HW5+Solution

HW5+Solution - IOE 543 Winter 2010 Homework 5 Solution...

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Unformatted text preview: IOE 543 Winter 2010 Homework 5 Solution Problem 5.1 (a) If you apply LPT rule, you will get the following schedule for each machine with Cmax=23. Machine 1: Job 13, Job 2, Job 1 Machine 2: Job 12, Job 3 Machine 3: Job 11, Job 4 Machine 4: Job 10, Job 5 Machine 5: Job 9, Job 6 Machine 6: Job 8, Job 7 (b) The optimal schedule for this problem is the following sequence with Cmax=18. Machine 1: Job 13, Job 5 Machine 2: Job 12, Job 4 Machine 3: Job 11, Job 7 Machine 4: Job 10, Job 6 Machine 5: Job 9, Job 8 Machine 6: Job 1, Job 2, Job 3 Problem 5.3 (a) According to the LPT rule, we have the following sequence on the three machines: Machine 1: J1, J3, J4, J5 Machine 2: J2, J6, J7, J8 Machine 3: Idle The objective function value is Cmax=31 (b) According to the LFJ rule, we have the following sequence on the three machines: Machine 1: J3, J4, J5 Machine 2: J6, J7, J8 Machine 3: J1, J2 The objective function value is Cmax=21 (c) Ratio=31/21=1.476 Problem 5.5 Jobs 1 2 3 4 5 6 7 8 9 10 11 12 13 Earliest 5 16 25 8 15 11 24 22 20 2 5 7 16 Latest 5 16 25 9 16 16 25 25 25 9 14 16 25 The critical jobs are {1,2,3} and the rest of the jobs are slack while the optimal makespan is Cmax=25 Problem 5.6 (a) Schedule under SPT: Machine 1: J1, J6, J11 Machine 2: J2, J7 Machine 3: J3, J8 Machine 4: J4, J9 Machine 5: J5, J10 Objective function value is 6 (b) Optimal schedule: Machine 1: J1, J2, J3 Machine 2: J4, J10 Machine 3: J5, J11 Machine 4: J6, J8 Machine 5: J7, J9 Objective function value is 0 since all jobs finish by time 15 ...
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