HW3+Solution

# HW3+Solution - IOE 543 Winter 2010 Homework 3 Solution...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ୫ୟ୶ ଴ is obtained by solving the relaxed version of this problem when preemption is allowed. The lower bound for node 0, L ୫ୟ୶ ଴ ൒ 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 ୫ୟ୶ ଵ ൒ 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 ୫ୟ୶...
View Full Document

## This note was uploaded on 03/16/2012 for the course IEOR 466 taught by Professor Richard during the Spring '12 term at Columbia.

### Page1 / 3

HW3+Solution - IOE 543 Winter 2010 Homework 3 Solution...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online