MT1_514_F05_sol

MT1_514_F05_sol - ISE 514 GEZA BOTTLIK MIDTERM NO. 1...

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ISE 514 Fall 2005 GEZA BOTTLIK MIDTERM NO. 1 Solution 09/29/05 Page 1
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ISE 514 Fall 2005 GEZA BOTTLIK MIDTERM NO. 1 Solution 09/29/05 Problem No. 2 A single machine is intended to process a number of jobs, each with a processing time and due date. Some jobs have to be finished before others can be started and the consequences of the finishing time of jobs vary with the job. Job 5 depends on both 6 and 3 being finished before it can be started. Job 2 is especially valuable and assessed a penalty of 3 times its completion. The penalty for Jobs 1, 4, and 5 is their tardiness. 6 and 3 have no penalty. Find the schedule with the least maximum penalty. (25 points) The processing times and due dates are as follows: Job 1 2 3 4 5 6 Processing time (hrs) 3 5 7 2 4 4 Due time 8 9 20 15 This is an application of Lawler. All but 3 and 6 are eligible to be last. Whichever is last will finish at 25. The penalties are: Job 1 2 4 5 Processing time (hrs) 3 5 2 4 Due time 8 9 20 15 Penalty at 25 17 75 5 10
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MT1_514_F05_sol - ISE 514 GEZA BOTTLIK MIDTERM NO. 1...

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