Chapter 15 Solutions:
Chapter 15 Solutions:
|1. |
| | | |column | |
| | | |reduction | |
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|2. |
| | | |column |
| | | |reduction |
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| |column | | |
| |reduction | | |
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| | | |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| | | |
|5. | a. Initial | | |
| |revised | | |
| |column | | |
| |reduction | | |
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| | | | |
| |column | | |
| |reduction | | |
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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 |
| | | |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 |Processing Time (Days) |Flow time |Due Date |Tardiness |
|sequence | | | | |
|A |14 |14 |20 |0 |
|C |7 |21 |15 |6 |
|D |6 |27 |17 |10 |
|B |10 |37 |16 |21 |
|( | |99 | |37 |
Solutions (continued)
[pic]
| |b. | |FCFS |SPT |EDD |CR |
| | |[pic] | | | | |
| | | |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 |
[pic]
| | | |FCFS |SPT |EDD |CR |
| | |[pic] | | | | |
| | | |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. | |
0 2 5 23 43 51 60 76 88
| | |e |b |g |h |d |c |a |f | |
| | | |e |b |g |h |
|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. | | | | | | | | |
0 16 36 79 114 156 216
| | |B |A |C |E |F |D | |
| | | |B |A | |C |E |F | |D |
0 16 46 73 79 130 158 182 216 228
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 | |
| | | | | | | | | | |
0 18 45 115 141 156
| | |b |a |c |d |e | |
| | | |b |a | |c |d |e |
0 18 51 96 115 145 169 179
| | | |
| |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. |
| | | |
141 156
0 9 18 31.5 45 80 115 128 148.5
| | |B |18 |A |27 |C 70 | |D |26 |
0 9 42 87 100 130 154 164
| | | |
| | |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: | |
|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 |Processing Time (Days) |Flow time |Due Date |Lateness |Tardiness |
|sequence | | | | | |
|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 |= |day| |Average flow |= |flow time | |
| | | | | |s | | | | | |
| | | | | |lat| | | | | |
| | | | | |e | | | | | |
| | |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 |Due date |Slack |Remaining number |Slack ( Remaining|Rank |
| | | |processing time | | |of operations |number of | |
| | | | | | | |operations | |
| | |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 |Job |Flow |Due date|Tardy |
| | | |ratio |time |time | | |
| | |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 |
[pic]
| | | |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 |Due |Tardiness |
| | | | |time |date | |
| | |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. Day Mon 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.
Day Mon 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.
Day Mon 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
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