scheduling operations ids 605 spring 1999. data collection for scheduling l jobs l activities l...
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Data Collection for Scheduling l Jobs l Activities l Employees l Equipment l Facilities Transparency 18.1TRANSCRIPT
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Scheduling OperationsScheduling Operations
IDS 605IDS 605Spring 1999Spring 1999
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Data Collection for SchedulingData Collection for Scheduling
Jobs Activities Employees Equipment Facilities
Transparency 18.1
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Managerial ConsiderationsManagerial Considerations
Meeting customer due date
Minimizing flow time Minimizing WIP
inventory Minimizing costs Minimizing idle time
» equipment» employee
Transparency 18.2
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Shop Floor Control - SchedulingShop Floor Control - Scheduling
Suppose we have a group of jobs waiting in front of a work center. How do we decide which job to run next (i.e., priority assignment)?
We will examine 4 common priority rules. We will evaluate the performance of the
rules via shop efficiency and customer service measurements.
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JOB PRODUCTION TIME UNTILNUMBER TIME (HRS.) DUE (HRS.)
M1 2 3M2 3.1 7M3 1.4 6M4 2.6 5M5 0.5 8M6 1.9 12
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First-Come First-Served (FCFS) First-Come First-Served (FCFS) SequenceSequence
Sequence: M1-M2-M3-M4-M5-M6
M1 M2 M3 M4 M5 M6
0 1 2 3 4 5 6 7 8 9 10 11
Average Flow Time:
2+5.1+6.5+9.1+9.6+11.5 = 43.8/6 = 7.30 hours
Average Number of Jobs in the System:
2(6)+3.1(5)+1.4(4)+2.6(3)+0.5(2)+1.9(1) = 43.8/11.5 =3.81 jobs
Average Job Lateness ~ MAX((c-d),0)
0+0+0.5+4.1+1.6+0 = 6.2/6 = 1.033
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SPT (a.k.a. SON) SequenceSPT (a.k.a. SON) Sequence
Sequence: M5-M3-M6-M1-M4-M2 0.5 1.4 1.9 2.0 2.6 3.1
M5 M3 M6 M1 M4 M2
0 1 2 3 4 5 6 7 8 9 10 11
Average Flow Time:
0.5 + 1.9 + 3.8 + 5.8 + 8.4 + 11.5 = 31.9/6 = 5.32 hours
Average Number of Jobs in the System:
.5(6)+1.4(5)+1.9(4)+2.0(3)+2.6(2)+3.1(1) = 31.9/11.5 =2.77 jobs
Average Job Lateness ~ MAX((c-d),0) M1 M4 M2
0+0+0+2.8+3.4+4.5 = 10.70/6 = 1.78 hours
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Earliest Due Date (EDD) Earliest Due Date (EDD) SequenceSequence
Sequence: M1-M4-M3-M2-M5-M6
M1 M4 M3 M2 M5 M6
0 1 2 3 4 5 6 7 8 9 10 11
Average Flow Time:
2.0 + 4.6 + 6.0 + 9.1 + 9.6 + 11.5 = 42.8/6 = 7.13 hours
Average Number of Jobs in the System:
2(6)+2.6(5)+1.4(4)+3.1(3)+0.5(2)+1.9(1) = 42.8/11.5 =3.72 jobs
Average Job Lateness ~ MAX((c-d),0) M2 M5
0+0+0+2.1+1.6+0 = 3.70/6 = 0.6 hours
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Critical Ratio (CR) SequenceCritical Ratio (CR) SequenceCRITICAL RATIOS:M1: 3/2 = 1.5 M4: 5/2.6 = 1.9M2: 7/3.1 = 2.3 M5: 8/0.5 = 16M3: 6/1.4 = 4.3 M6: 12/1.9 = 6.3
Sequence: M1-M4-M2-M3-M6-M5
M1 M4 M2 M3 M6 M5
0 1 2 3 4 5 6 7 8 9 10 11
Average Flow Time:
2.0 + 4.6 + 7.7 + 9.1 + 11.0 + 11.5 = 45.9/6 = 7.65 hours
Average Number of Jobs in the System:
2(6)+2.6(5)+3.1(4)+1.4(3)+1.9(2)+0.5(1) = 45.9/11.5 =3.99 jobs
Average Job Lateness ~ MAX((c-d),0) M2 M3 M5
0+0+0.7+3.1+0+3.5 = 7.30/6 = 1.22 hours
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SummarySummary
CRITERIA FCFS SPT EDD CR
Average Flow Time 7.3 5.32 7.13 7.65
Average Number of 3.81 2.77 3.72 3.99Jobs In System
Average Job 1.03 1.78 0.6 1.22Lateness
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Jobs to be ProcessedJobs to be Processedon Two Machineson Two Machines
Processing time (hours) Job Computing Printing A 1.5 1.0 B 1.0 .75 C .5 1.25 D 2.0 1.5 E .75 .5
Transparency 18.4
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Processing of Computer Jobs Processing of Computer Jobs Based on Sequencing by Based on Sequencing by
Johnson’s RuleJohnson’s Rule
Transparency 18.5 (Exhibit 18.6)
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Forward Schedule for Four Jobs Forward Schedule for Four Jobs with Finite Loadingwith Finite Loading
Transparency 18.6 (Exhibit 18.7)
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Backward Schedule for Jobs with Backward Schedule for Jobs with Infinite LoadingInfinite Loading
Transparency 18.7 (Exhibit 18.8)
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Gantt Load Chart forGantt Load Chart forForward ScheduleForward Schedule
Transparency 18.8 (Exhibit 18.10)