a study on two-machine flowshop scheduling problem with an availability constraint team: tianshu guo...
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A Study on Two-machine Flowshop Scheduling Problem with an Availability Constraint
Team: Tianshu Guo Yixiao Sha Jun Tian
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
Apply Johnson’s algorithm on F2/r-a(M1)/Cmax problem
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.
A.C. on M1 – Johnson’s Rule (H1)
1. Use Johnson’s Rule to schedule the jobs and compute the corresponding makespan MK1
2. Schedule jobs in non-increasing order of pi2/pi1 and find the corresponding makespan MK2 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
A.C. on M1 – An Improved Heuristic (H2)
Job M1 M2 M2/M1
A 4 20 5
B 50 60 1.2
C 80 120 1.5
An Example of H2
1. A -> B -> C
2. A -> C -> B
M1
M2
M1
M20 4 5
4134
4 24
54
114134 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
M20 4
4
30 40 64 144
24
64
124
264144
M1
M20 4
4 24
30
40
94
144
94
274
154
M1
M20 4
4
54 90 105 149
24
54
114149 269
M1
M20 4 8
4149
4 24
84
204 264
90
105
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 -> ∞
A.C. on M2 – Johnson’s Rule (H3)
1. Use Johnson’s Rule to schedule the jobs and compute the corresponding makespan MK1
2. Schedule jobs in non-increasing order of pi2/pi1 and find the corresponding makespan MK2
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 -> ∞
A.C. on M2 –Improved Heuristic (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 J3 be Jk and 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 -> ∞
Summary of H1 – H4Heuristic
r-a()
Makespan Error Bound
H1 M1 MK1 1
H2 M1 min {MK1, MK2, MK3}
1/2
H3 M2 MK1 1/2
H4 M2 min {MK1, MK2} 1/3
HI—improved version of H2
Reduce error bound to 1/3 C*
Where It 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.
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.
New σ4
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.
New σ5
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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