CHAPTER 15: SHORT-TERM SCHEDULING – Suggested Solutions
to Selected Questions
Summer II, 2009
Question 15.3
Original distance minimization problem:
Row subtraction is done next:
Site/Customer
A
B
C
D
Site/Customer
A
B
C
D
1
2
3
4
7
5
6
8
3
4
7
6
4
6
9
7
8
5
6
4
1
2
3
4
4
1
0
4
0
0
1
2
1
2
3
3
5
1
0
0
Column subtraction is done next:
Cover zeros with lines:
Site/Customer
A
B
C
D
Site/Customer
A
B
C
D
1
2
3
4
4
1
0
4
0
0
1
2
0
1
2
2
5
1
0
0
1
2
3
4
4
1
0
4
0
0
1
2
0
1
2
2
5
1
0
0
A minimum of four lines was used – implying existence of an optimal assignment.
Optimal assignment:
Taxi at post 1 to customer C
Taxi at post 2 to customer B
Taxi at post 3 to customer A
Taxi at post 4 to customer D
Total distance traveled = 4 + 4 + 6 + 4 = 18 miles.
Question 15.4 Because this is a maximization problem, create an opportunity loss table by subtracting
each number in the table from 11 (the highest number in the table):
Original problem:
Opportunity loss table:
Job/Machine
A
B
C
D
Job/Machine
A
B
C
D
1
2
3
4
7
10
11
9
9
9
5
11
8
7
9
5
10
6
6
8
1
2
3
4
4
1
0
2
2
2
6
0
3
4
2
6
1
5
5
3
Row subtraction is done next:
Column subtraction is done next:
Job/Machine
A
B
C
D
Job/Machine
A
B
C
D
1
2
3
4
3
0
0
2
1
1
6
0
2
3
2
6
0
4
5
3
1
2
3
4
3
0
0
2
1
1
6
0
0
1
0
4
0
4
5
3
Cover zeros with lines:
1
(a) An optimal assignment can be made:
Job/Machine
A
B
C
D
1
2
3
4
3
0
0
2
1
1
6
0
0
1
0
4
0
4
5
3
Job 1 to Machine D
Job 2 to Machine A
Job 3 to Machine C
Job 4 to Machine B
(b) Total production = 10 + 10 + 9 + 11 = 40
BUS P301:01
Question 15.9 Because this is a maximization problem, subtract each number from 95 to create an
opportunity loss table. The problem can then be solved using the minimization algorithm.
Original problem:
Opportunity loss table:
Professor/Course Stat Mgmt
Fisher
Golhar
Hug
Rustagi
90
70
85
55
65
60
40
80
Fin
Econ
Professor/Course
Stat
Mgmt
Fin
Econ
95
80
80
65
40
75
60
55
Fisher
Golhar
Hug
Rustagi
5
25
10
40
30
35
55
15
0
15
15
30
55
20
35
40
Row subtraction is done next:
Professor/Course Stat Mgmt
Fisher
Golhar
Hug
Rustagi
Column subtraction is done next:
Fin
Econ
Professor/Course
Stat
Mgmt
Fin
Econ
0
0
5
15
55
5
25
25
Fisher
Golhar
Hug
Rustagi
5
10
0
25
30
20
45
0
0
0
5
15
50
0
20
20
Mgmt
Fin
Econ
30
20
45
0
0
0
5
15
50
0
20
20
5
10
0
25
30
20
45
0
Cover zeros with lines:
Professor/Course Stat
Fisher
Golhar
Hug
Rustagi
5
10
0
25
(a) An optimal assignment can be made:
Fisher to Finance
Golhar to Economics
Hug to Statistics
Rustagi to Management
Total rating =
95
75
85
80
335
(b) Since Fisher is not teaching statistics, the answer does not change. Total rating
remains 335.
Question 15.12 This is a job-sequencing problem whose objective here is to minimize total lateness.
Comparing the scheduling efficiency of the several algorithms presented in terms of
lateness.
Original problem: Today is day 205.
2
Job
Due date
Remaining
Processing Time
A
B
C
D
212
209
208
210
6
3
3
8
BUS P301:01
(a)
First come, first served (FCFS):
Job
Processing Flow
Time
Time Start
A
B
C
D
(b)
6
3
3
8
20
Processing
Time
B
C
A
D
3
3
6
8
20
Processing
Time
D
A
C
B
8
6
3
3
20
Flow
Time Start End
3
6
12
20
41
205
208
211
217
Flow
Time Start
8
14
17
20
59
Processing
Time
Flow
Time
3
3
8
6
20
3
6
14
20
43
C
B
D
A
Days Late
0
4
8
14
Total: 26 days
Number of jobs in system= 47/20 = 2.35
Average lateness= 26/4 = 6.5
207
210
216
224
Due
Date
209
208
212
210
Days Late
0
2
Number of jobs in system= 41/20 = 2.05
4
14
Total: 20 days
Average lateness = 20/4 = 5.0
205
213
219
222
End
Due
Date
212
218
221
224
210
212
208
209
Days Late
2
6
Number of jobs in system= 59/20 = 2.95
13
15
Total: 36 days
Average lateness = 36/4 = 9.0
Start
205
208
211
219
Due
End Date
207
210
218
224
208
209
210
212
Days Late
0
1
Number of jobs in system= 43/20 = 2.15
8
12
Total: 21 days
Average= 21/4 = 5.25
Critical ratio:
Job
Due
Date
Remaining
Processing
Time
Critical Ratio
A
B
C
D
212
209
208
210
6
3
3
8
(212 – 205)/6 = 1.17
(209 – 205)/3 = 1.33
(208 – 205)/3 = 1.00
(210 – 205)/8 = 0.63
Critical ratio
3
212
209
208
210
Earliest due date (EDD):
Job
(e)
210
213
216
224
Longest processing time (LPT):
Job
(d)
205
211
214
217
Due
Date
Shortest processing time (SPT):
Job
(c)
6
9
12
20
47
End
Need date today s date
Days required to complete job
Job Sequence
Critical Ratio
D
C
A
B
0.63
1.00
1.17
1.33
BUS P301:01
Critical ratio:
Job
D
C
A
B
Processing
Time
Flow
Time
8
3
6
3
20
8
11
17
20
56
Due
Start End Date
205
213
216
222
212
215
221
224
210
208
212
209
Days Late
2
7
9
15
Total: 33 days
Number of jobs in system= 56/20 = 2.80
Average lateness = 33/4 = 8.25
A minimum total lateness of 20 days seems to be about the least we may achieve.
Scheduling
Rule
Average
Lateness
Average
Flow Time
Average number of Jobs
In System
FCFS
SPT
LPT
EDD
Critical ratio
6.5
5.0
9.0
5.25
8.25
11.75
10.25
14.75
10.75
14.00
2.35
2.05
2.95
2.15
2.80
SPT is best on all criteria.
Question 15.13 This is a job-sequencing problem comparing performance under different dispatching
rules.
(a)
Dispatching
Rule
EDD
SPT
LPT
FCFS
Flow Time
Utilization
Average Number
of Jobs
Average Late
Job Sequence
CX–BR–SY–DE–RG
BR–CX–SY–DE–RG
RG–DE–SY–CX–BR
CX–BR–DE–SY–RG
385
375
495
390
37.6%
38.6%
29.3%
37.2%
2.66
2.59
3.41
2.69
10
12
44
12
Starting day number: 241 (i.e., work can be done on day 241)
Method: SPT—Shortest processing time
Processing Time
Due Date
Order
Flow Time
Completion
Time
CX-01
25
270
2
40
BR-02
15
300
1
15
DE-06
35
320
4
105
SY-11
30
310
3
70
RG-05
40
360
5
145
Total
145
375
Average
75
Sequence: BR-02,CX-01,SY-11,DE-06,RG-05 Average # in system = 2.586 = 375/145
4
280
255
345
310
385
Late
10
0
25
0
25
60
12
BUS P301:01
Method: LPT—Longest processing time
Processing Time
Due Date
Order
Flow Time
Completion
Time
CX-01
25
270
4
130
BR-02
15
300
5
145
DE-06
35
320
2
75
SY-11
30
310
3
105
RG-05
40
360
1
40
Total
145
495
Average
99
Sequence: RG-05,DE-06,SY-11,CX-01,BR-02, Average # in system = 3.414 = 495/145
370
385
315
345
280
Late
100
85
0
35
0
220
44
Method: Earliest due date (EDD); earliest to latest date
Processing Time
Due Date
Slack
Order
CX-01
25
270
0
1
BR-02
15
300
0
2
DE-06
35
320
0
4
SY-11
30
310
0
3
RG-05
40
360
0
5
Total
145
Average
Sequence: CX-01,BR-02,SY-11,DE-06,RG-05 Average # in system = 2.655 = 385/145
Flow Time
Completion Time
Late
25
40
105
70
145
385
77
265
280
345
310
385
0
0
25
0
25
50
10
Flow Time
Completion Time
Late
25
40
75
105
145
390
78
265
280
315
345
385
0
0
0
35
25
60
12
Method: First come, first served (FCFS)
Processing Time
Due Date
Slack
Order
CX-01
25
270
0
1
BR-02
15
300
0
2
DE-06
35
320
0
3
SY-11
30
310
0
4
RG-05
40
360
0
5
Total
145
Average
Sequence: CX-01,BR-02,DE-06,SY-11,RG-05, Average # in system = 2.69 = 390/145
(b)
The best flow time is SPT
(c)
The best utilization is SPT
(d)
The best lateness is EDD
(e)
Either of these choices could be supported. LPT scores poorly on all three criteria.
Question 15.14
This is a job-sequencing problem.
Original problem: Today is day 130.
5
Production
Days
Needed
DateOrder
Due
Job
Date
Order
Received
A
B
C
D
E
110
120
122
125
130
20
30
10
16
18
180
200
175
230
210
BUS P301:01
(a)
(b)
(c)
FCFS (first come, first served):
Job Sequence
Date Order Received
A
B
C
D
E
110
120
122
125
130
EDD (earliest due date):
Job Sequence
Due Date
C
A
B
E
D
175
180
200
210
230
SPT (shortest processing time):
Job Sequence
Processing Time
C
D
E
A
B
10
16
18
20
30
(d) LPT (longest processing time):
Job Sequence
Processing Time
B
A
E
D
C
30
20
18
16
10
Scheduling
Rule
Average
Tardiness
Average
Flow Time
Average
Number
of Jobs in
System
FCFS
EDD
SPT
LPT
5.4
0.0
7.2
9.6
60.0
54.4
47.6
65.2
3.2
2.9
2.5
3.5
EDD is best for average Lateness and SPT for the other two measures.
6
BUS P301:01
Question 15.17 This problem calls for the sequencing of seven jobs on two machines. Use Johnson’s
rule.
(a, c)
The jobs should be processed in the sequence: V–Y–U–Z–X–W–T, for a total time of 57.
Job Shop Scheduling
T
U
V
W
X
Y
Z
Time
Flow Time
Printer
Binder
Order
Printer
Binder
15
7
4
7
10
4
7
3
9
10
6
9
5
8
seventh
third
first
sixth
fifth
second
fourth
54
15
4
39
32
8
22
57
28
14
51
45
19
36
57
(b, c)
(d)
7
Note: Y could also be placed first, with no change in total times.
Binding is idle from 0 to 4 and from 51 to 54 for a total of 7 hours.
BUS P301:01