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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.
- Spring '07