Flow shop Scheduling Problems with Transportation and Capacities Constraints Oulamara, A.; Soukhal,...
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![Page 1: Flow shop Scheduling Problems with Transportation and Capacities Constraints Oulamara, A.; Soukhal, A. 2001 IEEE SMC Conference Speaker: Chan-Lon Wang.](https://reader035.fdocuments.net/reader035/viewer/2022062714/56649d385503460f94a111c6/html5/thumbnails/1.jpg)
Flow shop Scheduling Problems with Transportation and Capacities Constraints
Oulamara, A.; Soukhal, A.
2001 IEEE SMC Conference
Speaker: Chan-Lon Wang
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Scheduling Classification
1. Open-shop scheduling
Each job visit each machine again Different jobs have different routes Processing time of job maybe is zero
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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
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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
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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
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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
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(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
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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
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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
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Problem Description-2
• The classical problem:– Unlimited intermediate buffer capacity– Infinite speed vehicles
• Constraints:– Transportation capacity– Transportation time
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Problem Description-3
• F2D|v=1, c=1, no wait |Cmax F3D| no wait |Cmax
• If the truck consider as a machine.
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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
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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
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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.
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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
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L3: Unlimited buffer space + Johnson’s order
•Unlimited buffer space regard as no buffer between machine.
•Johnson’s order is optimal.
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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
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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
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The Johnson Final Scheduling
The final completion time = 21
P1=17
P2=21
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Problem with unlimited buffer
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No buffer area between machines
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Conclusions
• With unlimited buffer between machines the problem is NP-hard
• With no buffer between machines the problem is NP-hard
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