CHAPTER 16: SCHEDULING

Transcription

1 CHAPTER 16: SCHEDULING Solutions: 1. Job A B C A B C row Worker reduction column reduction A B C Optimum: Worker 1, Job A B 3 C 2. Initial Job Initial revised A B C A B C A B C row Worker reduction column reduction A B C Optimum: B, 2 C, 3 A

2 3. Route A B C D E A B C D E Truck row reduction column reduction A B C D E A B C D E add and subtract Optimum: 1 A, 2 E, 3 D, 4 B, 5 C or 1 A, 2 D, 3 E, 4 B, 5 C 16-2

3 4. Job Job Initial + Dummy Machine A B C D row [no change due to dummy] reduction column reduction A B C D A B C D add and subtract Optimum: 1 B, 2 C, 3 D, 4 A Total cost = $ a. Initial revised Machine A B C D E A B C D E Job row reduction column reduction A B C D E Optimum: 1 A, 2 B, 3 C, 4 D, 5 E

4 5. b. Initial revised Machine A B C D E A B C D E Job row reduction column reduction A B C D E A B C D E add and subtract Optimum: 1 E, 2 B, 3 C, 4 D, 5 A 16-4

5 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 B C D SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy D C B A EDD: Job time Flow time Due date Days Job (days) (days) (days) tardy C Critical Ratio B D A Processing Time Job (Days) Due Date Critical Ratio Calculation A (20 0) / 14 = 1.43 B (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: 16-5

6 Processing Time Job (Days) Due Date Critical Ratio Calculation A B (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: Processing Time Job (Days) Due Date Critical Ratio Calculation A B (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 C D B

7 41 99 Average Tardiness day s; Average flow time 24.75days 4 4 b. 99 Average number of jobs 2.67 jobs 37 Flow time Average flow time Numberof jobs FCFS SPT EDD CR Average job tardiness Day stardy Numberof jobs Average number of c. SPT is superior. jobs at thecenter Flow time job times

8 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 b c d e SPT: Operation Flow time Due date Hours Job time (hr.) (hr.) (hr.) tardy c b a e d EDD: Operation Flow time Due date Hours Job time (hr.) (hr.) (hr.) tardy a b c e d

9 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 = (20 0) / 11 = 1.82 E (.10 x 75) +.5 = 8 15 (15 0) / 8 = 1.88 Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4 hours, the revised critical ratios are: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B 4 10 (10 4) / 4 = 1.5 C 2 12 (12 4) / 2 = 4 D (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: Processing Time Job (Hours) Due Date Critical Ratio Calculation A B C 2 12 (12 11) / 2 = 0.5 D (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 = 13. After the completion of Job C, the revised critical ratios are: 16-9

10 Processing Time Job (Hours) Due Date Critical Ratio Calculation A B C D (20 13) / 11 =.63 E 8 15 (15 13) / 8 =.25 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 Processing sequence Time (Days) Flow time Due Date Tardiness A B C E D Average tardiness ; Average flow time 84 Average number of jobs 2.63 jobs hours 5 Average flow time Average job tardiness Flow time Numberof jobs Hours late Numberof jobs FCFS SPT EDD CR Average number of jobs at thecenter Flow time job times

11 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 B ,2 (tie) C D ,2 (tie) E b. S/O: [Assume B D C A E] Time Flow time Due date Days Job (days) (days) (days) tardy B D C A E Time Flow time Due date Days Job (days) (days) (days) tardy A B C D E flow time: Average flow time = number of jobs flow time: Average number of jobs in the system = job times FCFS S/O

12 9. Time (hr.) Order Step 1 Step 2 A B C D E F G Sequence of assignment:.80 [C] last (i.e., 7th).90 [B] first 1.20 [A] 2nd 1.30 [G] 3rd 1.60 [E] 4th 1.50 [D] 6th 1.75 [F] 5th Thus, the sequence is b-a-g-e-f-d-c. 10. a. Job Machine A Machine B a b Thus, the sequence is e b g h d c a f. c d e f g h b

13 e B g h d c a f e b g h d c a f c. Original idle time for B: = 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 1 Machine 2 a a f f After splitting, we get the following Gantt chart: e b g h d c a 1 a 2 f 1 f 2 e b g h d c a 1 a 2 f 1 f After splitting, idle time is: = 16 hours, and the new makespan = 90. There is a savings of 2 hr. Time (minutes) 11. a. Job Center 1 Center 2 A B Thus, the sequence is B A C E F D. C D E F b B A C E F D B A C E F D

14 Idle time of 56 hours. 12. a. Job Station A Station B a b Thus, the sequence is b a c d e. c d e b a c d e b a c d e b. The Idle time for Station B is = = 37 minutes. c. Jobs B, A, C, D and E are candidates for splitting in order to reduce throughput time and idle time B 18 A 27 C 70 D 26 E 15 B 33 A 45 C 30 D 24 E Throughput time is 162 minutes, reducing this time by 17 minutes. The idle time for B of 20 minutes has decreased by 17 minutes

15 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 Polishing Using Johnson s Rule, we obtain: Cutting Polishing Job Start Finish Start Finish G A E D B C F 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 B A D G F E The Grinding flow time is 93 hours and Deburring flow time is 140 hours. The Total time is 37 hours

16 c. Johnson s Rule Grinding Deburring Job Start Finish Start Finish C B A F E D G 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: SPT: Job Flow Due Days Job Flow Due Days Job time time date tardy Job time time date tardy a d b e c f d a e c f b EDD: CR: Job Flow Due Days Job Flow Due Days Job time time date tardy Job time time date tardy a a c c b b e e f f d d

17 Critical Ratio Processing Time Job (Days) Due Date Critical Ratio Calculation A (10 0) / 4.5 = 2.22 B (17 0) / 6.0 = 2.83 C (12 0) / 5.2 = 2.31 D (27 0) / 1.6 = E (18 0) / 2.8 = 6.43 F (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: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B (17 4.5) / 6.0 = 2.08 C (12 4.5) / 5.2 = 1.44 D (27 4.5) / 1.6 = E (18 4.5) / 2.8 = 4.82 F (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 ( ). After the completion of Job C, the revised critical ratios are: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B (17 9.7) / 6.0 = 1.22 C D (27 9.7) / 1.6 = E (18 9.7) / 2.8 = 2.96 F (19 9.7) / 3.3 =

18 Job B has the lowest critical ratio, therefore it is scheduled next and it is completed after 15.7 days ( ). After the completion of Job B, the revised critical ratios are: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B C D ( ) / 1.6 = 7.06 E ( ) / 2.8 = 0.82 F ( ) / 3.3 = 1.0 Job E has the lowest critical ratio, therefore it is scheduled next and it is completed after 18.5 days ( ). After the completion of Job E, the revised critical ratios are: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B C D ( ) / 1.6 = 5.31 E F ( ) / 3.3 = 0.15 Job F has the lowest critical ratio therefore it is scheduled next and it is completed after 21.8 days ( ). 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 C B E F D

19 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 = days 91.5/23.4 = 3.91 jobs SPT 14.0/6 = 2.33 days 66.7/6 = days 66.7/23.4 = 2.85 EDD 3.3/6 = 0.55 days. 93.6/6 = days 93.6/23.4 = 4.00 CR 3.3/6 =.55 days 93.6/6 = 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 [Slight differences in ratios may arise due to rounding.] c. S/O job sequence is a-c-b-e-d-f 16. Job Remaining processing time Due date Slack Remaining number of operations Slack Remaining number of operations Rank a b c d e 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 b c d e f

20 SPT Job Job time Flow time Due date Tardy b e a c d f EDD Job Job time Flow time Due date Tardy e b a c f d Critical Ratio Processing Time Job (Days) Due Date Critical Ratio Calculation A (7 0) / 3.5 = 2.0 B (6 0) / 2.0 = 3.0 C (18 0) / 4.5 = 4.0 D (22 0) / 5.0 = 4.4 E (4 0) / 2.5 = 1.6 F (20 0) / 6 =

21 Job E has the lowest critical ratio, therefore it is scheduled first and completed after 2.5 hours. The revised critical ratios are: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A (7 2.5) / 3.5 = 1.29 B (6 2.5) / 2.0 = 1.75 C (18 2.5) / 4.5 = 3.44 D (22 2.5) / 5.0 = 3.90 E F (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 ( ). After the completion of Job A, the revised critical ratios are: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B (6 6) / 2.0 = 0 C (18 6) / 4.5 = 2.67 D (22 6) / 5.0 = 3.20 E F (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: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B C (18 8) / 4.5 = 2.22 D (22 8) / 5.0 = 2.80 E F (20 8) / 6 =

22 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: Processing Job Time (Hrs.) Due Date Critical Ratio Calculation A B C (18 14) / 4.5 = 0.89 D (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. Critical Job Flow Due Job ratio time time date Tardy e a b f c d Averageflow time Averagetardiness.67hours hours FCFS SPT EDD CR Average flow time Average job tardiness

23 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 C D B E 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 C A B D Average job tardiness = 60/5 = 12 minutes 19. Sequence Setup times Total A B C = 7 (best) A C B = 10 B A C = 16 B C A = 9 C A B = 9 C B A =

24 20. Sequence Setup times Total A B C = 5.6 A C B = 5.9 B A C = 6.2 B C A = 7.2 C A B = 6.4 C B A = 4.1 (best) 21. Sequence Setup times Total A B C D = 12 A B D C = 15 A D B C = 12 A D C B = 14 B A D C = 18 B C D A = 10 (best) C B A D = 16 C B D A = 11 C D A B = 14 C D B A = 15 D A B C = 14 D C B A = 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 Planned Actual Output Planned Actual Backlog

25 23. Period Input Planned Actual Deviation Cum. Dev Period Output Planned Actual Deviation Cum. Dev Backlog Day Mon Tue Wed Thu Fri Sat Staff needed Worker Worker (tie) Worker Worker Part-time worker No. working: Day Mon Tue Wed Thu Fri Sat Staff needed Worker Worker (tie) Worker (tie) Worker Worker (part-time worker) Worker (tie) (part-time worker) No. working:

26 26. Day Mon Tue Wed Thu Fri Sat Staff needed Worker Worker (tie) Worker Worker Worker Worker (tie) Worker (tie) Worker Worker (tie) Part-time worker No. working: