1. A Study on Two-machine Flowshop Scheduling Problem with an Availability Constraint Team: Tianshu Guo Yixiao Sha Jun Tian

2.  A machine may not always be available in the scheduling period  Stochastic – breakdown  Deterministic – preventive maintenance  Our problem  Deterministic Environment – unavailable time is known in advance  Two-machine Problem – one machine is always available  Resumable Jobs – if a job cannot finish before the unavailable period of a machine then the hob can continue after the machine is available again The General Problem

3.  Apply Johnson’s algorithm on F 2 /r-a(M 1 )/C max problem  Divide the n-hob set into two disjoint subsets, S 1 and S 2, where S 1 = { J i : p i1 p i2 }  Order the jobs in S 1 in the non-decreasing order of p i1 and those jobs in S 2 in the non-increasing order of p i2  Sequence jobs in S 1 first, followed by S 2  (C H1 – C*)/C* ≤ 1, and the bound is tight. A.C. on M1 – Johnson’s Rule (H1)

4.  1. Use Johnson’s Rule to schedule the jobs and compute the corresponding makespan MK1  2. Schedule jobs in non-increasing order of p i2 /p i1 and find the corresponding makespan MK2  Let t be the earliest time that M2 starts to be busy until MK2  Let J k be the job starts at t on M2  3. Same order as in 2 but make J k the first job in the sequence MK3  Let C H2 = min {MK1, MK2, MK3 }  (C H2 – C*)/C* ≤ 1/2 A.C. on M1 – An Improved Heuristic (H2)

5. JobM1M2M2/M1 A4205 B50601.2 C801201.5 An Example of H2 1.A -> B -> C2. A -> C -> B M1 M2 M1 M2 0 4 54 134 4 24 54 114 134 254 0 4 84 134 4 24 84 204 264  No availability constraint imposed  M1 not available from 30 to 40  M1 not available from 90 to 105 M1 M2 0 4 4 30 40 64 144 24 64 124264144 M1 M2 0 4 4 24 30 40 94 144 94 274 154 M1 M2 0 4 4 54 90 105149 24 54 114 149 269 M1 M2 0 4 84 149 4 24 84 204 264 90 105

6.  Apply Johnson’s algorithm on F 2 /r-a(M 2 )/C max problem  (C H3 – C*)/C* ≤ 1/2  Consider an instance with n jobs, p 11 = p 21 = p n-1,1 = 1, p 12 = p 22 = p n-1,2 = 1, and p n,1 = n, p k+1,2 = 1. Also s 2 = n, t 2 = 2n.  Apply H3 to this instance we may get a sequence J n, J 1, J 2,…,J n-1 with C H3 = 3n, while the optimal solution is J 1, J 2, …,J n-1,J n, with C* = 2n+1.  (C H3 – C*)/C* approaches ½ as n -> ∞ A.C. on M2 – Johnson’s Rule (H3)

7.  1. Use Johnson’s Rule to schedule the jobs and compute the corresponding makespan MK1  2. Schedule jobs in non-increasing order of p i2 /p i1 and find the corresponding makespan MK2  Let C H4 = min {MK1, MK2}, then (C H2 – C*)/C* ≤ 1/3  Consider an instance with n jobs, p 11 = p 21 = p n-1,1 = 1, p 12 = p 22 = p n-1,2 = 1, and p n,1 = n, p n,2 = n. Also s 2 = n, t 2 = 2n.  Apply H4 to this instance we may get a sequence J n, J 1, J 2,…,J n-1 with C H4 = 4n-1, while the optimal solution is J 1, J 2, …,J n-1,J n, with C* = 3n.  (C H3 – C*)/C* approaches ½ as n -> ∞ A.C. on M2 –Improved Heuristic (H4)

8.  In H2, we are unable to show the 1/2 bound is tight, but the following instance shows the bound cannot be better than 1/3  Consider an instance with n jobs, p 11 = p 12 = 1, p 21 = p 22 = p 31 = p 32 = k, and s 1 = 2k, t 1 = 3k.  Steps 1 and 2 yields J 1 -J 2 -J 3, with C = 3k+1+k = 4K +1. Then let J 3 be J k and apply step 3. The result: J 3 -J 1 -J 2, with C = 3k+1+k = 4k+1.  The optimal solution is J 2 -J 3 -J 1, with C* = 3k+1+1.  (C H2 – C*)/C* approaches 1/3 as n -> ∞ Summary of H1 – H4 Heuristicr-a()MakespanError Bound H1M1MK11 H2M1min {MK1, MK2, MK3}1/2 H3M2MK11/2 H4M2min {MK1, MK2}1/3

9. HI—improved version of H2 Reduce error bound to 1/3 C*

10. Where It Has Been Improved  Put 2 jobs with longest processing time on M2 in front instead of 1 job as new σ3.  With p k ≤ s 1, new scheme σ4 is derived from σ2 with J k moved to the slot right before s 1.  With p k ≥ s 1, new scheme σ5 combines this fact with Johnson’s Rule and σ2.

11. New σ3  Maximum 2 jobs that have long processing time on M2 in optimal schedule.  Thus we can possibly put these 2 jobs in the front of sequence.  This may result in improved error bound.

12. New σ4  Proved a hidden fact that there is an optional schedule with J k finishes before s 1, not otherwise.  Proven fact that the q k determines the error bound of σ2 algorithm.  Adjust position of J k is an option.

13. New σ5  S 1 ={ jobs with longer processing time on M2 but less than p k }  S 2 ={ jobs except for J k and S 1 }  Proved that there is an optimal solution with all jobs in S 1 are scheduled before J k followed by S 2.  Since J k finishes after t 1 on M1, use Johnson’s.  Order: σ2, J k, Johnson’s Rule