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A Stu dy o n T wo-machine F lowshop Scheduling P roblem with an A vailability C onstraint

A Stu dy o n T wo-machine F lowshop Scheduling P roblem with an A vailability C onstraint. Te am: Tia nshu G uo Yix iao S ha Ju n T ian. T he G eneral P roblem. A machine may not always be available in the scheduling period Stochastic – breakdown

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A Stu dy o n T wo-machine F lowshop Scheduling P roblem with an A vailability C onstraint

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  1. AStudy on Two-machine FlowshopScheduling Problem with an Availability Constraint Team: TianshuGuoYixiaoShaJun Tian

  2. The General Problem • 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

  3. A.C. on M1 – Johnson’s Rule (H1) Apply Johnson’s algorithm on F2/r-a(M1)/Cmaxproblem Divide the n-hob set into two disjoint subsets, S1 and S2, where S1 = { Ji : pi1 <= pi2} and S2= { Ji : pi1 > pi2} Order the jobs in S1 in the non-decreasing order of pi1 and those jobs in S2 in the non-increasing order of pi2 Sequence jobs in S1 first, followed by S2 (CH1 – C*)/C* ≤ 1, and the bound is tight.

  4. A.C. on M1 – An Improved Heuristic (H2) • 1. Use Johnson’s Rule to schedule the jobs and compute the corresponding makespanMK1 • 2. Schedule jobs in non-increasing order of pi2/pi1and find the corresponding makespanMK2 • Let t be the earliest time that M2 starts to be busy until MK2 • Let Jk be the job starts at t on M2 • 3. Same order as in 2 but make Jk the first job in the sequence MK3 • Let CH2= min {MK1, MK2, MK3 } • (CH2 – C*)/C* ≤ 1/2

  5. An Example of H2 A -> B -> C 2. A -> C -> B • No availability constraint imposed • M1 not available from 30 to 40 • M1 not available from 90 to 105 M1 M1 134 4 0 134 84 4 54 0 M2 M2 4 24 254 4 54 134 204 114 84 264 24 M1 M1 94 144 4 40 0 30 30 144 40 4 0 64 M2 M2 4 24 4 94 24 154 64 124 144 264 274 M1 M1 90 4 149 0 84 105 4 0 90 54 105 149 M2 M2 4 269 24 24 114 4 54 149 204 84 264

  6. A.C. onM2– Johnson’s Rule (H3) Apply Johnson’s algorithm on F2/r-a(M2)/Cmax problem (CH3 – C*)/C* ≤ 1/2 Consider an instance with n jobs, p11 = p21 = pn-1,1 = 1, p12= p22= pn-1,2= 1, and pn,1= n, pk+1,2= 1. Also s2 = n, t2 = 2n. Apply H3 to this instance we may get a sequence Jn, J1, J2,…,Jn-1 with CH3 = 3n, while the optimal solution is J1, J2, …,Jn-1 ,Jn, with C* = 2n+1. (CH3 – C*)/C* approaches ½ as n -> ∞

  7. A.C. onM2 –Improved Heuristic (H4) 1. Use Johnson’s Rule to schedule the jobs and compute the corresponding makespanMK1 2. Schedule jobs in non-increasing order of pi2/pi1and find the corresponding makespanMK2 Let CH4= min {MK1, MK2}, then (CH2– C*)/C* ≤ 1/3 Consider an instance with n jobs, p11 = p21 = pn-1,1 = 1, p12 = p22 = pn-1,2 = 1, and pn,1 = n, pn,2= n. Also s2 = n, t2 = 2n. Apply H4 to this instance we may get a sequence Jn, J1, J2,…,Jn-1 with CH4= 4n-1, while the optimal solution is J1, J2, …,Jn-1 ,Jn, with C* = 3n. (CH3 – C*)/C* approaches ½ as n -> ∞

  8. Summary of H1 – H4 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, p11 = p12= 1, p21= p22= p31= p32= k, and s1= 2k, t1= 3k. Steps 1 and 2 yields J1-J2-J3, with C = 3k+1+k = 4K +1. Then let J3be Jkand apply step 3 . The result: J3-J1-J2, with C = 3k+1+k = 4k+1. The optimal solution is J2-J3-J1, with C* = 3k+1+1. (CH2– C*)/C* approaches 1/3 as n -> ∞

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

  10. WhereIt Has Been Improved Put 2 jobs with longest processing time on M2 in front instead of 1 job as new σ3. With pk≤ s1, new scheme σ4 is derived from σ2 with Jk moved to the slot right before s1. With pk≥s1, new scheme σ5 combines this fact with Johnson’s Rule and σ2.

  11. Newσ3 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. Newσ4 Proved a hidden fact that there is an optional schedule with Jk finishes before s1, not otherwise. Proven fact that the qk determines the error bound of σ2 algorithm. Adjust position of Jk is an option.

  13. Newσ5 S1={ jobs with longer processing time on M2 but less than pk } S2={ jobs except for Jk and S1 } Proved that there is an optimal solution with all jobs in S1 are scheduled before Jk followed by S2. Since Jk finishes after t1 on M1, use Johnson’s. Order: σ2, Jk, Johnson’s Rule

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