Chapter 16 – Scheduling
16–71
Note 5: There is a tie for the two lowest consecutive requirements (1) between 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).
Wed-Thu adjacent requirements = (Tue) 2 + (Fri) 1 = 3.
Mon. In case of a tie, pick the pair with the lowest adjacent requirements (day to the left and day
to the right).
Fri-Sat adjacent requirements = (Thu) 1 + (Mon) 0 = 1.
Sat-Mon adjacent requirements = (Fri) 0 + (Tue) 1 = 1.
Both are tied for lowest adjacent requirements (1). Break the tie arbitrarily by circling Fri-Sat.
the nonzero values that are not circled.
Worker Schedule
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
On
Worker 5*
On
On
On
Worker 6**
On
On
*Worker 5 will work three days only.
**Worker 6 will work two days only.
Chapter 16 – Scheduling
Education.
26. 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
4
4
5
6
7
8
Day
Mon
Tue
Wed
Thu
Fri
Sat
Staff
needed
4
4
5
6
7
8
Worker 1
4
4
5
6
7
8
Note 1
Worker 2
4
4
4
5
6
7
Note 2
Worker 3
3
4
4
4
5
6
Note 3
Worker 4
3
4
3
3
4
5
Note 4
Worker 5
2
3
3
3
3
4
Note 5
Worker 6
2
3
2
2
2
3
Note 6
Worker 7
1
2
2
2
1
2
Note 7
Worker 8
0
1
1
1
1
2
Note 8
Worker 9
0
1
0
0
0
1
Note 9
No.
working:
4
4
5
6
7
8
Note 10
Note 1: Mon-Tue has the two lowest consecutive requirements (8). 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.
Tue-Wed has the lowest adjacent requirements (9); therefore, circle those days. Subtract one from
each day’s requirement, except for the circled days. Enter those values in the row for Worker 3.
Note 3: Mon-Tue has the two lowest consecutive requirements (7). Circle those days. Subtract
one from each day’s requirement, except for the circled days. Enter those values in the row for
Worker 4.
Worker 6.
Note 6: There is a tie for the two lowest consecutive requirements (4) 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) 2 = 5.
Chapter 16 – Scheduling
16–73
Note 7: There is a tie for the two lowest consecutive requirements (3) between Mon-Tue, Thu-Fri,
Fri-Sat, and Sat-Mon. In case of a tie, pick the pair with the lowest adjacent requirements (day to
Fri-Sat and Sat-Mon are tied for lowest adjacent requirements (3). Break the tie arbitrarily by
circling Fri-Sat. Subtract one from each day’s requirement, except for the circled days. Enter
those values in the row for Worker 8.
Note 8: Mon-Tue 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
Thu-Fri adjacent requirements = (Wed) 0 + (Sat) 1 = 1.
Wed-Thu and Thu-Fri are tied for lowest adjacent requirements (1). Break the tie arbitrarily by
circling Wed-Thu. When we subtract one from each day’s requirements, we are left with all zero
values. We have met all staffing requirements.
Worker Schedule
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
On
Worker 5
On
On
On
On
Worker 6
On
On
On
On
Worker 7
On
On
On
On
Worker 8
On
On
On
On
Worker 9*
On
On
Chapter 16 – Scheduling
16–74
Education.
Case: Hi-Ho Yo-Yo, Inc.
Given:
Pad printing runs one eight-hour shift per day.
Pad printing will run 23 days in July, beginning July 1.
Pad printing will work three Saturdays (July 9, 16, and 23) and will take a one-day holiday on July 4.
We have the following job information:
Job
Date Order
Received
Set-up
Time
Production
Time
Due Date
A
6/4
2 hours
6 days
7/11
B
6/7
4 hours
2 days
7/8
C
6/12
2 hours
8 days
7/25
D
6/14
4 hours
3 days
7/19
E
6/15
4 hours
9 days
7/29
Examine the following rules, summarize your findings, and advise Jeff Baker which rule to use: FCFS,
SPT, EDD, or CR:
Step 1:
Determine the job time for each job (a day is 8 hours):
Job
Set-up
Time
Production
Time
Job Time
A
2 hours
6 days
(2/8) + 6 = 6.25 days
B
4 hours
2 days
(4/8) + 2 = 2.5 days
C
2 hours
8 days
(2/8) + 8 = 8.25 days
D
4 hours
3 days
(4/8) + 3 = 3.5 days
E
4 hours
9 days
(4/8) + 9 = 9.5 days
Chapter 16 – Scheduling
16–75
Step 2:
Determine the time remaining until due date (working days) for each job:
List the working days in July. The calendar below shows the calendar date in parentheses followed by the
working day number. Note that there are 23 working days in July. Today is the start of Friday, July 1
(Working Day 1).
Time Remaining until Due Date = Working Day Due Date – Today’s Working Day Number (1)
For example, given that today is the start of Working Day 1 and Job A is due at the start of Working Day
7, we have Working Days 1, 2, 3, 4, 5, and 6 = 6 days to complete the job. Using the formula above, we
have the same answer: 7 – 1 = 6 days.
Mon
Tue
Wed
Thu
Fri
Sat
(7/1) 1
(7/5) 2
(7/6) 3
(7/7) 4
(7/8) 5
(7/9) 6
(7/11) 7
(7/12) 8
(7/13) 9
(7/14) 10
(7/15) 11
(7/16) 12
(7/18) 13
(7/19) 14
(7/20) 15
(7/21) 16
(7/22) 17
(7/23) 18
(7/25) 19
(7/26) 20
(7/27) 21
(7/28) 22
(7/29) 23
Job
Due Date
Working Days
Until Due Date
A
7/11
7 – 1 = 6
B
7/8
5 – 1 = 4
C
7/25
19 – 1 = 18
D
7/19
14 – 1 = 13
E
7/29
23 – 1 = 22
Step 3:
Prepare a table listing each job, the date the job was received, job time remaining, and time until due date:
Job
Date Order
Received
Job Time
Remaining
(Days)
Working Days
Until Due Date
A
6/4
6.25
6
B
6/7
2.5
4
C
6/12
8.25
18
D
6/14
3.5
13
E
6/15
9.5
22
Step 4:
Sequence the jobs using the different rules:
FCFS Sequence: A-B-C-D-E
SPT Sequence: B-D-A-C-E
EDD Sequence: B-A-D-C-E
CR Sequence: A-B-C-D-E (work shown below)
Chapter 16 – Scheduling
16–76
Education.
CR Calculations (round to 2 decimals):
Initial critical ratios at start of July 1 (Time 0):
Job
Job Time
Remaining
Due Date
(days)
Critical Ratio
Calculation
A
6.25
6
(6 – 0) / 6.25 = 0.96
B
2.5
4
(4 – 0) / 2.5 = 1.60
C
8.25
18
(18 – 0) / 8.25 = 2.18
D
3.5
13
(13 – 0) / 3.5 = 3.71
E
9.5
22
(22 – 0) / 9.5 = 2.32
After 0 + 6.25 = 6.25 days [Job A completed], the critical ratios are:
Job
Job Time
Remaining
Due Date
(days)
Critical Ratio
Calculation
A
—
—
—
B
2.5
4
(4 – 6.25) / 2.5 = –0.90
C
8.25
18
(18 – 6.25) / 8.25 = 1.42
D
3.5
13
(13 – 6.25) / 3.5 = 1.93
E
9.5
22
(22 – 6.25) / 9.5 = 1.66
Job B is scheduled next.
After 6.25 + 2.5 = 8.75 days [Job B completed], the critical ratios are:
Job
Job Time
Remaining
Due Date
(days)
Critical Ratio
Calculation
A
—
—
—
B
—
—
—
C
8.25
18
(18 – 8.75) / 8.25 = 1.12
D
3.5
13
(13 – 8.75) / 3.5 = 1.21
E
9.5
22
(22 – 8.75) / 9.5 = 1.39
After 8.75 + 8.25 = 17 days [Job C completed], the critical ratios are:
Job
Job Time
Remaining
Due Date
(days)
Critical Ratio
Calculation
A
—
—
—
B
—
—
—
C
—
—
—
D
3.5
13
(13 – 17) / 3.5 = –1.14
E
9.5
22
(22 – 17) / 9.5 = 0.53
Chapter 16 – Scheduling
Job D is scheduled next.
Only Job E remains. Job E is scheduled last.
CR Sequence: A-B-C-D-E
Step 5:
Determine the average flow time, average tardiness, and average number of jobs at the work center
for each rule:
FCFS:
Job time
Flow time
Due date
Days
Job
(days)
(days)
(days)
tardy
A
6.25
6.25
6
0.25
B
2.50
8.75
4
4.75
C
8.25
17.00
18
0.00
D
3.50
20.50
13
7.50
E
9.50
30.00
22
8.00
30.00
82.50
20.50
Average job flow time = Total job flow time / Number of jobs = 82.50 / 5 = 16.50 days.
Average job tardiness = Total job tardiness / Number of jobs = 20.50 / 5 = 4.10 days.
Average number of jobs = Total flow time / Makespan = 82.50 / 30.00 = 2.75 jobs.
SPT:
Job time
Flow time
Due date
Days
Job
(days)
(days)
(days)
tardy
B
2.50
2.50
4
0.00
D
3.50
6.00
13
0.00
A
6.25
12.25
6
6.25
C
8.25
20.50
18
2.50
E
9.50
30.00
22
8.00
30.00
71.25
16.75
Chapter 16 – Scheduling
16–78
Education.
EDD:
Job time
Flow time
Due date
Days
Job
(days)
(days)
(days)
tardy
B
2.50
2.50
4
0.00
A
6.25
8.75
6
2.75
D
3.50
12.25
13
0.00
C
8.25
20.50
18
2.50
E
9.50
30.00
22
8.00
30.00
74.00
13.25
CR:
Job time
Flow time
Due date
Days
Job
(days)
(days)
(days)
tardy
A
6.25
6.25
6
0.25
B
2.50
8.75
4
4.75
C
8.25
17.00
18
0.00
D
3.50
20.50
13
7.50
E
9.50
30.00
22
8.00
30.00
82.50
20.50
Rule
Avg. Flow
Time
Avg. Job
Tardiness
Avg. Number
of Jobs
FCFS
16.50 days
4.10 days
2.75 jobs
SPT
14.25 days
3.35 days
2.38 jobs
EDD
14.80 days
2.65 days
2.47 jobs
CR
16.50 days
4.10 days
2.75 jobs
Chapter 16 – Scheduling
Summary
1. As shown in the table above, FCFS and CR perform worst on all three performance measures.
2. SPT performs best on average flow time and average number of jobs.
3. EDD performs best on average job tardiness.
Recommendations
1. If Jeff Baker is more concerned about minimizing average flow time and average number of jobs
in the system, he should use the SPT rule.
2. If Jeff Baker is more concerned with minimizing average job tardiness, he should use the EDD
rule.