Flow shop Scheduling Problems with Transportation and Capacities Constraints

Oulamara, A.; Soukhal, A.

2001 IEEE SMC Conference

Speaker: Chan-Lon Wang

Scheduling Classification

1. Open-shop scheduling

Each job visit each machine again Different jobs have different routes Processing time of job maybe is zero

Open-shop scheduling• Example: Job: J1, J2, J3, J4

Machine: M1, M2, M3

M1 M2 M3J1 OUT

M1 M2 M3J2 OUT

M3 M2 M1J3 OUT

M2 M1 M3J4 OUT

Job-shop schedulingEach job visit each machine at most onceEach job has own routing

M1 M2 M3J1 OUT

M1 M2 M3J2 OUT

M3 M2 M1J3 OUT

M2 M1 M3J4 OUT

Flow-shop schedulingJobs are not preemptiveEach job has m tasks with processing timeEach job follow a same route

M1 M2 M3J1 OUT

M1 M2 M3J2 OUT

M1 M2 M3J3 OUT

M1 M2 M3J4 OUT

Scheduling Classification

Scheduling

Open shop

Job shop

Flow shop

Simple open shop scheduling

Flexible open shop scheduling

Simple Job shop scheduling

Flexible Job shop scheduling

Simple Flow shop scheduling

Flexible Flow shop scheduling

(5, 3)

(4, 4)

Car-1

Car-2 paintingP2

degreasingP1

Each machine center has one machine Ex: A car painting factory

Center 1

Center 2

5 9

8 13

The final completion time=13

Simple Flow Shop Problem

At least one machine center has more than one machine

Ex: two same machines in each center

Flexible Flow-Shop Problem

The final completion time=8

Center 2

Center 1

88

4

5

(5, 3)

(4, 4)

Car-1

Car-2 paintingP2

degreasingP1

Problem Description-1Graham et al.:

| |

F2D|v=1, c=2|Cmax F2:two machines, D:truck,

v: one truck, c=capacity of truck, Cmax=min makespan

Problem Description-2

• The classical problem:– Unlimited intermediate buffer capacity– Infinite speed vehicles

• Constraints:– Transportation capacity– Transportation time

Problem Description-3

• F2D|v=1, c=1, no wait |Cmax F3D| no wait |Cmax

• If the truck consider as a machine.

Flow-shop with capacity of truck limited to parts

• Flow-shop with unlimited buffer space– F2D|v=1, c=2|Cmax is NP-hard

• Flow-shop with limited buffer space at the output system– F2D|v=1, c=2, blacking(1, 2)|Cmax is NP-hard

Resolution methods

• Four greedy algorithms for solving – F2D|v=1, c=2|Cmax and

– F2D|v=1, c=2, blacking(1, 2)|Cmax

• L1: no wait + no-decreasing job sequence

• L2: no wait + no-increasing job sequence

• L3: unlimited buffer space + Johnson’s order

• L4: no wait + Gilmore & Gomory’s order

L1: no wait + no-decreasing job sequence

4 1 5 2 53 2 4 3 6

Car-1 Car-2 Car-3 Car-4 Car-5

•No wait regard as no buffer between machine.

•No decreasing regard as ascending.

•Ex: Job order: car2, car4, car1, car3, car5. For machine-1.

L2: no wait + no-increasing job sequence

•No wait regard as no buffer between machine.

•No increasing regard as descending.

•Ex: Job order: car3, car5, car1, car4, car2.

4 1 5 2 53 2 4 3 6

Car-1 Car-2 Car-3 Car-4 Car-5

L3: Unlimited buffer space + Johnson’s order

•Unlimited buffer space regard as no buffer between machine.

•Johnson’s order is optimal.

For two machines flow shop

Input: A set of n independent jobs Each with m tasks

Output: A schedule A nearly minimum completion time of the last job

Review of Johnson Algorithm

Review of Johnson Algorithm

The sequence

{U, V} = { }

={Car-2, car-4, car-5, car-3, car-1}

U= {t1i< t2i}= {Car-2, Car-4, Car-5} =>ascending by Sum

V={t1j>=t2j}= {Car-1, Car-3} =>descending by Sum

Makespan=21

Ex: A car painting factoryCar-1 Car-2 Car-3 Car-4 Car-5

4 1 5 2 53 2 4 3 6

Sum 7 3 9 5 11

The Johnson Final Scheduling

The final completion time = 21

P1=17

P2=21

Problem with unlimited buffer

No buffer area between machines

Conclusions

• With unlimited buffer between machines the problem is NP-hard

• With no buffer between machines the problem is NP-hard