4QA3 F12 Week 11 Lecture Notes

Of o ofmaximum n average tardiness tardy jobs

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Unformatted text preview: tardiness = 17.4, Number of tardy jobs=4 4QA3 F12 A. Gandomi 16 AVERAGE Average AVERAGE No. Of O. OFMaximum N Average Tardiness Tardy Jobs Tardiness MAXIMUM Flow Time Rule RULE COMPLETION TIME TARDINESS JOBS TARDY FCFS 24.2 3 42 TARDINESS 53.6 EDD CR SPT 4QA3 F12 47.0 57.8 27.0* 6.6* 17.4 8.6 A. Gandomi 4 4 1* 18* 31 43 17 ●  ●  ●  ●  ●  A machine can process one job at a time, A job can be processed by one machine at a time, Jobs have equal importance and the same arrival time, Setup time is independent of processing sequence, Preemption is not allowed. So, once a job is started on a machine, the job must be completed before another job can be processed by that machine. 4QA3 F12 A. Gandomi 18 ●  For the single machine static sequencing, the following objectives are equivalent: o  o  o  o  o  o  o  o  o  ●  Total completion time Mean completion time Total 5low time (if ready times are all zero) Mean 5low time (if ready times are all zero) Total waiting time Mean waiting time Total lateness Mean lateness Inventory carrying costs Shortest processing time (SPT) rule minimizes all of the above objectives. 4QA3 F12 A. Gandomi 19 ●  Lateness and tardiness are closely related but are not equivalent. If a schedule minimizes maximum lateness, the schedule also minimizes maximum tardiness. However, the converse is not true. ●  EDD minimizes maximum lateness and, thus, maximum tardiness. ●  Moore’s algorithm minimizes number of tardy jobs. ●  Minimizing total/average tardiness is dif5icult and not covered. 4QA3 F12 A. Gandomi 20 ●  ●  An algorithm due to Moore (1968) that minimizes the number of tardy jobs for the single machine problem. Let [1], [2], … [n] be any permutation of the integers 1, 2, …, n. 1.  Order the jobs according to the EDD rule: d[1] ≤ d[ 2 ] ≤ ... ≤ d[ n ] 2.  Find the 5irst tardy job, say d[i]. If none exists go to step 4. 3.  Find and remove job [k] (k=1, 2, 3, …, i) with the longest processing time. Return to step 2. 4.  Form an optimal sequence by taking the current sequence and appending to it the rejected jobs. 4QA3 F12 A. Gandomi 21 Job A B C D E Due Date 7 9 8 17 4 18 6 19 6 21 Job A C D E 4QA3 F12 Processing Time Processing Time Due Date 7 9 4 18 6 19 6 21 A. Gandomi Completion Time 7 15 19 25 31 Completion Time 7 11 17 23 22 Job Processing Time Due Date C 4 18 D 6 19 E 6 21 ●  Completion Time 4 10 16 The optimal sequence is C- D- E- B- A or C- D- E- A- B. In each case, the number of tardy jobs is 2. 4QA3 F12 A. Gandomi 23 ●  The algorithm is applicable for single machine problems with any precedence constraint. The objective function is assumed to be of the form: min max g ( F ) ii 1≤i≤n gi can be any non- decreasing function of the 5low time Fi , including: g i ( Fi ) = max(Fi − d i ,0) g i ( Fi ) = Fi − d i = Li Minimizing maximum lateness 4QA3 F12 Minimizing maximum tardiness A. Gandomi 24 ●  General scheme: The algorithm 5irst assigns a job to the last position, then a job to the position next to last, and so on. ●  Candidate job for a position: Due to precedence...
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