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Data Collection for Scheduling l Jobs l Activities l Employees l Equipment l Facilities Transparency 18.1
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Scheduling OperationsScheduling Operations
IDS 605IDS 605Spring 1999Spring 1999
Data Collection for SchedulingData Collection for Scheduling
Jobs Activities Employees Equipment Facilities
Transparency 18.1
Managerial ConsiderationsManagerial Considerations
Meeting customer due date
Minimizing flow time Minimizing WIP
inventory Minimizing costs Minimizing idle time
» equipment» employee
Transparency 18.2
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.
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
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
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
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
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
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
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
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)
Forward Schedule for Four Jobs Forward Schedule for Four Jobs with Finite Loadingwith Finite Loading
Transparency 18.6 (Exhibit 18.7)
Backward Schedule for Jobs with Backward Schedule for Jobs with Infinite LoadingInfinite Loading
Transparency 18.7 (Exhibit 18.8)
Gantt Load Chart forGantt Load Chart forForward ScheduleForward Schedule
Transparency 18.8 (Exhibit 18.10)