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p1.pptx
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 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)
4. 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)
5. 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
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 64 124 264
144
M1
M2
0 4
4 24
30 40 94 144
94 274
154
M1
M2
0 4
4
54 90 105 149
24 54 114 149 269
M1
M2
0 4 84 149
4 24 84 204 264
90 105
6. 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)
7. 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)
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, 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 – H4
Heuristic 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
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 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. 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 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. 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