IP(1) with relating the binary variable to be fractional. More precisely, LP(1) is as follows: max 19x1 + 23x2 + 30x3 + 40x4 s.t. 6x1 + 8x2 + 10x3 + 13x4 ≤ 25, (LP(1)) 0 ≤ xi ≤ 1, for i = 1, . . . , 4. We next require to compute an optimal solution for LP(1). There is a simple way to solve LP(1) since there is a single constraint. In fcat, we want to select as much as possible of the items with the greatest profit per weight ratio. In Table 1, the profit per weight ratio of each item is given. Table 1: Profit pe...
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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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