CHAPTER 15: SHORT-TERM SCHEDULING – Suggested 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 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: Professor/Course Stat
Fisher Golhar Hug Rustagi
Opportunity loss table: Mgmt
Fin
Econ
Professor/Course
Stat
Mgmt
Fin
Econ
65 60 40 80
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
90 70 85 55
Row subtraction is done next: Professor/Course Stat
Fisher Golhar Hug Rustagi
Column subtraction is done next:
Mgmt
Fin
Econ
Professor/Course
Stat
Mgmt
Fin
Econ
30 20 45 0
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
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
B C A D
3 3 6 8 20
0 4 8 14 Total: 26 days
Number of jobs in system= 47/20 = 2.35
Average lateness= 26/4 = 6.5
3 6 12 20 41
205 208 211 217
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
Flow Time
8 6 3 3 20
8 14 17 20 59
Start
End
Due Date
205 213 219 222
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
Earliest due date (EDD): Job
Processing Time
Flow Time
3 3 8 6 20
3 6 14 20 43
C B D A
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
A B C D
212 209 208 210
6 3 3 8
Critical ratio
3
212 209 208 210
Flow Time Start End
Processing Time
D A C B
(e)
210 213 216 224
Days Late
Longest processing time (LPT): Job
(d)
205 211 214 217
Due Date
Shortest processing time (SPT): Processing Job Time
(c)
6 9 12 20 47
End
Need date
Critical Ratio
(212 – 205)/6 = 1.17 (209 – 205)/3 = 1.33 (208 – 205)/3 = 1.00 (210 – 205)/8 = 0.63
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
25 40 105 70 145 385 77
265 280 345 310 385
Flow Time
Completion Time
25 40 75 105 145 390 78
265 280 315 345 385
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
175
A B E
180 200 210
D
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 rule. (a, c)
Johnson’s
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
