# presentation 1

```A Study on Two-machine Flowshop
Scheduling Problem with an
Availability Constraint
Team: Tianshu Guo
Yixiao Sha
Jun Tian
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
 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
A.C. on M1 – Johnson’s Rule (H1)
 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 – An Improved Heuristic
(H2)
 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
An Example of H2
Job
M1
M2
M2/M1
A
4
20
5
B
50
60
1.2
C
80
120
1.5
1. A -> B -> C
2. A -> C -> B
 No availability constraint imposed
M1
0 4
M2
4 24
54
134
54
114 134
M1
254
M2
04
4 24
 M1 not available from 30 to 40
M1
0 4 30 40 64
M2
4 24
64
M1
144
124 144
264
M2
134
84
204
84
0 4 30 40
94
4 24
94
144
154
 M1 not available from 90 to 105
M1
0 4
M2
4 24
54 90 105
54
M1
149
114 149
269
M2
04
4 24
84 90 105
84
264
274
149
204
264
A.C. on M2 – 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 -> ∞
A.C. on M2 –Improved Heuristic (H4)
 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 -> ∞
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
 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 -> ∞
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
```