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)