Job-shop Scheduling n jobs m machines No recirculation – Jobs do not revisit the same machine (i, j) is referred to as an operation in which job j is processed.

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Presentation transcript:

Job-shop Scheduling n jobs m machines No recirculation – Jobs do not revisit the same machine (i, j) is referred to as an operation in which job j is processed on machine i Processing time is p ij Objective is to minimize Cmax – makespan – max completion time Jobs have a machine sequence Find the sequence for each machine 1

Training matrix analogous to Job shop Scheduling There are n trainees There are m departments Each trainee has a pre-determined sequence based on their qualification Each trainee spends a different number of weeks in each department based on their final placements. Find a schedule that minimizes the completion time of all required training for this new batch of trainees. 2

Hospital sequencing Medical depts have equipment, doctors which are like the machines Jobs are patients Each patient has a pre-determined sequence for testing and consultation based on their ailment Each patient spends a different amount of time in each medical department (with equipment, doctors) based on their ailment. Find a schedule that minimizes the completion time of all patient diagnostics. 3

Job-shop Scheduling Jobsm/c seqpij p11=10, p21=8, p31= p22 = 8, p12=3, p42=5, p32= p13=4, p23=7, p43=3 4

Job-shop Scheduling Conjunctive (solid arcs) and disjunctive (dotted arcs) Conjunctive – machine sequence for a job Disjunctive – job sequence for a machine 1 sink because the completion time of the last job is important 5 1,12,13,1 U2,21,24,23,2V 1,32,34,3 Sink Source P23 =

Job-shop Scheduling IP solution Let y ij be starting time of job j on machine i Minimize Cmax s.t. y kj – y ij >= p ij for all (i,j) (k,j) Cmax - y ij >= p ij for all (i,j) y ij -y il >= p il or y il –y ij >= p ij for all (i,l) and (i,j) i = 1……m y ij >= 0 Solve using branch and bound – Computationally prohibitive for large n and m 6

Shifting Bottleneck Heuristic Very efficient heuristic for n job m machine job shop with jobs having a pre-determined sequence Has been proven to be very close to optimal, which has been verified numerous times with the branch and bound optimal search. Proven to be very fast compared to B&B 7

Shifting Bottleneck Heuristic – completion time M is the set of all machines Mo is the set of machines for which the sequence has been determined An iteration results in selecting a machine from M-Mo for inclusion in Mo. – Each machine in M-Mo is considered as a single machine problem with release and due dates for which the maximum lateness is to be minimized (Lmax) – Then the machine with the largest Lmax is chosen and is termed as a bottleneck. This is included in Mo – Update Cmax = Cmax + Lmax – Re-sequence all machines in Mo-the last machine added. – Continue until M-Mo is a null set. Release date of job j on machine i is the longest path from source to node (i,j) Due date of job j on machine i is the longest path from node (i,j) to sink – p ij and the resultant is subtracted from Cmax of set Mo 8

Shifting Bottleneck Heuristic Gantt Chart 9 M J1 J2 J M1 M2 M3 M4 J See handout for solution

Composite Dispatching rules ATC Apparent tardiness cost ATCS Apparent tardiness cost with set up See page

Shifting Bottleneck Heuristic – weighted tardiness M is the set of all machines Mo is the set of machines for which the sequence has been determined An iteration results in selecting a machine from M-Mo for inclusion in Mo. – Each machine in M-Mo is considered as a single machine problem with release and due dates for which a priority index is calculated I ij (t) – I ij (t) is computed using the ATC rule (Apparent Tardiness Cost) – Sequence jobs on the machine with the highest to lowest I ij (t) – Calculate weighted tardiness – Then the machine with the largest weighted tardiness is chosen and is termed as a bottleneck. This is included in Mo – Re-sequence all machines in Mo-the last machine added. – Continue until M-Mo is a null set. Release date of job j on machine i is the longest path from source to node (i,j) Due date of job j on machine i is the longest path from node (i,j) to sink – p ij and the resultant is subtracted from Cmax of set Mo – If a path does not exist then make it infinity. 11

Jobswj rj dj m/c seqpij p11=5, p21=10, p31= p32 = 4, p12=5, p22= p33=5, p23=3, p13=7 12 Shifting Bottleneck Heuristic – weighted tardiness

3 sinks because the completion time of all jobs is important 13 1,12,13,1 U3,21,22,2V 3,32,31,3 Sink Source P23 = V V Sink See handout for solution

14 Shifting Bottleneck Heuristic – weighted tardiness

Software for Job shop scheduling LEKIN Summary – Single machine scheduling With or without setup time – Parallel machines (machines can process all jobs) – Flow shop (All jobs have to flow first on one machine and then on another) – Job-shop Minimize Completion time Minimize Weighted Tardiness – Flexible flow shop Methods – Dispatching rules, Tabu, Simulated Annealing, Shifting bottleneck heuristics for completion time and weighted tardiness objective. 15