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Unformatted text preview: IOE 543 Winter 2010 Homework 3 Solution Problem 3.6 Please see the figure on the next page. First, consider the node 0 at level 0. The lower bound on the L is obtained by solving the relaxed version of this problem when preemption is allowed. The lower bound for node 0, L 2 , is given by the optimal sequence 1,2 1 ,4,2 2 ,3,6 1 ,7,6 2 ,5 which is found by preemptive EDD. (Note that we preempt the jobs 2 and 6). Then for level 1 we have 7 nodes. However, the nodes that are starting with jobs 4,5,6,7 can be eliminated because it is easy to see that they are not released at time t=0 and these nodes cannot be optimal. Let us calculate the lower bound for node 1. The schedule starts with job 1, and to determine the rest of the sequence, we again use preemptive EDD. The optimal sequence and the lower bound for node 1 are: 1,2 1 ,4,2 2 ,3,6 1 ,7,6 2 ,5 and L 2 . By the same argument we have the lower bounds for the following nodes: Node 2: 2,1,4,3,6 1 ,7,6 2 ,5 with L...
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