Flow shop Scheduling Problems with Transportation and Capacities Constraints Oulamara, A.; Soukhal, A 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: J 1, J 2, J 3, J 4 Machine: M 1, M 2, M 3 M1M1 M2M2 M3M3 J1J1 OUT M1M1 M2M2 M3M3 J2J2 M3M3 M2M2 M1M1 J3J3 M2M2 M1M1 M3M3 J4J4

Job-shop scheduling  Each job visit each machine at most once  Each job has own routing M1M1 M2M2 M3M3 J1J1 OUT M1M1 M2M2 M3M3 J2J2 M3M3 M2M2 M1M1 J3J3 M2M2 M1M1 M3M3 J4J4

Flow-shop scheduling  Jobs are not preemptive  Each job has m tasks with processing time  Each job follow a same route M1M1 M2M2 M3M3 J1J1 OUT M1M1 M2M2 M3M3 J2J2 M1M1 M2M2 M3M3 J3J3 M1M1 M2M2 M3M3 J4J4

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 painting P2 degreasing P1 Each machine center has one machine Ex: A car painting factory Center 1 Center 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 (5, 3) (4, 4) Car-1 Car-2 painting P2 degreasing P1

Problem Description-1 Graham et al.:  |  |  F2  D|v=1, c=2|C max F2:two machines, D:truck, v: one truck, c=capacity of truck, C max =min makespan

Problem Description-2 The classical problem: –Unlimited intermediate buffer capacity –Infinite speed vehicles Constraints: –Transportation capacity –Transportation time

Problem Description-3 F2  D|v=1, c=1, no wait |C max  F3  D| no wait |C max If the truck consider as a machine.

Flow-shop with capacity of truck limited to parts Flow-shop with unlimited buffer space –F2  D|v=1, c=2|C max is NP-hard Flow-shop with limited buffer space at the output system –F2  D|v=1, c=2, blacking(1, 2)|C max is NP-hard

Resolution methods Four greedy algorithms for solving –F2  D|v=1, c=2|C max and –F2  D|v=1, c=2, blacking(1, 2)|C max 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 Car-1Car-2Car-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, car Car-1Car-2Car-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

The sequence {U, V} = {} ={Car-2, car-4, car-5, car-3, car-1} U= {t 1i ascending by Sum V={t 1j >=t 2j }= {Car-1, Car-3} =>descending by Sum Makespan=21 Ex: A car painting factory Car-1Car-2Car-3 Car-4 Car Sum

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