Chapter 16 – Scheduling
16–61
Education.
Determine the average flow time and the average job tardiness for each method:
FCFS:
Job
Job Time
(hours)
Flow Time
(hours)
Due Date
(hours)
Tardy
(hours)
a
3.5
3.5
7
0.0
b
2.0
5.5
6
0.0
c
4.5
10.0
18
0.0
d
5.0
15.0
22
0.0
e
2.5
17.5
4
13.5
f
6.0
23.5
20
3.5
23.5
75.0
17.0
SPT:
Job
Job Time
(hours)
Flow Time
(hours)
Due Date
(hours)
Tardy
(hours)
b
2.0
2.0
6
0.0
e
2.5
4.5
4
0.5
a
3.5
8.0
7
1.0
c
4.5
12.5
18
0.0
d
5.0
17.5
22
0.0
f
6.0
23.5
20
3.5
23.5
68.0
5.0
EDD:
Job
Job Time
(hours)
Flow Time
(hours)
Due Date
(hours)
Tardy
(hours)
e
2.5
2.5
4
0.0
b
2.0
4.5
6
0.0
a
3.5
8.0
7
1.0
c
4.5
12.5
18
0.0
f
6.0
18.5
20
0.0
d
5.0
23.5
22
1.5
23.5
69.5
2.5
Chapter 16 – Scheduling
16–62
Education.
CR:
Job
Job Time
(hours)
Flow Time
(hours)
Due Date
(hours)
Tardy
(hours)
e
2.5
2.5
4
0.0
a
3.5
6.0
7
0.0
b
2.0
8.0
6
2.0
f
6.0
14.0
20
0.0
c
4.5
18.5
18
0.5
d
5.0
23.5
22
1.5
23.5
72.5
4.0
Average job flow time = Total job flow time / Number of jobs
Average job tardiness = Total job tardiness / Number of jobs
Avg. Job Tardiness
FCFS
75.0/6 =
12.50 hours
17.0/6 =
2.83 hours
SPT
68.0/6 =
11.33 hours
5.0/6 =
0.83 hours
EDD
69.5/6 =
11.58 hours
2.5/6 =
0.42 hours
CR
72.5/6 =
12.08 hours
4.0/6 =
0.67 hours
Chapter 16 – Scheduling
16–64
Education.
d. Sequence the jobs using SPT:
SPT Sequence: E-C-A-B-D
Order
Job Time
(min.)
Flow Time
(min.)
Due Date
(min.)
Tardy
(min.)
E
4
4
220
0
C
30
34
180
0
A
64
98
160
0
B
72
170
200
0
D
80
250
190
60
60
Average job tardiness = Total job tardiness / Number of jobs = 60/5 = 12.00 minutes.
Conclusion: The SPT rule does produce better results in terms of average job tardiness.
The average job tardiness is 3.20 minutes lower for SPT (15.20 – 12.00 = 3.20).
19. Given:
The following table contains order-dependent setup times for three jobs:
Following Job’s Setup Time
(hrs.)
Setup Time
(hrs.)
A
B
C
Preceding Job
A
2
—
3
5
B
3
8
—
2
C
2
4
3
—
List each sequence and its total setup time. Select the sequence with the lowest total setup time:
Sequence
Setup times
Total
A–B–C
2 + 3 + 2 =
7 (best)
A–C–B
2 + 5 + 3 =
10
B–A–C
3 + 8 + 5 =
16
B–C–A
3 + 2 + 4 =
9
C–A–B
2 + 4 + 3 =
9
C–B–A
2 + 3 + 8 =
13
Conclusion: The sequence A-B-C will minimize total setup time at 7 hours.
Chapter 16 – Scheduling
16–65
Education.
20. Given:
The following table contains order-dependent setup times for three jobs:
Following Job’s Setup Time
(hrs.)
Setup Time
(hrs.)
A
B
C
Preceding Job
A
2.4
—
1.8
2.2
B
3.2
0.8
—
1.4
C
2.0
2.6
1.3
—
List each sequence and its total setup time. Select the sequence with the lowest total setup time:
Sequence
Setup times
Total
A–B–C
2.4 + 1.8 + 1.4 =
5.6
A–C–B
2.4 + 2.2 + 1.3 =
5.9
B–A–C
3.2 + 0.8 + 2.2 =
6.2
B–C–A
3.2 + 1.4 + 2.6 =
7.2
C–A–B
2.0 + 2.6 + 1.8 =
6.4
C–B–A
2.0 + 1.3 + 0.8 =
4.1 (best)
Conclusion: The sequence C-B-A will minimize total setup time at 4.1 hours.
16–66
Education.
21. Given:
The following table contains order-dependent setup times for four jobs. For safety reasons, Job C
cannot follow Job A, nor can Job A follow Job C:
Following Job’s Setup Time (hrs.)
Setup Time
(hrs.)
A
B
C
D
Preceding Job
A
2
—
5
X
4
B
1
7
—
3
2
C
3
X
2
—
2
D
2
4
3
6
—
List each sequence and its total setup time. Select the sequence with the lowest total setup time:
Sequence
Setup times
Total
A–B–C–D
2 + 5 + 3 + 2 =
12
A–B–D–C
2 + 5 + 2 + 6 =
15
A–D–B–C
2 + 4 + 3 + 3 =
12
A–D–C–B
2 + 4 + 6 + 2 =
14
B–A–D–C
1 + 7 + 4 + 6 =
18
B–C–D–A
1 + 3 + 2 + 4 =
10 (best)
C–B–A–D
3 + 2 + 7 + 4 =
16
C–B–D–A
3 + 2 + 2 + 4 =
11
C–D–A–B
3 + 2 + 4 + 5 =
14
C–D–B–A
3 + 2 + 3 + 7 =
15
D–A–B–C
2 + 4 + 5 + 3 =
14
D–C–B–A
2 + 6 + 2 + 7 =
17
16–67
Education.
22. Given:
We have the information on planned and actual inputs and outputs for a service center. The
beginning backlog is 12 hours of work. The figures shown are standard hours of work:
Period
1
2
3
4
5
Input
Planned
24
24
24
24
20
Actual
25
27
20
22
24
Output
Planned
24
24
24
24
23
Actual
24
22
23
24
24
Determine the backlog each period:
Backlog = Backlog from Previous Period + (Actual Input – Actual Output)
Period
1
2
3
4
5
Input
Planned
24
24
24
24
20
Actual
25
27
20
22
24
Output
Planned
24
24
24
24
23
Actual
24
22
23
24
24
Backlog
12
13
18
15
13
13
Chapter 16 – Scheduling
16–68
Education.
23. Given:
We have the information on planned and actual inputs and outputs for a work center. The
beginning backlog is 7 hours of work. The figures shown are standard hours of work:
Period
1
2
3
4
5
6
Input
Planned
200
200
180
190
190
200
Actual
210
200
179
195
193
194
Output
Planned
200
200
180
190
190
200
Actual
205
194
177
195
193
200
Determine the cumulative deviation and the backlog for each period:
Deviation = Actual – Planned
Backlog = Backlog from Previous Period + (Actual Input – Actual Output)
Period
1
2
3
4
5
6
Input
Planned
200
200
180
190
190
200
Actual
210
200
179
195
193
194
Deviation
+10
0
-1
+5
+3
-6
Cumulative
Deviation
+10
+10
+9
+14
+17
+11
Output
Planned
200
200
180
190
190
200
Actual
205
194
177
195
193
200
Deviation
+5
-6
-3
+5
+3
0
Cumulative
Deviation
+5
-1
-4
+1
+4
+4
Backlog
7
12
18
20
20
20
14
Backlog calculations:
Backlog (Period 1) = 7 + (210 – 205) = 12
Backlog (Period 2) = 12 + (200 – 194) = 18
Chapter 16 – Scheduling
16–69
Education.
24. Given:
We are given the staffing requirements shown below. Workers must have two consecutive days
off (not including Sunday). Determine the minimum number of workers needed and a schedule
for staffing.
Day
Mon
Tue
Wed
Thu
Fri
Sat
Staff
needed
2
3
1
2
4
3
Day
Mon
Tue
Wed
Thu
Fri
Sat
Staff
needed
2
3
1
2
4
3
Worker 1
2
3
1
2
4
3
Note 1
Worker 2
1
2
1
2
3
2
Note 2
Worker 3
1
2
0
1
2
1
Note 3
Worker 4
0
1
0
1
1
0
Note 4
No.
working:
2
3
1
2
4
3
Note 5
Note 1: Wed-Thu has the two lowest consecutive requirements (3). Circle those days. Subtract
one from each day’s requirement, except for the circled days. Do not subtract from a value of 0.
Enter those values in the row for Worker 2.
Note 2: There is a tie for the two lowest consecutive requirements (3) between Mon-Tue, Tue-
Wed, Wed–Thu, and Sat-Mon. In case of a tie, pick the pair with the lowest adjacent requirements
(day to the left and day to the right).
Mon-Tue adjacent requirements = (Sat) 2 + (Wed) 1 = 3.
Tue-Wed adjacent requirements = (Mon) 1 + (Thu) 2 = 3.
Note 3: Wed-Thu has the two lowest consecutive requirements (1). Circle those days. Subtract
one from each day’s requirement, except for the circled days. Enter those values in the row for
Worker 4.
Note 4: Sat-Mon has the two lowest consecutive requirements (0). Circle those days. When we
subtract one from each day’s requirements, we are left with all zero values. We have met all
Mon
Tue
Wed
Thu
Fri
Sat
Worker 1
On
On
On
On
Worker 2
On
On
On
On
Worker 3
On
On
On
On
Worker 4*
On
On
On
Chapter 16 – Scheduling
25. Given:
We are given the staffing requirements shown below. Workers must have two consecutive days
off (not including Sunday). Determine the minimum number of workers needed and a schedule
for staffing.
Day
Mon
Tue
Wed
Thu
Fri
Sat
Staff
needed
3
4
2
3
4
5
Day
Mon
Tue
Wed
Thu
Fri
Sat
Staff
needed
3
4
2
3
4
5
Worker 1
3
4
2
3
4
5
Note 1
Worker 2
2
3
2
3
3
4
Note 2
Worker 3
1
3
2
2
2
3
Note 3
Worker 4
1
3
1
1
1
2
Note 4
Worker 5
0
2
0
1
1
1
Note 5
Worker 6
0
1
0
1
0
0
Note 6
No.
working:
3
4
2
3
4
5
Note 7
Note 1: Wed-Thu has the two lowest consecutive requirements (5). Circle those days. Subtract
one from each day’s requirement, except for the circled days. Do not subtract from a value of 0.
Enter those values in the row for Worker 2.
Note 2: There is a tie for the two lowest consecutive requirements (5) between Mon-Tue, Tue-
Wed, and Wed-Thu. In case of a tie, pick the pair with the lowest adjacent requirements (day to
the left and day to the right).
Note 3: There is a tie for the two lowest consecutive requirements (4) between Mon-Tue, Wed-
Thu, Thu-Fri, and Sat-Mon. In case of a tie, pick the pair with the lowest adjacent requirements
(day to the left and day to the right).
Mon-Tue adjacent requirements = (Sat) 3 + (Wed) 2 = 5.
Wed-Thu adjacent requirements = (Tue) 3 + (Fri) 2 = 5.
Note 4: There is a tie for the two lowest consecutive requirements (2) between Wed-Thu and
Thu-Fri. In case of a tie, pick the pair with the lowest adjacent requirements (day to the left and
day to the right).
Wed-Thu adjacent requirements = (Tue) 3 + (Fri) 1 = 4.
Thu-Fri adjacent requirements = (Wed) 1 + (Sat) 2 = 3.