30-10-2025
Scheduling and Sequencing
Sushil Punia
VGSOM, IIT KGP
Agenda
•
What is Scheduling?
•
How to do job shop scheduling?
•
Learning various Sequencing Methods
•
How to do personnel scheduling?
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Scheduling
in OM
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Objectives of Scheduling
Measuring Scheduling Performance:
Meet due dates.
Minimize work-in-process (WIP) inventory.
Minimize the average time spent through the
system by job.
Minimize machine/worker idle time
Reduce setup times
Minimize production and worker costs.
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Sequencing Jobs
Assume that there is a collection of jobs that must be
processed on a machine and that each job has
associated with it a processing time and a due date.
• Specifies the order in which jobs should be
performed at work centers
• Priority rules are used to dispatch or sequence
jobs
• FCFS: First come, first served
• SPT: Shortest processing time
• EDD: Earliest due date
• LPT: Longest processing time
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Sequencing Example
Applying the sequencing rules to these five
jobs
Job
Job Work (Processing) Time
(Days)
Job Due Date
(Days)
A
B
C
D
E
6
2
8
3
9
8
6
18
15
23
6
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Some Important Metrics
Flow time of a Job i
The flow time of job i is the time that elapses from the initiation of the first job on the
first machine to the completion of job i.
The sum of waiting to be processed, transportation time between operations, and any
waiting time related to equipment breakdowns, unavailable parts, quality problems, etc.
Average Flow Time of Jobs = Sum of Flow Time of all Jobs / Number of Jobs
Makespan time of Jobs
Total time needed to complete a group of jobs from the beginning (processing) of the first
job on the first machine to the completion of the last job (on the last machine).
Or the Flow time of the last job.
Independent of priority rules for single work center scheduling.
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Some Important Metrics
Lateness of a Job
The difference between the actual completion date and the due date. Positive
or Negative.
Tardiness is lateness for late jobs only. For early jobs, tardiness is zero.
Number of tardy jobs
Count of jobs with positive lateness
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Sequencing Example
FCFS: Sequence A-B-C-D-E
Processing
Time
Due
Date
A
6
8
B
2
6
C
8
18
D
3
15
E
9
23
Job
Sequence
Make Span
(Processing)
Time
Flow
Time
Job Due
Date
Job
Tardiness
A
6
6
8
0
B
2
8
6
2
C
8
16
18
0
D
3
19
15
4
E
9
28
23
5
28
77
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Processing
Time
Sequencing Example
SPT: Sequence B-D-A-C-E
Due
Date
A
6
8
B
2
6
C
8
18
D
3
15
E
9
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Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
B
2
2
6
D
3
5
15
A
6
11
8
3
C
8
19
18
1
E
9
28
23
5
28
65
Job
Tardiness
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Processing
Time
Sequencing Example
EDD: Sequence B-A-D-C-E
Due
Date
A
6
8
B
2
6
C
8
18
D
3
15
E
9
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Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Tardiness
B
2
2
6
0
A
6
8
8
0
D
3
11
15
0
C
8
19
18
1
E
9
28
23
5
28
68
6
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Processing
Time
Sequencing Example
LPT: Sequence E-C-A-D-B
Due
Date
A
6
8
B
2
6
C
8
18
D
3
15
E
9
23
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
E
9
9
23
C
8
17
18
A
6
23
8
15
D
3
26
15
11
B
2
28
6
22
28
103
Job
Tardiness
48
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Performance Comparison
Average
Completion or
Flow Time (Days)
Average
Tardiness
(Days)
No. of Tardy
jobs
FCFS
15.4
2.2
03
SPT
13.0
1.8
03
EDD
13.6
1.2
02
LPT
20.6
9.6
03
Rule
1) Average Flow Time = Total Flow Time/ Number of Jobs
2) Average Tardiness = Total Tardiness/Number of Jobs
3) Number of tardy jobs.
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Comparison of Sequencing Rules
No one sequencing rule excels on all criteria.
1. SPT does well in minimizing flow time and the
number of jobs in the system
But SPT moves long jobs to the end, which may result in
dissatisfied customers
2. FCFS does not do especially well (or poorly) on any
criteria, but is perceived as fair by customers
3. EDD minimizes maximum lateness
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Critical Ratio (CR)
• An index number found by dividing the time
remaining until the due date by the work time
remaining on the job
• Jobs with low critical ratios are scheduled
ahead of jobs with higher critical ratios
• Performs well on average job lateness criteria
CR =
Time remaining
=
Workdays remaining
Due date – Today’s date
Work (lead) time remaining
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Critical Ratio Example
Currently Day 25
JOB
A
B
C
JOB
A
B
C
DUE DATE
30
28
27
CRITICAL RATIO
(30 - 25)/4 = 1.25
(28 - 25)/5 = .60
(27 - 25)/2 = 1.00
WORKDAYS REMAINING
4
5
2
PRIORITY ORDER
3
1
2
With CR < 1, Job B is late. Job C is just on schedule and
Job A has some slack time.
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Critical Ratio Technique
Helps determine the status of specific jobs
Establishes relative priorities among jobs on
a common basis
Adjusts priorities automatically for changes in
both demand and job progress
Provides a balance between SPT (which
focuses only on processing times) and EDD
(which focuses only on due dates)
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Hodgson/Moore rule
A machine shop processes custom orders from a variety of clients. One of the
machines, a grinder, has six jobs remaining to be processed. The processing times
and promised due dates (both in hours) for the six jobs are given here.
We see that the first tardy job is job 1, and there are a total of four tardy jobs. We
now consider jobs 2, 3, and 1 and reject the job with the longest processing time.
This is clearly job 1.
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Hodgson/Moore rule
The first tardy job in the current sequence is now job 4. We consider the sequence
2, 3, 5, 4, and reject the job with the longest processing time, which is job 5. The
current sequence is now
Clearly, there are no tardy jobs at this stage. The optimal sequence is 2, 3, 4, 6, 5,
1 or 2, 3, 4, 6, 1, 5. In either case, the number of tardy jobs is exactly 2.
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Scheduling in Services
• The objective is to meet staffing requirements
with the minimum number of workers
• Schedules need to be smooth and keep
personnel happy
• Many techniques exist, from simple algorithms
to complex linear programming solutions
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Cyclical Scheduling Example
1) Determine the staffing requirements
2) Identify two consecutive days with the
lowest total requirements and assign these
as days off
3) Make a new set of requirements by
subtracting the days worked by the first
employee
4) Apply step 2 to the new row
5) Repeat steps 3 and 4 until all requirements
have been met
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Cyclical Scheduling Example
DAY
M
T
W
T
F
S
S
Staff required
5
5
6
5
4
2
3
M
T
W
T
F
S
S
Employee 1
5
5
6
5
4
2
3
Employee 2
4
4
5
4
3
2
3
Employee 3
3
3
4
3
2
2
3
Employee 4
2
2
3
2
2
2
2
Employee 5
1
1
2
2
2
1
1
Employee 6
1
1
1
1
1
0
0
Capacity (Employees)
5
5
6
5
4
2
3
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Thanks
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0
