1. Production SchedulingP.C. Chang, IEM, YZU. 1 Hybrid Algorithm for mean T Hybrid Algorithm : Dominance matrix to decompose problem in to sub-problems. Branch and Bound Original problem for DP n! = C(n,1) + C(n,2) + ………….+ C(n,n) 30~35 jobs → out of computation time Baker p.49 N! N2!N2!N1!N1!

2. Production SchedulingP.C. Chang, IEM, YZU. 2 Decomposition Ex. Divide 50 persons into 4 subgroups by according to their height. Below 160 161~170 Mutual exclusive S Decompose S 1,S 2,………,Sn and 171~180 For any, put i in Then the objective function Above 181

3. Production SchedulingP.C. Chang, IEM, YZU. 3 Hybrid Algorithm SPTEDD Notation :

4. Production SchedulingP.C. Chang, IEM, YZU. 4 Hybrid Algorithm Property 3.2: (i before j) Bj tj Property 3.3: (j before i)

5. Production SchedulingP.C. Chang, IEM, YZU. 5 Hybrid Matrix Job j Job i P 11 P 12 …P 1n Q1Q1 ∣A1∣∣A1∣ P 21 P 22 …P 2n Q2Q2 ∣A2∣∣A2∣...... Pn 1 P n2 …P nn QnQn ∣An∣∣An∣ R1R1 R2R2 RnRn ∣B1∣∣B1∣∣B2∣∣B2∣∣Bn∣∣Bn∣

6. Production SchedulingP.C. Chang, IEM, YZU. 6 Steps: Baker p.51

7. Production SchedulingP.C. Chang, IEM, YZU. 7 Ex. (See Baker p.52) j12345678910 tjtj 3275808693107109115136144 djdj 216536812886633445716424

8. Production SchedulingP.C. Chang, IEM, YZU. 8 Final table 12345678910 ∣A∣∣A∣ 100110111117 210111111119 300000001012 400100111116 510110111118 600100011115 700100001114 800000000011 900000000011 00000000000 ∣B∣∣B∣ 2063145769 2-5-1-4-6-7-XXX-10 389

9. Production SchedulingP.C. Chang, IEM, YZU. 9 jtjdj 380681-502=179 8115716-502=214 9136424-502= -78 By DP 2-5-1-4-6-7-9-3-8-10

10. Production SchedulingP.C. Chang, IEM, YZU. 10 HW. j12345678 tjtj 12114710279130839688 djdj 260269400266337336683719 Use Hybrid to solve

11. Production SchedulingP.C. Chang, IEM, YZU. 11 Midterm

12. Production SchedulingP.C. Chang, IEM, YZU. 12 Single Machine Heuristics 1.Earliest Due Date (EDD) Rule 2.Cost Over Time (COVERT) Rule 3.Shortest Processing Time (SPT) Rule 4.Largest Penalty per Unit Length (LPUL) Rule 5.Largest Weight (WT) and LPUL Rule 6.Shortest Processing Time and LPUL Rule 7.Shortest Weighted Processing Time (SWPT) Rule 8.Critical Ratio (CR) Rule Sule p.14

13. Production SchedulingP.C. Chang, IEM, YZU. 13 COVERT Notations: TT - Total processing times of all jobs RT - Remaining total processing times of the set of unscheduled jobs ST - Starting time for the next scheduled job,0 for the first job CF - Coefficient PR - Priority

14. Production SchedulingP.C. Chang, IEM, YZU. 14 COVERT Rules Step l: calculate PR for all jobs PiPi ST ST+P i PR=0 PR=1 DiTT Step 2: calculate CF for job i, yet to be scheduled. Step 3: Job with the Max CF is scheduled next. Repeat step1 RT

15. Production SchedulingP.C. Chang, IEM, YZU. 15 Ex. job jpjpj djdj wjwj PR 13749144/56 22736557/66 31111 42837556/65 TT=93 RT=93 Iteration 1. Step l : Caculate PR

16. Production SchedulingP.C. Chang, IEM, YZU. 16 Ex......... Step 2: Step3: job 3 is scheduled. Iteration 2. job jpjpj djdj wjwj PR 137491? 227365? 3---- 428375?

17. Production SchedulingP.C. Chang, IEM, YZU. 17 Backward & Forward Method Improve ： Swapping to jump out the local optimal Initial scheduled Use any one of the Lower Bound Heuristic method Not the optimum

18. Production SchedulingP.C. Chang, IEM, YZU. 18 BF Method Lag = X X X X X 5 4 3 2 1 4: 5 – 1 3: 5 – 2, 4 – 1 2: 5 – 3, 4 – 2, 3 – 1 1: 5 – 4, 4 – 3, 3 – 2, 2 –1 5 jobs

19. Production SchedulingP.C. Chang, IEM, YZU. 19 Ex. Step1. TT = 37+27+1+28 = 93 Calculate the penalty for each job if it is to be complete at T=93. Job1: (93-49)*1=44 Job2: (93-36)*5=285 Job3: (93-1)*1=92 Job4: (93-37)*5=280  job 1 has the least cost, so it is scheduled in the last position job JPiDi Li (weight) 137491 227365 3111 428375 (1) Backward Phase:

20. Production SchedulingP.C. Chang, IEM, YZU. 20 EX. Step2. T= 93-37 = 56 Calculate the penalty for each job if it is to be complete at T=56. Job2: (56-36)*5=100 Job3: (56-1)*1=55 Job4: (56-37)*5=95  job 3 has the least cost, so it is scheduled in position 3

21. Production SchedulingP.C. Chang, IEM, YZU. 21 EX. Step3. T= 56-1 = 55 Calculate the penalty for each job if it is to be complete at T=55. Job2: (55-36)*5=95 Job4: (55-37)*5=90  job 4 has the least cost, so it is scheduled in position 2 Step4. The final sequence 2-4-3-1 The total penalty for the sequence 2-4-3-1 is (18*5)+(55*1)+(44*1)= 189

22. Production SchedulingP.C. Chang, IEM, YZU. 22 Ex. (2) Forward Phase: Initial sequence : 2-4-3-1 penalty = 189 Lag = 3 : 1-4-3-2 penalty = 490 – no change Lag = 2 : 3-4-2-1 penalty = 144 – change New sequence : 3-4-2-1 penalty = 144 Lag = 3 : 1-4-2-3 penalty = 644 – no change Lag = 2 : 2-4-3-1 penalty = 189 – no change Lag = 2 : 3-1-2-4 penalty > 144 – no change Lag = 1 : 4-3-2-1 penalty = 172 – no change Lag = 1 : 3-2-4-1 penalty = 139 – change New sequence : 3-2-4-1. We can get the final sequence is 3-2-4-1

23. Production SchedulingP.C. Chang, IEM, YZU. 23 HW. Use Forward + Covert to solve the following problem : (or you want to compare your solution with BF) job JPiDi Li (weight) 125393 26162 313251 421504 59212 630801