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