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  1. Material Management Class Note # 5 SCHEDULING Prof. Yuan-Shyi Peter Chiu Feb. 2012

  2. § S1:Introduction

  3. § S1:Introduction

  4. § S5:Characteristics of Job Shop Scheduling

  5. § S5:Objectivesof Job Shop Management IT IS IMPOSSIBLE TO OPTIMIZE ALL 7 OBJECTIVES SIMULTANEOUSLY. CUSTOMER Service Quality vs. Plant Efficiency / Cost

  6. § S6:Flow Shop vs. Job Shop (1) FLOW SHOP job M/C M/C M/C M/C M/C ... 1 2 3 4 m     . . .  Assembly Line different M/C = different operations. n   . . .  n ◇ (2) JOB SHOP job M/C M/C M/C M/C M/C ... 1 2 3 4 m {1, 3, 4, 1, m} {2, 4, 2, m-1} PROBLEMS are extremely complex. All-purpose solution algorithms for solving general job shop problems do not exist.

  7. § S6:Flow Shop vs. Job Shop (3)Parallel processing vs. Sequential processing JOB M/C M/C M/C M/C ... 1 2 3 m Sequential   . . .  n JOB M/C M/C M/C M/C    1 3 2 m 1 3 Parallel processing ... 2 m 1 3  . . .  n 3

  8. § S7:Indicators of Performance Evaluation ◇ (1) FLOW TIME job 1 t1 = 25 sec. job 2 t2 = 13 sec. job 3 t3 = 21 sec. . . . (2) MEAN FLOW TIME

  9. § S7:Indicators of Performance Evaluation (3) MAKESPAN:F[n] TIME REQUIRED TO COMPLETE ALL n jobs Minimizing to the Makespan is a common objective in multiple - m/c sequencing problem. (4) TARDINESS max ( F[i] - D[i] , 0 ) where F[i] : Completion time of job i. D[i] : Job i ’s Due Date. Example : Job F[i] D[i] Tardiness 1 3 10 0 2 7 12 0 3 14 13 1 4 16 13 3

  10. (6) MINIMIZINGAVG. TARDINESS MAX. TARDINESS ARE COMMON SCHEDULING OBJECTIVES. § S7:Indicators of Performance Evaluation ◇ (5) LATENESS F[i] - D[i] Job F[i] D[i] Lateness 1 3 10 -7 2 7 12 -5 3 14 13 1 4 16 13 3

  11. § S8:Notation ◇ ti ; di  Wi = Fi- ti Fi = Wi + ti   Li = Fi- di Ti = max[Li , 0] Ei = max[-Li , 0] Tmax = max {T1 , T2 , … , Tn} → SPT minimizes Mean flow Time. Mean waiting Time. Mean lateness. → EDD minimizes maximum lateness Lmax ~ Tmax. Lmax = max {L1 , L2 , L3 , … , Ln} 3 1 0 -3 -3 -2 -1 0

  12. • Common Scheduling Rules for single machine: • FCFS • Shortest Processing Time (SPT) • (3) Earliest Due Date (EDD) • (4) CRITICAL RATIO (CR)

  13. (1) Plane 1 2 3 4 5 t[i]26 11 19 16 23 F[i] 95 11 46 27 69 makespan = 95 SPT = {2, 4, 3, 5, 1} = 49.6 (2) Plane 1 2 3 4 5 t[i]26 11 19 16 23 # of Passengers 180 12 45 75 252 # of Passengers per minute6.9 1.1 2.4 4.7 10.9 49 95 84 65 23 F[i] 432 564 552 507 252 # of Passengers {5, 1, 4, 3, 2} Example 8.2 – An example of priority rules

  14. 100% 90% % Passengers 76% 45% 23 49 65 time 20 40 60 80 100 (3) ARRIVAL TIME = DUE TIME. (4) PRIORITY e.g. continuing flights. low fuel level. carrying precious or perishable cargo.

  15. § S9:EDD Scheduling minimizes the maximum lateness. ◇ § S10:Moore’s (1968) algorithm minimizes the number of Tardy jobs. Step1:Sequence by earliest due date i.e. d[1] ≦ d[2] ≦ d[3] ≦ …≦ d[n] Step2:Find the 1st tardy job in the current sequence, say job i. IF None exists go to Step 4 Step3:Consider jobs [1], [2], …, [i] Reject the job with largest tj .and Return to Step2. Step4:Current sequence + rejected job(In any order.) Applications:chef, runway, preparing exam. garage, dock.

  16. Ti = { max ( F[i] - D[i] , 0 ) } > 0 Job# 2 3 5 4 6 di 6 9 20 23 30 ti 3 4 10 8 6 Fi 3 7 17 25 31 × 5 × Example 8.3 Job# 2 3 1 5 4 6 Di 6 9 15 20 23 30 ti 3 4 10 10 8 6 Fi 3 7 17 27 35 41 × 1 ×

  17. Example 8.3 2 3 4 6 di 6 9 23 30 ti 3 4 8 6 Fi 3 7 15 21 Done! 2 - 3 - 4 - 6 - 5 - 1 1 - 5 2 3 4 6 5 1 di 6 9 23 30 20 15 ti 3 4 8 6 10 10 Fi 3 7 15 21 31 41

  18. §.S10.1: Class Problems Discussion Chapter 8 :# 3, 4, 5 p.413 30, 32(a),(b), 34 p.449-450 Preparation Time : 15 ~ 20 minutes Discussion : 15 minutes

  19. § S11:Lawler’s Algorithm : Precedence ◇ min. max. gi(Fi) 1 ≦ i ≦ N • gi(Fi) = Fi – di = Li min. max. Lateness. • gi(Fi) = max (Fi – di , 0) Tardiness. … next to LAST LAST

  20. JOBS: 1 2 3 5 4 6 Example 8.4 JOB 1 2 3 4 5 6 ti 2 3 4 3 2 1 di 3 6 9 7 11 7

  21. JOBS 3 5 6 6 3 Example 8.4 ◇ v = { 3, 5, 6 } Total Processing time of all jobs is 15 min iv min iv {gi(Fi)}= {Fi-di} = min {15-9 , 15-11 , 15-7} = min {6 , 4 , 8}= 4 ∴ JOB # 5 is scheduled last. (6th) JOBS v = { 3, 6 } Total Processing time is 13 min iv {gi(Fi)}= min {13-9 , 13-7}= min {4 , 6}= 4 ∴ JOB # 3 is scheduled last. (5th) So far, { … , 3 , 5}

  22. JOBS 1 2 4 2 6 2 v = { 2, 6 } Example 8.4 ◇ Total Processing time is 9 min iv {gi(Fi)}= min {9-6 , 9-7}= 2 ∴ JOB # 6 is scheduled last. (4th) So far, { … , 6 , 3 , 5} JOBS v = { 2, 4 } Total Processing time is 8 min iv {gi(Fi)}= min {8-6 , 8-7}= 1 ∴ JOB # 4 is scheduled 3rd ∴ So far, { … , 4 , 6 , 3 , 5} JOBS → ∴ { , 2 , 4 , 6 , 3 , 5} JOBS → ∴ { 1 , 2 , 4 , 6 , 3 , 5}

  23. Example 8.4 Job# ti Fi di Ti (Tardiness)1 2 2 3 0 2 3 5 6 0 4 3 8 7 1 6 1 9 7 2 3 4 13 9 4 5 2 15 11 4 ∴ maximum tardiness is 4 days.

  24. §.S12: Class Problems Discussion Chapter 8 : # 6, 7, 8, 9p.419 10# 37,p.451 Preparation Time : 10 ~ 15 minutes Discussion : 10 minutes

  25. M/C 2 M/C m ◇ § S13:n job on m M/C’s (1) n JOBS Must BE PROCESSED ON 1 M/C’s n …….. 3 2 1 1 ……..(n-2) (n-1) n ∴ there are n! possible ways. (2) “n” JOBS Must BE PROCESSED ON “m” M/C’s n …….. 3 2 1 n! .……. n! …….. n!   ∴ there are (n!)m possible ways. #12 p.428 M/C M/C1 … … … …

  26. § S14:Permutation schedules – characteristics • It provides better system performance in terms of both total flow time ( makespan ) and average flow time ( ). • mean IDLE time? • (B) SCHEDULING n Jobs on 2 M/C’s : if each Jobs must be processed in the order M/C-1 then M/C-2. • Results: The permutation schedule will minimize • “makespan” and minimizes “ ”. • (C) Theorem 8.2: (p.422) • The optimal solution for scheduling n jobs on 2 M/C’s is always a permutation schedule. F F

  27. § S15:2 jobs on 2 M/C’s • All Possible Schedules for Two Jobs on Two Machines. • (B) Assumes that both jobs must be processed first on M/C#1 then on M/C#2. makespanidleflow time 9 6 10 10 Machine 1 I J Machine 2 I J Machine 1 J I Machine 2 J I Machine 1 J I Machine 2 I J Machine 1 I J Machine 2 J I 4 5 9 1 5 6 1 5 6 10 4 5 9 10

  28. § S16:Permutation Schedules - Definition *Permutation Schedules= Same Sequence on both (all) M/C’s Total # of permutation schedules is exactly n!

  29. § S17: Johnson’s Rule ◇ TO MINIMIZE THE MAKESPAN (1) Definition: M/C-A M/C-B Jobs must be processed first on M/C-A then M/C-B Ai : Processing Time of Job i on M/C-A Bi : Processing Time of Job i on M/C-B Job i procedes Job i+1 if min (Ai , Bi+1) < min (Ai+1,Bi) 2 M/C’s

  30. (2) Working Procedures for Johnson’s rule: 1. List the values of Ai & Bi in 2 columns. 2. Find the smallest remaining element in the 2 columns. If it appears in column A then schedule that job next. If it appears in column B then schedule that job last. 3. Cross off the jobs as they are scheduled. Stop when all jobs have been scheduled. An easy way to implement Johnson’s Rule

  31. Example 8.5 Five jobs are to be scheduled on two machines. The processing times are Jobs Machine A Machine B 1 5 2 2 1 6 3 9 7 4 3 8 5 10 4

  32. 1 4 9 18 28 A #2 #4 #1 #3 #5 B #2 #4 #1 #3 #5 0 1 7 15 17 25 28 32 1 4 13 18 28 A #2 #4 #3 #5 #1 B #2 #4 #3 #5 #1 0 1 7 15 22 23 27 28 30 Example 8.5 (A) if using SPT in Ai then obtain the above 2-4-1-3-5 _____________________________________________________ IF BY JOHSON’S ROLE TO MINIMIZE THE MAKESPAN OR TOTAL FLOW TIME. Rule: Job i precedes job i+1 if MIN(Ai,Bi+1)<(Ai+1,Bi) in 2-4-3-5-1 order (B) makespan=30

  33. § S18:n jobs on 3 M/C’s ◇ (A)Objective : To minimize total flow time (“make span.”) it is still true that a permutation schedule is optimal. (B) But it is not necessarily optimal for the case of F’ . (C) 3 M/C’s can be reduced to 2 M/C’s (then using Johnson’s Rule to solve it) if min Ai≧max Bi or min Ci≧max Bi “either one” of these conditions be satisfied. then define

  34. M/C’s Jobs A B C 1 4 5 8 2 9 6 10 3 8 2 6 4 6 3 7 5 5 4 11 ◇ Example 8.6 Consider the following job times for a three-machine problem. Assume that the jobs are processed in the sequence A-B-C Min Ci=6 Max Bi=6 ∴ Min Ci ≧ Max Bi let

  35. M/C’s Jobs A’ B’ 1 9 13 2 15 16 3 10 8 4 9 10 5 9 15 ◇ Example 8.6 Using Johnson’s Rule 5-1-4-2-3

  36. 5-4-1-2-3 ◇ Example 8.6 5-1-4-2-3

  37. §.S18.1: Class Problems Discussion Chapter 8 : # 13, 14p.428 38,40p.451 ◆ Preparation Time : 10 ~ 15 minutes Discussion : 10 minutes

  38. and ◇ § S19:more on 3 M/C’s problems • If neither ﹛min Ai ≧ max Bi Nor min Ci ≧ max Bi ﹜ • Then Using Will usually give “REASONABLE” but possibly “SUBOPTIMAL” result. Simplifying 3 M/C’s problems 2 M/C’s. (B) TO MINIMIZE “MAKESPAN” OR “TOTAL FLOW TIME” A PERMUTATION SCHEDULE IS OPTIMAL ON 3 M/C’s

  39. § S20:Akers’ procedures for solving 2 Jobs on m M/C’s problems ◇ • Draw a cartesian coordinate system. • Job 1 on X-axis • Job 2 on Y-axis • mark off the operation times • on X & Y axis • (2) BLOCK OUT AREAS for each M/C’s • (3)Determine a path from origin to the end of the final block. • move: • The path with minimum vertical distance is “Optimal”.

  40. JOB 1 JOB 2 Oper. Time Oper. Time (A) Sanding 3 A 2 (B) Lacquering 4 B 5 (C) Polishing 5 C 3 Example 8.7 ◇ A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamps must be sanded, lacquered, and polished. Each operation requires a different machine. There are currently shipments of two models awaiting processing. The times required for the three operations for each of the two shipments are

  41. 1C 1B 2C 1B 2C 2B 2B 1C 2B 1A 1B 2A 2A 1B 1A Example 8.7 Fig. 8-8 p.426 maximize the diagonal movement = min. horizontal vertical

  42. Fig. 8-9 p.426 Example 8.7

  43. Example 8.8

  44. 16 (16,14) 14 12 (7,11) (16,11) 10 8 6 4 2 2 4 6 8 10 12 14 16 18 Example 8.8 ◇ A1 →B1 →C1 A2 A2 ↓ C1 D2 ↓ C2←D1←D1 ←D2 C2 B2 Time=16+1+3=20 C Job 2 (BOB) B D A Job 1 (Reggie) A2 →D2 →B2 →C2 →C2 →C2 →C1 →D1 A1 A1 A1 B1 Time=16+4+3=23

  45. §.S21: Class Problems Discussion Chapter 8 : # 15,16 p.428 39p.452 Preparation Time : 25 ~ 30 minutes Discussion : 20 minutes

  46. § S22:Parallel processing on identical M/C’s. SPT → minimize mean flow time. LPT → minimize total flow time, or makespan. ( longest ) The End

  47. Scheduling Preview : Chap. 8 [ pp.413~442 ] Problem: # 5, #32(a),(b), #7, #8, #10, #37, #40, #39, #14, #15, #16