2. 1 By: Prof. Y. Peter Chiu Scheduling ~ HOMEWORK SOLUTION ~

3. 2 HW # 4 Four trucks, 1, 2, 3, and 4, are waiting on a loading dock at XYZ Company that has only a single service bay. The trucks are labeled in the order that they arrived at the dock. Assume the current time is 1:00 p.m. The times required to unload each truck and the times that the goods they contain are due in the plant are given in the table below: Determine the schedules that result for each of the rules: (1) FCFS ; (2) SPT ; (3) EDD ; and (4) CR. In each case compute the mean flow time, average tardiness, and number of tardy jobs. Truck#Unloading time Time Material Is Due 120 mins 1:25 p.m. 214 mins 1:45 p.m. 335 mins 1:50 p.m. 410 mins 1:30 p.m.

4. 3 HW # 4 FCFS Mean flow time = 202/4 = 50.5 Avg. Tardiness = 68/4 = 17 Number of Tardy trucks = 2 Truck # tiFidiTi 120 250 21434450 335695019 410793049 Total20268

5. 4 HW # 4 SPT F’=157/4=39.25 Avg. Ti=48/4=12 # of Tardy Trucks = 2 Truck#tiFidiTi 410 300 21424450 120442519 335795029 Total15748

6. 5 HW # 4 EDD F’=43.25 T’=7.25 # of Tardy job = 1 Truck#tiFidiTi 120 250 41030 0 21444450 335795029 Total17329

7. 6 HW # 4 CR : Current Time = 1:00pm Truck# titi didi CR 1202525/20=1.25* 2144545/14=3.21 3355050/35=1.43 4103030/10=3 Pick!

8. 7 HW # 4 CR : Current Time = 1:00pm+20min = 1:20pm Truck# titi didi d i- CurTime CR 214452525/14=1.79 335503030/35=0.86* 410301010/10=1 Pick!

9. 8 HW # 4 Current Time = 1:20pm+35min = 1:55pm Late Jobs use “SPT” order, pick #4, then #2 Truck# titi didi di-CurTime CR 21445-10 41030-25 Pick! Truck#tiFidiTi 120 250 F’=54.75 33555505 T’=18.50 410653035 #of Tardy Jobs=3 214794534

10. 9 HW # 5 p.414 Five jobs must be scheduled for batch processing on a mainframe computer system. The processing times and the promised times for each of the jobs are listed here. job 1 2 3 4 5 processing time 40 min 2.5 hr 20 min 4 hr 1.5 hr promised time 11:00 A.M. 2:00 P.M. 2:00 P.M. 1:00 P.M. 4:00 P.M. Assume that the current time is 10:00 A.M. (a) If the jobs are scheduled according to SPT, find the tardiness of each job and the mean tardiness of all jobs. (b) Repeat the calculation in part (a) for EDD scheduling.

11. 10 HW # 5 SPT : Current Time 10:00 AM 3 – 1 – 5 – 2 – 4 Job titi Fi didi Ti 32010:202400 14011:00600 51:3012:003600 22:3015:0024060 44:0019:00180360 T’=84 mins 420

12. 11 HW # 5 EDD : 1 – 4 – 3 – 2 – 5 (Tie uses ‘SPT’ ), otherwise : 1 – 4 – 2 – 3 – 5 ; T’=136 Job titi Fi didi Ti 14010:40600 44:0014:40180100 32015:0024060 22:3017:30240210 51:3019:00360180 T’=110 550

13. 12 HW # 6 Consider the information given in Problem 4. Determine the sequence that the truck should be unload in order to minimize (a) Mean flow time. (b) Maximum lateness. (c) Number of tardy jobs. Four trucks, 1, 2, 3, and 4, are waiting on a loading dock at XYZ Company that has only a single service bay. The trucks are labeled in the order that they arrived at the dock. Assume the current time is 1:00 p.m. The times required to unload each truck and the times that the goods they contain are due in the plant are given in the table below: Truck#Unloading time Time Material Is Due 120 mins 1:25 p.m. 214 mins 1:45 p.m. 335 mins 1:50 p.m. 410 mins 1:30 p.m.

14. 13 HW # 6 From #4 (a) SPT minimizes F’ : 4 – 2 – 1 – 3 (b) EDD minimizes maximum lateness : 1 – 4 – 2 – 3 (c) By Moore’s Algorithm : 1 – 4 – 2 – 3

15. 14 HW # 7 On May 1, a lazy MBA student suddenly realizes that he has done nothing on seven different homework assignments and projects that are due in various courses. He estimates the time required to complete each project (in days) and also notes their due dates: Project 1 2 3 4 5 6 7 Time (days) 4 8 10 4 3 7 14 Due date 4/20 5/17 5/28 5/28 5/12 5/7 5/15 Because projects 1, 3, and 5 are from the same class, he decides to do those in the sequence that they are due. Furthermore, project 7 requires results from projects 2 and 3, so project 7 must be done after 2 and 3 are completed. Determine the sequence in which he should do the projects in order to minimize the maximum lateness.

16. 15 HW # 7

17. 16 PROJ# 1234567 Time 481043714 Due Date -111627 11614 HW # 7

18. 17 PROJ# titi Fi didi Fi-di 144-1115 671165 5314113 2822166 31032275 714461432 44502723 HW # 7

19. 18 HW # 8

20. 19 HW # 9 (a)To minimize mean flow time SPT ： 6-4-2-1-3-5 (b) To minimize “Number of Tardy Jobs” Moore’s Algorithm

21. 20 Jobti Delivery titi didi 11:12151:273:30 2:4015552:00 32:12152:273:00 4:3015455:00 53:06153:214:00 6:2515406:00 HW # 9

22. 21 HW # 9 (b) Moore’s Algorithm starting from EDD Jobti’Fi’diLi 255 2:00-1:05 32:273:223:0022 11:274:493:301:19 53:218:104:004:10* 4458:555:003:55 6409:356:003:35 Delete Job # 3

23. 22 Jobti’Fi’di 255 2:00 11:272:223:30 53:215:434:00 4455:00 6406:00 Jobti’Fi’di 255 2:00 11:272:223:30 4453:075:00 6403:476:00 HW # 9 (b) Moore’s Algorithm Delete Job # 5

24. 23 2-1-4-6-3*-5* (by SPT for tardy jobs) or 2-1-4-6-5*-3* # of Tardy Jobs = 2 (c) To minimize maximum lateness EDD ： 2-3-1-5-4-6 Max Li=4:10 HW # 9 (b) Moore’s Algorithm

25. 24 HW # 10 Seven jobs are to be processed through a single machine. The processing times and due dates are given here. Project 1 2 3 4 5 6 7 Processing time 3 6 8 4 2 1 7 Due date 4 8 12 15 11 25 21 Determine the sequence of the jobs in order to minimize: (a) Mean flow time. (b) Number of tardy jobs. (c) Maximum lateness & compute the maximum lateness. (d) What is the makespan for any sequence?

26. 25 HW # 10 (a)SPT minimize F’ 6 – 5 – 1 – 4 – 2 – 7 – 3 (b) To minimize Number of Tardy Jobs Use Moore’s Algorithm first EDD ： 1 – 2 – 5 – 3 – 4 – 7 – 6

27. 26 Job titi Fi didi Li 1334 26981 5211 0 3819127 4423158 7730219* 6131256 Delete Job # 2 HW # 10 (b) By Moore’s Algorithm

28. 27 Job titi Fidi 1334 52511 381312 44 15 77 21 61 25 Delete Job # 3 HW # 10 (b) By Moore’s Algorithm

29. 28 ∴ 1 – 5 – 4 – 7 – 6 – 2* - 3* ( tardy jobs by SPT) or 1 – 5 – 4 – 7 – 6 – 3* - 2* Number Of Tardy Jobs = 2 Job titi Fidi 1334 52511 44915 771621 611725 HW # 10 (b) By Moore’s Algorithm

30. 29 (C)EDD minimize Maximum lateness 1 – 2 – 5 – 3 – 4 – 7 – 6 maximum lateness = 9 (d) HW # 10

31. 30 #37 To minimize maximum lateness subject to the precedence constraints. Lawler’s Algorithm Job#titi DiDi 145 210 327 419 5822 6325 7218 8620 1→ 2→ 5→ 6 4→ 7→ 8 3

32. 31 By Chiu’s algorithm V 1 ={3,6,8} max D i → 6 V 2 ={3,5,8} max D i → 5 V 3 ={2,3,8} → 8 V 4 ={2,3,7} → 7 V 5 ={2,3,4} → 2 V 6 ={1,3,4} → 4 V 7 ={1,3} → 3 Last → 1 The Optimal sequence is 1-3-4-2-7-8-5-6 #37

33. 32 The End