Chapter 15 Solutions

Chapter 15 Solutions:

1. / Job
A / B / C / A / B / C
1 / 5 / 8 / 6 / row / 1 / 0 / 3 / 1
/ Worker / 2 / 6 / 7 / 9 / reduction / 2 / 0 / 1 / 3
/ 3 / 4 / 5 / 3 / 3 / 1 / 2 / 0
column
reduction
A / B / C
/ 1 / 0 / 2 / 1 / Optimum:
/ 2 / 0 / 0 / 3 / Worker / 1, / Job / A
3 / 1 / 1 / 0 / 2 / B
3 / C
2. / Initial
Job / Initial revised
A / B / C / A / B / C / A / B / C
1 / 5 / 8 / 6 / 1 / 4 / 1 / 3 / row / 1 / 3 / 0 / 2
/ Worker / 2 / 6 / 7 / 9 / 2 / 3 / 2 / 0 / reduction / 2 / 3 / 2 / 0
3 / 4 / 5 / 3 / 3 / 5 / 4 / 6 / 3 / 1 / 0 / 2
column
/ reduction
A / B / C
Optimum: / 1 / 2 / 0 / 2
/ 1–B, 2–C, 3–A / 2 / 2 / 2 / 0
/ 3 / 0 / 0 / 2

Solutions (continued)

3. / Route
A / B / C / D / E / A / B / C / D / E
1 / 4 / 5 / 9 / 8 / 7 / 1 / 0 / 1 / 5 / 4 / 3
2 / 6 / 4 / 8 / 3 / 5 / 2 / 3 / 1 / 5 / 0 / 2
Truck / 3 / 7 / 3 / 10 / 4 / 6 / row / 3 / 4 / 0 / 7 / 1 / 3
/ 4 / 5 / 2 / 5 / 5 / 8 / reduction / 4 / 3 / 0 / 3 / 3 / 6
5 / 6 / 5 / 3 / 4 / 9 / 5 / 3 / 2 / 0 / 1 / 6
column
/ reduction
A / B / C / D / E / A / B / C / D / E
/ 1 / 0 / 2 / 6 / 4 / 1 / 1 / 0 / 1 / 5 / 4 / 1
/ 2 / 3 / 2 / 6 / 0 / 0 / 2 / 3 / 1 / 5 / 0 / 0
/ 3 / 3 / 0 / 7 / 0 / 0 / add and / 3 / 4 / 0 / 7 / 1 / 1
/ 4 / 2 / 0 / 3 / 2 / 3 / subtract 1 / 4 / 3 / 0 / 3 / 3 / 4
/ 5 / 2 / 2 / 0 / 0 / 3 / 5 / 3 / 2 / 0 / 1 / 4
Optimum: / 1–A, / 2–E, / 3–D, / 4–B, / 5–C
or / 1–A, / 2–D, / 3–E, / 4–B, / 5–C

Solutions (continued)

4. / Initial + Dummy Machine
A / B / C / D
1 / 12 / 8 / 11 / 0
Job / 2 / 13 / 10 / 8 / 0 / row / [no change due to dummy]
/ 3 / 14 / 9 / 14 / 0 / reduction
4 / 10 / 7 / 12 / 0 / column
reduction
A / B / C / D / A / B / C / D
/ 1 / 1 / 0 / 2 / 0 / 1 / 2 / 1 / 3 / 0
/ Job / 2 / 3 / 3 / 0 / 1 / add and / 2 / 3 / 3 / 0 / 0
/ 3 / 3 / 1 / 5 / 0 / subtract 1 / 3 / 4 / 2 / 6 / 0
/ 4 / 0 / 0 / 4 / 1 / 4 / 0 / 0 / 4 / 0
Optimum: / 1–B, / 2–C, / 3–D, / 4–A
5. / a.Initial revised
Machine
A / B / C / D / E / A / B / C / D / E
/ 1 / 14 / 18 / 20 / 17 / 18 / 1 / 0 / 4 / 6 / 3 / 4
2 / 14 / 15 / 19 / 50 / 17 / 2 / 0 / 1 / 5 / 36 / 3
Job / 3 / 12 / 16 / 15 / 14 / 17 / row / 3 / 0 / 4 / 3 / 2 / 5
/ 4 / 11 / 13 / 14 / 12 / 14 / reduction / 4 / 0 / 2 / 3 / 1 / 3
5 / 10 / 16 / 15 / 14 / 13 / 5 / 0 / 6 / 5 / 4 / 3
column
/ reduction
A / B / C / D / E
1 / 0 / 3 / 3 / 2 / 1
/ Optimum:
1–A, 2–B, 3–C,
4–D, 5–E / 2 / 0 / 0 / 2 / 35 / 0
/ 3 / 0 / 3 / 0 / 1 / 2
/ 4 / 0 / 1 / 0 / 0 / 0
/ 5 / 0 / 5 / 2 / 3 / 0

Solution (continued)

5. / b.Initial revised
Machine
A / B / C / D / E / A / B / C / D / E
/ 1 / 50 / 18 / 20 / 17 / 18 / 1 / 33 / 1 / 3 / 0 / 1
/ 2 / 14 / 15 / 19 / 50 / 17 / 2 / 0 / 1 / 5 / 36 / 3
Job / 3 / 12 / 16 / 15 / 14 / 17 / row / 3 / 0 / 4 / 3 / 2 / 5
/ 4 / 11 / 13 / 14 / 12 / 14 / reduction / 4 / 0 / 2 / 3 / 1 / 3
5 / 10 / 16 / 15 / 14 / 13 / 5 / 0 / 6 / 3 / 4 / 3
column
/ reduction
A / B / C / D / E / A / B / C / D / E
/ 1 / 34 / 1 / 1 / 0 / 0 / 1 / 38 / 0 / 0 / 0 / 0
/ 2 / 0 / 0 / 2 / 35 / 1 / 2 / 0 / 0 / 2 / 36 / 2
/ 3 / 0 / 3 / 0 / 1 / 3 / add and / 3 / 0 / 3 / 0 / 2 / 4
/ 4 / 0 / 1 / 0 / 0 / 1 / subtract 1 / 4 / 0 / 1 / 0 / 1 / 2
/ 5 / 0 / 5 / 2 / 3 / 1 / 5 / 0 / 5 / 2 / 4 / 2
Optimum: 1–E, 2–B, 3–C, 4–D, 5–A

Solutions (continued)

6. / a. / FCFS: A–B–C–D
SPT: D–C–B–A
EDD: C–B–D–A
CR: A–C–D–B
FCFS: / Job time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
A / 14 / 14 / 20 / 0
B / 10 / 24 / 16 / 8
C / 7 / 31 / 15 / 16
D / 6 / 37 / 17 / 20
37 / 106 / 44
SPT: / Job time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
D / 6 / 6 / 17 / 0
C / 7 / 13 / 15 / 0
B / 10 / 23 / 16 / 7
A / 14 / 37 / 20 / 17
37 / 79 / 24
EDD: / Job time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
C / 7 / 7 / 15 / 0
B / 10 / 17 / 16 / 1
D / 6 / 23 / 17 / 6
A / 14 / 37 / 20 / 17
84 / 24

Critical Ratio

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / 14 / 20 / (20 – 0) / 14 = 1.43
B / 10 / 16 / (16 – 0) /10 = 1.60
C / 7 / 15 / (15 – 0) / 7 = 2.14
D / 6 / 17 / (17 – 0) / 6 = 2.83

Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are:

Solutions (continued)

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 10 / 16 / (16 – 14) /10 = 0.20
C / 7 / 15 / (15 – 14) / 7 = 0.14
D / 6 / 17 / (17 – 14) / 6 = 0.50

Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are:

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 10 / 16 / (16 – 21) /10 = –0.50
C / – / – / –
D / 6 / 17 / (17 – 21) / 6 = –0.67

Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.

The critical ratio sequence is A–C–D–B and the makespan is 37 days.

Critical Ratio sequence / Processing Time (Days) / Flow time / Due Date / Tardiness
A / 14 / 14 / 20 / 0
C / 7 / 21 / 15 / 6
D / 6 / 27 / 17 / 10
B / 10 / 37 / 16 / 21
 / 99 / 37

Solutions (continued)

b. / / FCFS / SPT / EDD / CR
26.50 / 19.75 / 21.00 / 24.75
11.0 / 6.00 / 6.00 / 9.25
2.86 / 2.14 / 2.27 / 2.67

c. SPT is superior.

Solutions (continued)

7. / FCFS: a–b–c–d–e
SPT: c–b–a–e–d
EDD: a–b–c–e–d
CR: a–e–b–c–d
FCFS: / Operation / Flow time / Due date / Hours
Job / time (hr.) / (hr.) / (hr.) / tardy
a / 7 / 7 / 4 / 3
b / 4 / 11 / 10 / 1
c / 2 / 13 / 12 / 1
d / 11 / 24 / 20 / 4
e / 8 / 32 / 15 / 17
32 / 87 / 26
SPT: / Operation / Flow time / Due date / Hours
Job / time (hr.) / (hr.) / (hr.) / tardy
c / 2 / 2 / 12 / 0
b / 4 / 6 / 10 / 0
a / 7 / 13 / 4 / 9
e / 8 / 21 / 15 / 6
d / 11 / 32 / 20 / 12
32 / 74 / 27
EDD: / Operation / Flow time / Due date / Hours
Job / time (hr.) / (hr.) / (hr.) / tardy
a / 7 / 7 / 4 / 3
b / 4 / 11 / 10 / 1
c / 2 / 13 / 12 / 1
e / 8 / 21 / 15 / 6
d / 11 / 32 / 20 / 12
32 / 84 / 23

Critical Ratio

Job / Processing Time (Hours) / Due Date / Critical Ratio Calculation
A / (.14 x 45) + .7 = 7 / 4 / (4 – 0) / 7 = .57
B / (.25 x 14) + .5 = 4 / 10 / (10 – 0) / 4 = 2.5
C / (.10 x 18) + .2 = 2 / 12 / (12 – 0) / 2 = 6
D / (.25 x 40) + 1 = 11 / 20 / (20 – 0) / 11 = 1.82
E / (.10 x 75) + .5 = 8 / 15 / (15 – 0) / 8 = 1.88

Solutions (continued)

Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4 hours, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 4 / 10 / (10 – 4) / 4 = 1.5
C / 2 / 12 / (12 – 4) / 2 = 4
D / 11 / 20 / (20 – 4) / 11 = 1.45
E / 8 / 15 / (15 – 4) / 8 = 1.38

Job B has the lowest critical ratio therefore it is scheduled next and it is completed after 11 hours (7 + 4). After the completion of Job B, the revised critical ratios are:

Job / Processing Time (Hours) / Due Date / Critical Ratio Calculation
A / – / – / –
B / – / – / –
C / 2 / 12 / (12 – 11) / 2 = 0.5
D / 11 / 20 / (20 – 11) / 11 = .81
E / 8 / 15 / (15 – 11) / 8 = .5

Job C and Job E are tied for the lowest critical ratio and Job C is arbitrarily selected and is scheduled next. Job C is completed in 2 hours bringing the total completion time to 11 + 2 = 13. After the completion of Job C, the revised critical ratios are:

Job / Processing Time (Hours) / Due Date / Critical Ratio Calculation
A / – / – / –
B / – / –
C / – / – / –
D / 11 / 20 / (20 – 13) / 11 = .63
E / 8 / 15 / (15 – 13) / 8 = .25

Solutions (continued)

Job E has the lowest critical ratio therefore it is scheduled next. The critical ratio final sequence is A–B–C–E–D. Total completion of all six jobs (makespan) is 32 hours.

Critical Ratio sequence / Processing Time (Days) / Flow time / Due Date / Tardiness
A / 7 / 7 / 4 / 3
B / 4 / 11 / 10 / 1
C / 2 / 13 / 12 / 1
E / 8 / 21 / 15 / 6
D / 11 / 32 / 20 / 12
 / 32 / 84 / 23
/ FCFS / SPT / EDD / CR
17.40 / 14.80 / 16.80 / 16.8
5.20 / 5.4 / 4.60 / 4.6
2.72 / 2.31 / 2.63 / 2.63

Solutions (continued)

8. / a. / (1) FCFS: A–B–C–D–E
(2) S/O: B–D–C–A–E OR D–B–C–A–E [see below]
Time / Due date / Remaining number
Job / (days) / (days) / Slack / of operations / Ratio / Rank
A / 8 / 20 / 12 / 2 / 6.0 / 4
B / 10 / 18 / 8 / 4 / 2.0 / 1,2 (tie)
C / 5 / 25 / 20 / 5 / 4.0 / 3
D / 11 / 17 / 6 / 3 / 2.0 / 1,2 (tie)
E / 9 / 35 / 26 / 4 / 6.5 / 5

b. S/O: [Assume B–D–C–A–E]

Time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
B / 10 / 10 / 18 / 0
D / 11 / 21 / 17 / 4
C / 5 / 26 / 25 / 1
A / 8 / 34 / 20 / 14
E / 9 / 43 / 35 / 8
43 / 134 / 27
Time / Flow time / Due date / Days
Job / (days) / (days) / (days) / tardy
A / 8 / 8 / 20 / 0
B / 10 / 18 / 18 / 0
C / 5 / 23 / 25 / 0
D / 11 / 34 / 17 / 17
E / 9 / 43 / 35 / 8
43 / 126 / 25
FCFS / S/O
Average flow time = / flow time: / 25.20 / 26.80
number of jobs
Average number of jobs in the system = / flow time: / 2.93 / 3.12
job times

Solutions (continued)

9. / Time (hr.)
Order / Step 1 / Step 2 / Sequence of assignment:
A / 1.20 / 1.40 / .80 / [C] / last (i.e., 7th)
B / 0.90 / 1.30 / .90 / [B] / first
C / 2.00 / 0.80 / 1.20 / [A] / 2nd
D / 1.70 / 1.50 / 1.30 / [G] / 3rd
E / 1.60 / 1.80 / 1.60 / [E] / 4th
F / 2.20 / 1.75 / 1.50 / [D] / 6th
G / 1.30 / 1.40 / 1.75 / [F] / 5th
10. / a. / Job / Machine A / Machine B
/ a / 16 / 5 / 7
/ b / 3 / 2 / 13 / Thus, the sequence is e–b–g–h–d–c–a–f.
/ c / 9 / 6 / 6
/ d / 8 / 7 / 5
/ e / 2 / 1 / 14
/ f / 12 / 4 / 8
/ g / 18 / 14 / 3
/ h / 20 / 11 / 4
b.

025234351607688

e / b / g / h / d / c / a / f
/ e / b / g / h / d / c / a / f

0216294354616776818892

c. Original idle time for B: 2 + 9 + 7 = 18 hrs., and original makespan is 92.

The last two tasks in the sequence are a and f. Splitting both of their completion times evenly, we get the following results.

Machine 1Machine 2

a182.5

a282.5

f162

f262

Solutions (continued)

After splitting, we get the following Gantt chart:

02523435160 68 76 82 88

e / b / g / h / d / c / a1 / a2 / f1 / f2
/ e / b / g / h / d / c / / a1 / / a2 / / f1 / / f2

021629435461676870.57678.58490

After splitting, idle time is: 2 + 1 + 5.5 + 3.5 + 4 = 15 hours, and the new makespan = 90.

There is a savings of 3 hr.

Time (minutes)
11. / a. / Job / Center 1 / Center 2
/ A / 20 / 2 / 27
/ B / 16 / 1 / 30 / Thus, the sequence is B–A–C–E–F–D.
/ C / 43 / 3 / 51
/ D / 60 / 12 / 6
/ E / 35 / 28 / 4
/ F / 42 / 24 / 5
b.

0163679114156216

B / A / C / E / F / D
/ B / A / C / E / F / D

01646 73 79130158182 216228

Solutions (continued)

12. / a. / Job / Station A / Station B
/ a / 27 / 2 / 45
/ b / 18 / 1 / 33 / Thus, the sequence is b–a–c–d–e.
/ c / 70 / 30 / 3
/ d / 26 / 24 / 4
/ e / 15 / 10 / 5

01845115141156

b / a / c / d / e
/ b / a / c / d / e

0185196115145169179

b. / The Idle time for Station B is = 18 + 19 = 37 minutes.
c. / Jobs B, A, C, D and E are candidates for splitting in order to reduce throughput time and idle time.

141156

091831.54580115128148.5

B / 18 / A / 27 / C 70 / D / 26 / E / 15
/ B 33 / A 45 / C 30 / D 24 / E 10

094287100130154164

Throughput time is 164 minutes, reducing this time by 15 minutes.
The idle time for B of 22 minutes has decreased by 15 minutes.

Solutions (continued)

13. / Determine job times from the schedule table, and then use Johnson’s Rule to sequence the jobs. The job times are:
Job / A / B / C / D / E / F / G
Cutting / 2 / 4 / 5 / 4 / 2 / 3 / 1
Polishing / 3 / 3 / 2 / 5 / 3 / 1 / 4

Using Johnson’s Rule, we obtain:

Cutting / Polishing
Job / Start / Finish / Start / Finish
G / 0 / 1 / 1 / 5
A / 1 / 3 / 5 / 8
E / 3 / 5 / 8 / 11
D / 5 / 9 / 11 / 16
B / 9 / 13 / 16 / 19
C / 13 / 18 / 19 / 21
F / 18 / 20 / 21 / 22

Note: The order of Jobs A and E can be reversed with no effect on times.

14. / a.,b.
SPT / Grinding / Deburring
Job / Start / Finish / Start / Finish
C / 0 / 1 / 1 / 6
B / 1 / 3 / 6 / 10
A / 3 / 6 / 10 / 16
D / 6 / 10 / 16 / 19
G / 10 / 16 / 19 / 21
F / 16 / 24 / 24 / 31
E / 24 / 33 / 33 / 37
93 / 140

The Grinding flow time is 93 hours and Deburring flow time is 140 hours. The Total time is 37 hours.

Solutions (continued)

c. Johnson’s Rule

Grinding / Deburring
Job / Start / Finish / Start / Finish
C / 0 / 1 / 1 / 6
B / 1 / 3 / 6 / 10
A / 3 / 6 / 10 / 16
F / 6 / 14 / 16 / 23
E / 14 / 23 / 23 / 27
D / 23 / 27 / 27 / 30
G / 27 / 33 / 33 / 35
107 / 147

The Grinding flow time is 107 hours and Deburring flow time is 140 hours. The Total time is 35 hours.

d. The tradeoff is between shorter flow time in the Grinding and Deburring departments and shorter total processing time. Ed would be indifferent if the benefit to be gained by shorter total processing time was equal to the cost of additional flow time in the Grinding and Deburring departments.

15. / a. / FCFS:
Job / Flow / Due / Days / Job / Flow / Due / Days
Job / time / time / date / tardy / Job / time / time / date / tardy
a / 4.5 / 4.5 / 10 / 0 / d / 1.6 / 1.6 / 27 / 0
b / 6.0 / 10.5 / 17 / 0 / e / 2.8 / 4.4 / 18 / 0
c / 5.2 / 15.7 / 12 / 3.7 / f / 3.3 / 7.7 / 19 / 0
d / 1.6 / 17.3 / 27 / 0 / a / 4.5 / 12.2 / 10 / 2.2
e / 2.8 / 20.1 / 18 / 2.1 / c / 5.2 / 17.4 / 12 / 5.4
f / 3.3 / 23.4 / 19 / 4.4 / b / 6.0 / 23.4 / 17 / 6.4
23.4 / 91.5 / 10.2 / 23.4 / 66.7 / 14.0
EDD: / CR:
Job / Flow / Due / Days / Job / Flow / Due / Days
Job / time / time / date / tardy / Job / time / time / date / tardy
a / 4.5 / 4.5 / 10 / 0 / a / 4.5 / 4.5 / 10 / 0
c / 5.2 / 9.7 / 12 / 0 / c / 5.2 / 9.7 / 12 / 0
b / 6.0 / 15.7 / 17 / 0 / b / 6.0 / 15.7 / 17 / 0
e / 2.8 / 18.5 / 18 / .5 / e / 2.8 / 18.5 / 18 / 0.5
f / 3.3 / 21.8 / 19 / 2.8 / f / 3.3 / 21.8 / 19 / 2.8
d / 1.6 / 23.4 / 27 / 0 / d / 1.6 / 23.4 / 27 / 0
23.4 / 93.6 / 3.3 / 23.4 / 93.6 / 3.3

Solutions (continued)

Critical Ratio

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / 4.5 / 10 / (10 – 0) / 4.5 = 2.22
B / 6.0 / 17 / (17 – 0) / 6.0 = 2.83
C / 5.2 / 12 / (12 – 0) / 5.2 = 2.31
D / 1.6 / 27 / (27 – 0) / 1.6 = 16.88
E / 2.8 / 18 / (18 – 0) / 2.8 = 6.43
F / 3.3 / 19 / (19 – 0) / 3.3 = 5.76

Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4.5 days. The revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 6.0 / 17 / (17 – 4.5) / 6.0 = 2.08
C / 5.2 / 12 / (12 – 4.5) / 5.2 = 1.44
D / 1.6 / 27 / (27 – 4.5) / 1.6 = 14.06
E / 2.8 / 18 / (18 – 4.5) / 2.8 = 4.82
F / 3.3 / 19 / (19 – 4.5) / 3.3 = 4.39

Job C has the lowest critical ratio, therefore it is scheduled next and it is completed after 9.7 days (4.5 + 5.2). After the completion of Job C, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 6.0 / 17 / (17 – 9.7) / 6.0 = 1.22
C / – / – / –
D / 1.6 / 27 / (27 – 9.7) / 1.6 = 10.81
E / 2.8 / 18 / (18 – 9.7) / 2.8 = 2.96
F / 3.3 / 19 / (19 – 9.7) / 3.3 = 2.82

Job B has the lowest critical ratio, therefore it is scheduled next and it is completed after 15.7 days (9.7 + 6). After the completion of Job B, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / – / – / –
C / – / – / –
D / 1.6 / 27 / (27 – 15.7) / 1.6 = 7.06
E / 2.8 / 18 / (18 – 15.7) / 2.8 = 0.82
F / 3.3 / 19 / (19 – 15.7) / 3.3 = 1.0

Job E has the lowest critical ratio, therefore it is scheduled next and it is completed after 18.5 days (15.7 + 2.8). After the completion of Job E, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / – / – / –
C / – / – / –
D / 1.6 / 27 / (27 – 18.5) / 1.6 = 5.31
E / – / – / –
F / 3.3 / 19 / (19 – 18.5) / 3.3 = 0.15

Job F has the lowest critical ratio therefore it is scheduled next and it is completed after 21.8 days (18.5 + 3.3). The final critical ratio sequence is A–C–B–E–F–D. Total completion of all six jobs (makespan) is 23.4 days.

Critical Ratio sequence / Processing Time (Days) / Flow time / Due Date / Lateness / Tardiness
A / 4.5 / 4.5 / 10 / –5.5 / 0
C / 5.2 / 9.7 / 12 / –2.3 / 0
B / 6.0 / 15.7 / 17 / –1.3 / 0
E / 2.8 / 18.5 / 18 / 0.5 / 0.5
F / 3.3 / 21.8 / 19 / 2.8 / 2.8
D / 1.6 / 23.4 / 27 / –3.6 / 0
 / 93.6 / –9.4 / 3.3

Solutions (continued)

Rule / Average / = / days late / Average flow / = / flow time / Average / = / flow time
tardiness / no. of jobs / time / no. of jobs / no. of jobs / job time
FCFS / 10.2/6 = / 1.7 days / 91.5/6 = / 15.25 days / 91.5/23.4 = / 3.91 jobs
SPT / 14.0/6 = / 2.33 days / 66.7/6 = / 11.117 days / 66.7/23.4 = / 2.85
EDD / 3.3/6 = / 0.55 days. / 93.6/6 = / 15.60 days / 93.6/23.4 = / 4.00
CR / 3.3/6 = / .55 days / 93.6/6 = / 15.60 days / 93.6/23.4 = / 4.00

b. There are several ways to show this. One is to calculate the ratio of average flow time to average number of jobs for each rule and then observe that they are equal. Here the ratios are approximately 3.90. [Slight differences in ratios may arise due to rounding.]

16.
Job / Remaining processing time / Due date / Slack / Remaining number of operations / Slack  Remaining number of operations / Rank
a / 5 / 8 / 3 / 2 / 1.50 / 4
b / 6 / 5 / –1 / 4 / –0.25 / 1
c / 9 / 10 / 1 / 4 / 0.25 / 2
d / 7 / 12 / 5 / 3 / 1.67 / 5
e / 8 / 10 / +2 / 2 / +1.00 / 3
Using the S/O rule, the sequence is B–C–E–A–D
17. / FCFS
Job time / Due date / Flow time / Tardy
Job / (hr.) / (hr.) / (hr.) / (hr.)
a / 3.5 / 7 / 3.5 / 0
b / 2.0 / 6 / 5.5 / 0
c / 4.5 / 18 / 10.0 / 0
d / 5.0 / 22 / 15.0 / 0
e / 2.5 / 4 / 17.5 / 13.5
f / 6.0 / 20 / 23.5 / 3.5
23.5 / 75.0 / 17.0

Solutions (continued)

SPT
Job / Job time / Flow time / Due date / Tardy
b / 2.0 / 2.0 / 6 / 0
e / 2.5 / 4.5 / 4 / 0.5
a / 3.5 / 8.0 / 7 / 1
c / 4.5 / 12.5 / 18 / 0
d / 5.0 / 17.5 / 22 / 0
f / 6.0 / 23.5 / 20 / 3.5
23.5 / 68.0 / 5.0
EDD
Job / Job time / Flow time / Due date / Tardy
e / 2.5 / 2.5 / 4 / 0
b / 2.0 / 4.5 / 6 / 0
a / 3.5 / 8.0 / 7 / 1
c / 4.5 / 12.5 / 18 / 0
f / 6.0 / 18.5 / 20 / 0
d / 5.0 / 23.5 / 22 / 1.5
23.5 / 69.5 / 2.5

Critical Ratio

Job / Processing Time (Days) / Due Date / Critical Ratio Calculation
A / 3.5 / 7 / (7 – 0) / 3.5 = 2.0
B / 2.0 / 6 / (6 – 0) / 2.0 = 3.0
C / 4.5 / 18 / (18 – 0) / 4.5 = 4.0
D / 5.0 / 22 / (22 – 0) / 5.0 = 4.4
E / 2.5 / 4 / (4 – 0) / 2.5 = 1.6
F / 6.0 / 20 / (20 – 0) / 6 = 3.33

Solutions (continued)

Job E has the lowest critical ratio, therefore it is scheduled first and completed after 2.5 hours. The revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / 3.5 / 7 / (7 – 2.5) / 3.5 = 1.29
B / 2.0 / 6 / (6 – 2.5) / 2.0 = 1.75
C / 4.5 / 18 / (18 – 2.5) / 4.5 = 3.44
D / 5.0 / 22 / (22 – 2.5) / 5.0 = 3.90
E / – / – / –
F / 6.0 / 20 / (20 – 2.5) / 6 = 2.92

Job A is scheduled next because Job A has the lowest critical ratio. Job A will be completed after 6 hours (2.5 + 3.5). After the completion of Job A, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / 2.0 / 6 / (6 – 6) / 2.0 = 0
C / 4.5 / 18 / (18 – 6) / 4.5 = 2.67
D / 5.0 / 22 / (22 – 6) / 5.0 = 3.20
E / – / – / –
F / 6.0 / 20 / (20 – 6) / 6 = 2.33

Since Job B has the lowest critical ratio, it is scheduled next and it is completed after 8 hours (6 + 2). After the completion of Job B, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / – / – / –
C / 4.5 / 18 / (18 – 8) / 4.5 = 2.22
D / 5.0 / 22 / (22 – 8) / 5.0 = 2.80
E / – / – / –
F / 6.0 / 20 / (20 – 8) / 6 = 2.00

Solutions (continued)

Since Job F has the lowest critical ratio, it is scheduled next and it will be completed after 14 hours (8 + 6). After the completion of Job F, the revised critical ratios are:

Job / Processing Time (Hrs.) / Due Date / Critical Ratio Calculation
A / – / – / –
B / – / – / –
C / 4.5 / 18 / (18 – 14) / 4.5 = 0.89
D / 5.0 / 22 / (22 – 14) / 5.0 = 1.60
E / – / – / –
F / – / – / –

Since Job C has the lowest critical ratio, it is scheduled next. Job C will be completed after 18.5 hours. The final critical ratio sequence of all jobs is E–A–B–F–C–D. Total completion of all six jobs (makespan) is 23.5 hours.

Job / Critical ratio / Job
time / Flow time / Due date / Tardy
e / 1.6 / 2.5 / 2.5 / 4 / 0
a / 2.0 / 3.5 / 6.0 / 7 / 0
b / 3.0 / 2.0 / 8.0 / 6 / 2
f / 3.3 / 6.0 / 14.0 / 20 / 0
c / 4.0 / 4.5 / 18.5 / 18 / .5
d / 4.4 / 5.0 / 23.5 / 22 / 1.5
23.5 / 72.5 / 4.0
FCFS / SPT / EDD / CR
Average flow time / 12.5 / 11.33 / 11.58 / 12.08
Average job tardiness / 2.83 / 0.83 / 0.42 / .67

Solutions (continued)

18. / a.
Order / Job time
A / 16 x / 4 = / 64
B / 6 x / 12 = / 72
C / 10 x / 3 = / 30
D / 8 x / 10 = / 80
E / 4 x / 1 = / 4
DD
Job / Job time / Flow time / Due date / Tardiness
A / 64 / 64 / 160 / 0
C / 30 / 94 / 180 / 0
D / 80 / 174 / 190 / 0
B / 72 / 246 / 200 / 46
E / 4 / 250 / 220 / 30
250 / 828 / 76

b. Average job tardiness = 76/5 = 15.2 minutes

c. Average number of jobs in the system = 828/250 = 3.31

d. / SPT
Job / Job time / Flow time / Due date / Tardiness
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 = 60/5 = 12 minutes

19. / 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

Solutions (continued)

20. / 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)
21. / 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

22.Each period’s backlog is equal to actual input – actual output. That amount is added to (or subtracted from) the previous backlog to obtain the current (shown) backlog for the period.

Period
Input / 1 / 2 / 3 / 4 / 5
Planned / 24 / 24 / 24 / 24 / 20
Actual / 25 / 27 / 20 / 22 / 24
Planned / 24 / 24 / 24 / 24 / 23
Actual / 24 / 22 / 23 / 24 / 24
Backlog / 12 / 13 / 18 / 15 / 13 / 13

Solutions (continued)

23.DayMon Tue Wed Thu Fri Sat

Staff needed 2 3 1 2 4 3

Worker 1 2 3 1 2 4 3

Worker 2 1 2 1 2 3 2 (tie)

Worker 3 0 2 1 1 2 1

Worker 4 0 1 0 0 1 1 Part-time worker

No. working: 2 3 1 2 4 3

24.

DayMon Tue Wed Thu Fri Sat

Staff needed 3 4 2 3 4 5

Worker 1 3 4 2 3 4 5

Worker 2 2 3 2 3 3 4 (tie)

Worker 3 1 3 2 2 2 3 (tie)

Worker 4 0 2 1 2 2 2

Worker 5 0 2 0 1 1 1 (part-time worker)

Worker 6 0 1 0 1 0 0 (tie) (part-time worker)

No. working: 3 4 2 3 4 5

Solutions (continued)

25.

DayMon Tue Wed Thu Fri Sat

Staff needed 4 4 5 6 7 8

Worker 1 4 4 5 6 7 8

Worker 2 4 4 4 5 6 7 (tie)

Worker 3 3 4 4 4 5 6

Worker 4 3 4 3 3 4 5

Worker 5 2 3 3 3 3 4

Worker 6 2 3 2 2 2 3 (tie)

Worker 7 1 2 2 2 1 2 (tie)

Worker 8 0 1 1 1 1 2

Worker 9 0 1 0 0 0 1 (tie) Part-time worker

No. working: 4 4 5 6 7 8