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Job Shop Scheduling

Job Shop Scheduling. Definitions. The most common formulation of the job shop problem specifies that each job has exactly m operations, one on each machine. (i, j, k) means operation j of job i requires machine k. Job Shop Scheduling. operation. 4. 3. 1. 2. 3. 2. 1. 4. 4. 5. 6.

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Job Shop Scheduling

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  1. Job Shop Scheduling

  2. Definitions • The most common formulation of the job shop problem specifies that each job has exactly m operations, one on each machine. • (i, j, k) means operation j of job i requires machine k.

  3. Job Shop Scheduling operation 4 3 1 2 3 2 1 4 4 5 6 4 0 13 routing 1 3 2 7 8 9 1 2 3 10 11 12

  4. Job Shop Scheduling M1 4 3 1 2 3 2 1 4 4 5 6 4 0 13 1 3 2 7 8 9 1 2 3 10 11 12 Disjunctive, we don’t know the sequence, but all of them are processed on M1.

  5. Job Shop Scheduling M2 4 3 1 2 3 2 1 4 4 5 6 4 0 13 1 3 2 7 8 9 1 2 3 10 11 12

  6. Job Shop Scheduling M3 4 3 1 2 3 2 1 4 4 5 6 4 0 13 1 3 2 7 8 9 1 2 3 10 11 12

  7. Job Shop Scheduling Set to be sequenced on M1 Set to be sequenced on M2 Set to be sequenced on M3 Head Tail Be changed with different sequence - Carlier

  8. Job Shop Scheduling For M1 T H T H 5 5 9 1 9 0 4 5 8 H T H T 4 6 5 9 12 1 5 6 7 Use B&B to find a bound: h(s) = 0 + 12 + 0 = 12

  9. Job Shop Scheduling For M2 T H T H 4 2 3 3 2 8 4 7 9 3 5 8 H T H T 8 4 4 10 1 9 3 7 h(s) = 0 + 9 + 2 = 11

  10. Job Shop Scheduling For M3 T H T H 5 3 7 0 7 9 3 8 H T H T 4 3 6 11 5 9 3 6 7 h(s) = 0 + 12 + 0 = 12

  11. Schrage Algorithm • Arrange 7 jobs on M2 Step 1. t = 0, job6 is ready, schedule job6, t = 0 + 6 = 6, S={6}. Step 2. t = 6, no job is ready, so job1 with the smallest r is scheduled, t = Max(6 , 10) + 5 = 15, S={6,1}. Step 3. t = 15, job2 and 3 are ready, schedule job2 first, t = 15 + 6 = 21, S= {6,1,2}.

  12. Schrage Algorithm Step 4. t = 21, job3 and 4 are ready, schedule job3, t = 21 + 7 = 28, S= {6,1,2,3}. Step 5. t = 28, job4 is ready, schedule job4, t = 28 + 4 = 32, S= {6,1,2,3,4}. Step 6. t = 32, job5 and job 7 are ready, and q5>q7 , so job 5 is scheduled, t = 32 + 3 = 35, S= {6,1,2,3,4,5}. Step 7. t = 35, job7 is schedule, t = 35 + 3 = 37, S= {6,1,2,3,4,5,7}.

  13. 6 6 h = 0+23=23 23 1 h = 10+12=22 0 12 10 5 h = 15+32=47 2 32 13 h = 21+31=52 6 11 31 0 3 * 7 20 25 h = 28+25= 53 max ! 4 30 11 h = 32+11=43 4 30 2 5 h = 34+2=36 3 7

  14. Carlier • Schrage’s schedule L = 53 • Carlier : Lower bound • Theorm : • Let L be the makespan of the schrage schedule • If this schedule is not optimal , there is a critical job c and a • critical set J such that: Pc > L – h (J) upper bound lower bound (b) If this schedule is optimal ,there exists J such that h(J) =L

  15. Carlier L=53 Schrage’s schedule Opt*(s) h(J) h(J)=49

  16. Carlier <Improvement> • Move critical job C job 1 ∵ The tail of job 1 is only 7 ∴ put job 1 after the critical sequence so as to reduce makespan 1-2-3-4  3-2-4-1 ∴ makespan = 50  it’s still not an optimal solution • For optimum  change Head and Tail  Artificial Head and Tail

  17. Carlier S C Before J C After J Next page !

  18. Carlier

  19. Carlier

  20. Carlier S C Before J C After J C Before J C After J

  21. Hw. Use Carlier to find an solution.

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