HW2+Solution

# HW2+Solution - IOE 543 Winter 2010 Homework 2 Solution...

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Unformatted text preview: IOE 543 Winter 2010 Homework 2 Solution Problem 3.1 (a) In order to apply WSPT rule, we have to calculate the values of w j / p j and this is given as: Job 1 2 3 4 5 6 7 Wj/Pj 0 3 2 1.6 2 2.125 1.7778 Optimal schedule: 2‐>6‐>3‐>5‐>7‐>4‐>1 2‐>6‐>5‐>3‐>7‐>4‐>1 The objective value is ∑w C j j =1610 (b) If we change p2 from 6 to 7, there is no change to the optimal schedule. (c) The new objective value will become ∑w C j j =1689. Problem 3.2 The ρ-factor of the chain 1‐>2 is (0+18)/(3+6)=2 and determined by job 2. Similarly, theρ -factor of the chain 3‐>4‐>5 is max{12/6, (12+8)/(6+5),(12+8+8)/(6+5+4)}=12/6=2 and determined by job 3 and for the chain 6‐>7, 17/8=2.125 which is determined by job 6. Since the largestρ-factor is 2.125, job 6 has to be processed first. Now, theρ-factor of the remaining chain is 1.78(16/9 determined by job 7), and 1.78<2. We have to process either job 1‐>2 or job 3 after processing job 6. In other words, the next sequence is either 6‐1‐2‐3 or 6‐3‐1‐2. Theρ-factor of the second chain is now (8+8)/(5+4)=1.78 and determined by job 5. Finally we can choose either sequence 4‐>5 or 7. The optimal schedules are: 6‐1‐2‐3‐4‐5‐7 6‐1‐2‐3‐7‐4‐5 6‐3‐1‐2‐4‐5‐7 6‐3‐1‐2‐7‐4‐5 Problem 3.4 Any schedule in which job 3 is first and job 1 is completed at or before 48 is an optimal sequence. In other words, if we start with job 3 and don’t schedule job 1 as the last job or as the 6th job followed by job 5, then the sequence is optimal. There are overall 4(5!+4!)=576 such sequences. The objective value is 144 Problem 3.5 Since job 1 must precede 7 and 6, so job 1 is guaranteed to finish before time 48. Thus, any sequence in which the precedence constraints are followed and job 3 is first is an optimal sequence. There are 7 possibilities for the ordering of jobs 1, 4, 5, 6, 7, which are: 1‐5‐4‐7‐6 5‐1‐4‐7‐6 1‐5‐7‐4‐6 5‐1‐7‐4‐6 1‐5‐7‐6‐4 5‐1‐7‐6‐4 5‐4‐1‐7‐6 Job 2 may be placed anywhere relative to these jobs (6 possibilities), job 3 must precede all other jobs. So overall there are 6*7=42 optimal sequences. Extra Credits Problem 3.15 In this problem, the preemptive SRPT is optimal. Note that from the graph, there are two schedules S1 and S2. The only difference between S1 and S2 is that at time t, job j is processed in S1 and job i is processed in S2. Without loss of generality, at time t, suppose the remaining processing time for Job j is pj and for Job i is pi and pj >pi. Note that at t+ pi+ pj, both jobs are already completed. In addition, between t and t+ pi+ pj, no other jobs is released or is scheduled to be processed. For schedule S1, For schedule S2, ∑C j = ∑C j = ∑C k + 2t + 2 p j + pi ∑C k + 2t + 2 pi + p j k =i , j k =i , j Since pj >pi, so schedule S2 is better than schedule S1. ...
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