CHAPTER 13
SCHEDULING OPERATIONS
Teaching Notes
This chapter presents classical material on the scheduling of operations. The chapter
discusses the scheduling of batch operations including such topics as Gantt charting, finite
capacity scheduling, loading, theory of constraints and dispatching rules. The chapter is
concluded with descriptions of planning and control systems used to schedule various types of
batch operations.
In teaching this chapter I usually begin by contrasting the different types of scheduling
problems in operations: line, batch and project. I also discuss the relationship among the facility
planning, aggregate planning and scheduling problems. Then I move into methods for batch
scheduling, usually teaching by means of a series of example problems. Depending on how
much time is available, I try to cover Gantt charting, finite capacity scheduling, loading, and
dispatching rules. In the process of working these problems I also illustrate various scheduling
principles such as the relationship between scheduling and aggregate planning, the presence of
multiple objectives, and ways to deal with uncertainty. I finish with some comments on the
theory of constraints as it applies to batch processes.
The articles by Breen (2002), Rudberg and Olhager (2002), Shih and Lee-Chein (2001),
Odwazny (2004) and Holmes and Hendricks (2005) may be helpful in teaching the chapter.
Answers to Questions
1.
2.
a.
Hospital - Nurses assigned to shifts in each department, use of operating rooms,
use of special equipment such as X-ray, EEG, EKG, brain scanners, and labs.
b.
University - What classes are needed during which hours, assigning faculty to
classes, scheduling library services, counseling and registration.
c.
Movie making - What props and actors and actresses are needed, when, what
order scenes are filmed in.
d.
Make-to-order factory - When and what kind of product should be produced with
what equipment, when and where the product is to be delivered.
a.
Hospital - Efficient yet appropriate availability of special equipment and services.
b.
University - Ability for students to take related classes during the same quarter
while professors have satisfactory schedules. Efficient yet adequate availability of
services and administration.
c.
Movie making - Scheduling all scenes requiring same location, props, actors,
actresses at the same time.
13 - 1
d.
Make-to-order factory - Ability to deliver the specified product on-time.
3.
It is important to view a batch process operation as a network of interconnected queues
because each machine or work center is linked to other machines and work centers in a
network fashion. The flows between work centers are clarified when the job shop is
viewed as a network of queues.
4.
Scheduling of patients in a doctor's clinic is similar to scheduling jobs in a factory in that
patients are comparable to jobs, and waiting rooms, examining rooms and consulting
rooms are comparable to various work centers. To facilitate the "flow" of patients,
dispatching rules might be used. The clinic is different from a factory because people
instead of jobs are waiting. Some people may not appreciate waiting longer than others in
an effort to improve the clinic's efficiency; patients may insist on a first come first served
system except for emergencies.
5.
Gantt charting and FCS schedule jobs one at a time according to priorities on the
resources available while infinite capacity loading works from job due dates backwards to
determine the capacity required to meet the due dates. Infinite capacity loading assumes
an average waiting time and infinite capacity while Gantt charting and FCS does not.
FCS differs from Gantt charting in that each work center may have multiple machines or
resources.
6.
Backward loading to infinite capacity determines the capacity required in each work
center for each time period.
7.
Lead time, which is a function of capacity and priority decisions, can be managed by
controlling input, output, capacity and priorities.
Lead time cannot be constant because of job priorities, variability in demand, processing
and supply that lead to changing input, load factors, processing time, and starting dates.
8.
The m x n scheduling algorithms utilize a highly restrictive set of assumptions; constant
processing time, no passing of jobs, no lot splitting, etc. Real sequencing problems have
a great deal of variability in processing times as well as multiple objectives. Optimal
rules should be more widely used provided they can better model real sequencing
problems.
9.
The purpose of shop floor control systems is to implement the schedule and then correct
it if necessary to meet the plan.
Scheduling needs monitoring and feedback to be effective, i.e. to meet changing needs.
Therefore, controls are required to provide the necessary information. Effective
scheduling cannot be done without shop floor control systems.
10.
The main goal of Theory of Constraints is to make money. This is done by identifying
the bottlenecks in the process and focusing on eliminating these constraints one by one by
subordinating all other parts/elements in the system to eliminating these bottlenecks.
13 - 2
11.
A bottleneck is any resource whose capacity is less than the demand placed on it, and less
than the capacity of all other resources. A bottleneck will therefore constrain the output
of the entire plant.
12.
All non-bottleneck resources should be scheduled to make sure that the bottleneck is not
starved for materials and can keep busy processing orders needed for sale. A queue
should be formed in front of the bottleneck to insure that it stays busy. Non-bottleneck
recourses do not need to operate at full capacity, provided they process enough to keep
the bottleneck busy. Thus non-bottleneck work centers may have idle time in their
schedules. Non-bottleneck recourses should not produce inventory just to increase
resource utilization. They should be idle when their capacity is not needed to supply the
bottleneck.
13.
Measures that can be taken to provide more capacity at the bottleneck work center
include: reduction of setup time, use of bottleneck resources 24 hours a day without
breaks, and add resources to the bottleneck via additional labor or machines.
13 - 3
Answers to Problems
1.
a.
Registration (R) and Fees (F):
Sequence: E, D, C, B, A
0
4
7
12
19
31
R | E____|___D__|___C____|_____B______|__________A___________|
F|
 _____E____|_______D_______|______C______|__B__|
____A__|
0
10
18
27
29
36
Total time required = 36 minutes
b.
Yes, a better sequence can be constructed. The sequence D, E, C, A, B requires a
total time of 33 minutes.
0
3
7
12
24
31
R | D__|___E___|___C____|________A_________|__________B______|
F|
____D_________|____E____|______C_______|______A______|__B__|
0
11
17
26
31
33
2.
c.
Students going through the registration process will probably protest the use of a
process sequence other than a first-come first-served rule especially when their
waiting times will be long.
a.
Job Sequence: 1, 2, 3, 4, 5, 6; requires 56 minutes.
0
10
16
23
31 34
40
A |_____1____|___2__|___3___|____4___|_5_|___6__|
B|
___1__|_____2______|___3____|__4_|_____5___|___6___|
0
16
28
35 39
48
56
b.
Yes, a better sequence can be constructed. The sequence 5, 2, 6, 3, 1, 4 requires a
total processing time of 49 minutes.
0 3
9
15
22
32
40
A |_5_|__2___|__6__|___3___|_____1_____|____4___|
B | ____5____|______2_____|_____6___|____3__|__1_|__4__|
0
12
24
32
39 45
49
13 - 4
3.
Backward Load:
Workcenter A
Workcenter B
8_




8_


7_¦
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7_¦
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6_¦
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6_¦
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5_¦
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1_¦ 1 ¦ 2 ¦___¦
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1_¦
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1
2
3
4
Workcenter C
Days
2
3




7_¦
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6_¦
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5_¦
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3_¦ 4 ¦
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1_¦ 2 ¦
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1
4
2
3
Days
Days
4.
0
3
7 8
13 14
18
A|
[1]___|____[2]____|[4]|____________|[3]|___________|
B | ________|_[1]_|_______[3]________ |_[4]_|___________|
C |[2]__|___[4]_____|__|_[1]__|_________________|___[3]___|
0
2
4
6 7
9
12
14 15
18
a.
Makespan = 18 hours
b.
Machine idle time:
Machine A - 9 hours
Machine B - 8 hours
Machine C - 7 hours
Total - 24 hours
c.
Job
1
2
3
4
d.
Job
1
2
3
4
Due Date (Hrs.)
24
16
32
24
Delivery Time (Hrs.)
9
7
18
14
Job Idle Time
0
0
6
_5
11 hours total
13 - 5
¦
¦___¦_4_¦_ _¦___¦
¦___¦___¦_ _¦___¦
1
8_
4
e.
Sequence: 3, 2, 4, 1 has a makespan of 14 hours.
0
A
B
C
3
7
8
9
10
12
14
¦____[1]___¦____[2]_______¦_[3] ¦ [4] ¦_ _¦____ __¦_______¦
¦_______[3]__________¦__[1]____¦________¦__[4]___¦_______¦
¦_[2]___¦_______[4]____¦_____________¦___[3]______¦__[1]___¦
0
2
6
8
9
12
14
Machine idle time:
Machine A - 5 hours
Machine B - 4 hours
Machine C - 3 hours
Total 12 hours
Job
Due
Date
Delivery
Time(hrs)
Job Idle
Time(hrs)
1
2
3
4
24
16
32
24
14
7
12
12
5
0
0
3
8 hrs.
The makespan is reduced from 18 hours to 14 hours, machine idle time from 24
hours to 12 hours, and job idle time from 11 hours to 8 hours.
5.
a.
Finite capacity schedule by means of a Gantt Chart
A1
A2
B1
B2
C1
C2
0
2
3
4 5
6
7
8
9
|___[1]______|
|_[4]_|
|______[2]________|_[3]_|
|___[1]___|
|__[4]___|
|__________[3]____________|
|__[2]___|
|__[1]___|
|_______[4]______|
|___[3]_____|
0
2
3
4
5
6
7
8 9
12
b.
Make Span = 12 hours
Work Center Idle Time:
A - 15 hours
B - 14 hours
C - 13 hours
42 hours
Job
1
2
3
4
Due Date
(hr)
24
16
32
24
Delivery
Time (hr)
9
7
12
9
Job Idle
Time (hr)
0
0
0
0
Work Center Idle Time Percentage = 58%
The FCS has a shorter makespan but a larger idle time percentage than the Gantt
chart of problem 4. FCS also leads to shorter delivery time for jobs 3 and 4, and
no idle time for all jobs. There is also no bottleneck work center.
13 - 6
6.
a.
Jobs are sequenced in order 1, 2, 3, 4, 5, 6
0
6 10 13 16 18
22 23
27
35
A1|___[1]__|___[4]______|
A2|_[2]|___[3]___|[5] |___[6]_____|
B1
|______[2]_______|______[5]_____ __|_______[6]______ |
B2
|___[1]__ |____[3]_______|__[4]__|
16 18
23
27
35
Machine center idle (min)
A1 – 17
A2 – 13
B1 – 6
B2 – 18
Job
1
2
3
4
5
6
b.
Job waiting time (min) Delivery time (min)
0
16
0
18
9
23
15
27
15
27
21
35
In comparison to the Gantt Chart from problem 2, which requires 56 min (1, 2, 3,
4, 5, 6 sequence) to complete the task, the FCS requires only 35 min. Also the waiting
time of the jobs is less for FCS in comparison to Gantt Chart, however machine
utilization is better for the Gantt Chart, when we compare machine center idle time
for both cases.
Machine center idle for problem 2 (min)
A–0
B – 10
Job
1
2
3
4
5
6
Job waiting time (min) Delivery time (min)
0
16
10
28
21
35
27
39
36
48
42
56
13 - 7
7.
a.
Backward Load
Incoming Mail
Underwriting
Policy Control
32
32
32
28
28
28
24
24
24
5
4
20
20
20
3
16
16
16
12
12
12
2
4
8
8
8
1
4
4
4
3
2
1
4
1
3
5
1
1
b.
2
2
3
2
4
5
1
2
3
4
5
1
2
3
4
5
Priority used: 1 - 2 - 3 - 4 - 5
0
3
5 6
8 9
12
I | 1 | 2 |3| 4 |5|
U
|
1
| 3 |
P
|
1
9
20
4
|
17
|
2
| 3 |
27
31
4
|
37
5
|
43
Both the backward load and Gantt chart show the problem of congestion in each
department as the waiting applications pass through. The congestion moves from
Incoming Mail to Underwriting to Policy Control. The Gantt chart probably gives better
information because it gives a more realistic idea of the due date that can be expected.
13 - 8
8.
a.
Gantt Chart: Priority Order for Samples: 2, 3, 1, 4, 5
A
B
C
Samples
1
2
3
4
5
Completion Date (hr.)
12
10
8
18
14
Due Date (hr.)
10
6
8
12
14
0 1 2
4 5 6
8
10
14
|1 | |
3
|2 | 4 | 5 |
| 2 | 1 |
|
5
|
| 3 |
2
|5 | 3 |
1
|
|
4
9__

8_


8_
7_¦___¦
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6_¦ 4 ¦
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3_¦ 3 ¦
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3_¦ 3 ¦ 3 ¦___¦
2_¦
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2_¦
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2_¦
1_¦ 2 ¦
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1_¦ 2 ¦ 1 ¦
¦
1_¦ 2 ¦ 1 ¦ 4 ¦
¦ 1 ¦ 5 ¦
1
¦
2
¦
0*
0
¦ 4 ¦
1
2
2
5
6
¦
0*
8
10 11
A
B
D
| 5
12
14
C
*Day 0* represents Past Due Hours
13 - 9
¦
1
Backward Load
Days
¦
¦ 2 ¦ 1 ¦
¦
Days
Days
b.
|4 |
|

8_
8.
17 18
¦
2

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7_¦ 5 ¦
¦
6_¦ 4 ¦
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1_¦ 2 ¦
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1
|

8_¦- --¦-
2
| 3 | Days2
| | 1
8.
c.
Finite Capacity Schedule: Earliest due date first - 2, 3, 1, 4, 5
A1
A2
B1
B2
C1
C2
D1
D2
0
|1
|
|
|
|
|
|
|
2
4
6
8
| |
3
|2 |
4 | | 5 |
2 |
|4
| 1
|
|
5
|
|
2
| | 3 |
3 | 5 |
1
|
4
|
| 4 | |3|
2
5 |
|1|
Samples
Completion Date (hr.)
Due Date (hr.)
1
7
10
2
10
6
3
8
8
10
|
|
|
|
4
10
12
5
9
14
8.
d.
Considering the solutions in parts (a) through (c), it can be inferred that stations
A, C, and D are the bottlenecks. Increasing the capacity available at these stations
will increase throughput. Adding capacity for station B will not increase
throughput since currently it does not constrain production. The problem
involving blood sample 2 is that it requires past due testing hours in station B
which could be a result of past scheduling or resource problems.
9.
a.
b.
c.
DCEAB
DCBAE
BAECD
10.
a.
b.
Job 1 (1 hour processing time for Job 1 versus 2 hours for Job 4)
Job 1: 10/7 = 1.43
Job 4: 12/8 = 1.50
Choose Job 1
11.
a. For problem 2 machine B is the bottleneck, since jobs 3, 4, 5 and 6 all must wait for
machine B to become available. Another machine of type B (or more hours of capacity)
would reduce the make span and thus increase shop throughput.
b. For problem 4 machine B in clearly the bottleneck since job 3 has to wait 6 hours
before starting on machine B thereby also delaying the completion of this job. If there
were a second machine B (or more hours), job 3 could start immediately and then move
on to machine A and C where idle time exists.
c. In problem 7, department P is the bottleneck since policies 2, 3, 4 and 5 all wait for it to
become available. Additional capacity at department P would reduce the make span and
thus improve the flow of all life insurance policies.
13 - 10
12.
a. For problem 2 the theory of constraints could be used in two ways. First, we could
find a better sequence of jobs than 1, 2, 3, 4, 5 and thus reduce job interference. Second,
we could make more capacity available at the bottleneck and thus further improve the
make span.
b. For problem 4 the theory of constraints could be used in much the same way as
problem 2, find a better sequence and then add capacity to the bottleneck.
c. For problem 7, the same approach holds true.
13 - 11