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15
C H A P T E R
Short-Term Scheduling
DISCUSSION QUESTIONS
1. The objective of scheduling is to optimize the use of
LO 15.1: Explain the relationship between short-term scheduling,
capacity planning, aggregate planning, and a master schedule
2. Four criteria for scheduling are: minimizing completion time,
LO 15.1: Explain the relationship between short-term scheduling,
capacity planning, aggregate planning, and a master schedule
AACSB: Reflective thinking
3. Loading is the assignment of jobs to work processing centers.
method are loading techniques.
4. Five priority sequencing rules are:
◼ First come, first served (FCFS) or first in, first out (FIFO):
Jobs are sequenced in the order in which they arrive at the
order of the processing time required at the workstation,
with the job requiring the least processing time at the
left to be done to time left to do the work).
5. SPT minimizes the average flow time, average lateness, and
average number of jobs in the system. It maximizes the number of
jobs completed at any point. The disadvantage is that long jobs are
6. A due date may range from a (meaningless) promise to a
contractual obligation. It is a target with or without penalties.
7. Flow time is the length of time a job is in the system; lateness
is completion time minus due date.
LO 15.4: Name and describe each of the priority sequencing rules
Interesting to see student assumption about sequence, damage, etc.
LO 15.4: Name and describe each of the priority sequencing rules
AACSB: Application of knowledge
9. Johnson’s rule is used to sequence several jobs through two
work centers.
10. Four effectiveness measures for dispatching rules: average
LO 15.4: Name and describe each of the priority sequencing rules
11. The assignment method involves adding and subtracting
appropriate numbers in the problem’s table in order to find the
12. Advantages of finite capacity scheduling are:
• Wide selection of scheduling options and priority rules
13. Input/output control keeps track of planned versus actual
ETHICAL DILEMMA
CHAPTER 15 SH O R T - T ER M SC H E D U L I N G 253
Rotating shifts is very difficult, and the shorter the rotation
cycle the worse—a 1-week cycle is a killer. Given the box the
manager has gotten into, the manager may want to let the
empowered employees move to a 1-week rotation and simulta-
MYOMLAB)
ACTIVE MODEL 15.1: Job Shop Sequencing
1. Which schedule (rule) minimizes the average completion time,
2. Use the scrollbar to change the processing time for job C and
use the scrollbar to modify the due date for job C. Does the same
3. Which schedule (rule) minimizes the average lateness for this
example?
EDD
4. Use the scrollbar to change the due date for job C. Does the
15.2
Site/Customer
A
B
C
D
1
7
3
4
8
2
5
4
6
5
3
6
7
9
6
Site/Customer
A
B
C
D
1
4
0
1
5
Site/Customer
A
B
C
D
1
4
0
0
5
2
1
0
1
1
3
0
1
2
0
4
4
2
2
0
15.4 (a)
Original data
Job/Machine
A
B
C
D
1
7
9
8
10
2
10
9
7
6
3
11
5
9
6
4
9
11
5
8
First, we create a minimizing table by subtracting every number from 11
Job/Machine
A
B
C
D
1
4
2
3
1
3
0
6
2
5
4
2
0
6
3
then we do row subtraction:
2
0
1
3
4
3
0
6
2
5
4
2
0
6
3
1
3
1
0
0
2
0
1
1
4
Because it takes four lines to cover all zeros, an optimal
assignment can be made at zero squares.
(b) Total production: 40 [= 10 + 10 + 9 + 11]
15.5
15.6 Convert the minutes into $:
Marketing
Finance
Operations
Human
Resources
Chris
$80
$120
$125
$140
Steve
$20
$115
$145
$160
Juana
$40
$100
$85
$45
Rebecca
$65
$35
$25
$75
Minimum-cost solution =
Chris Finance $120
Steve Marketing $20
Juana Human Resources $45
15.7 Original problem:
Squad 3 to case B
Squad 4 to case A
Squad 5 to case E
Total person-days projected using this assignment = 3 + 6 +
3 + 8 + 8 = 28 days.
3
C
4
B
Assignment
Rating
C53 at plant 1
10 cents
C81 at plant 3
4 cents
Squad\Case
A
B
C
D
E
1
14
7
3
7
27
5
13
25
24
26
8
(a) Row subtraction is done next:
3
7
0
1
2
18
4
1
5
0
5
14
5
5
17
16
18
0
Column subtraction is done next:
2
13
1
6
0
24
3
6
0
1
2
18
Copyright ©2017 Pearson Education, Inc.
can be made at zero squares.
15.10
Period
1
2
3
4
Total
Planned input
80
80
100
100
360
Actual input
85
85
85
85
340
Deviation
+5
+5
–15
–15
–20
Planned output
90
90
90
90
360
Actual output
85
85
80
80
330
Deviation
–5
–5
–10
–10
–30
Subtracting the smallest number from each row:
zeros:
The solution is now optimal, and the assignments can be
Assignment
Cost
Divorce case to Attorney 3
1,200
Felony case to Attorney 1
500
Discrimination case to Attorney 2
1,000
$2,700
Assignment
Rating
Fisher—finance
95
Golhar—economics
75
Copyright ©2017 Pearson Education, Inc.
Subtracting the smallest number from each row:
15.13*
Subtracting the smallest number from every column:
258 CHAPTER 15 SH O R T - T ER M SC H E D U L I N G
Subtracting the smallest number from every row:
Draw the minimum number of lines needed to cover the
CHAPTER 15 SH O R T - T ER M SC H E D U L I N G 259
Draw the minimum number of lines needed to cover the
zeros:
The following optimal assignments can now be made:
Assignments
Cost
Component G99 to Plant 1
0.55
Component E2 to Plant 2
0.07
Component C81 to Plant 3
0.04
Component D5 to Plant 4
0.30
Component D44 to Plant 5
0.10
SPT
Job
Due Date
Duration (days)
Flow
Start
End
Lateness
E
314
3
3
275
277
0
D
314
5
8
278
282
0
A
313
8
16
283
290
0
B
312
16
32
291
306
0
C
325
40
72
307
346
21
Totals:
72
131
21
LPT
Job
Due Date
Duration (days)
Flow
Start
End
Lateness
C
325
40
40
275
314
0
B
312
16
56
315
330
18
A
313
8
64
331
338
25
D
314
5
69
339
343
29
E
314
3
72
344
346
32
Totals:
72
301
104
