Automated Scheduling, School of Computer Science and IT, University of Nottingham 1. Job Shop Scheduling. Disjunctive Graph.


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1 Job hop cheduling Contents 1. Problem tatement 2. Disjunctive Graph. he hifting Bottleneck Heuristic and the Makespan Literature: 1. cheduling, heory, Algorithms, and ystems, Michael Pinedo, Prentice Hall, 199, or new: econd Addition, 22 Chapter or 2. Operations cheduling with Applications in Manufacturing and ervices, Michael Pinedo and Xiuli Chao, McGraw Hill, 2 Chapter 1 2 Problem tatement Disjunctive Graph Job shop environment: Example of a job shop problem: machines and jobs m machines, n jobs objective function each job follows a predetermined route routes are not necessarily the same for each job machine can be visited once or more than once (recirculation) NP hard problems,2,1,2 1, 2,, Jm C max How to construct a feasible schedule? (i, j) processing of job j on machine i processing time of job j on machine i G = (N, A B) (i, j) N all the operations that must be performed on the n jobs A conjunctive (solid) arcs represent the precedence relationships between the processing operations of a single job B disjunctive (broken) arcs connect two operations which belong to two different jobs, that are to be processed on the same machine, they go in opposite directions Disjunctive arcs form a clique for each machine. Clique is a maximal subgraph in which all pairs of nodes are connected with each other. elect D  a subset of disjunctive arcs (one from each pair) such that the resulting directed graph G(D) has no cycles. Graph G(D) contains conjunctive arcs + D. D represents a feasible schedule. A cycle in the graph corresponds to a schedule that is infeasible. Example invalid h,j Operations in the same clique have to be done on the same machine. i,k h,k i,j machine i machine h j k k j Automated cheduling, chool of Computer cience and I, University of Nottingham 1
2 1,1,2,2 1, 2,, he hifting Bottleneck Heuristic and the Makespan Idea: A classic idea in nonlinear programming is to hold all but one variable fixed and then optimise over that variable. hen hold all but a different one fixed, and so on. Furthermore, if we can do the onevariable optimisation in order of decreasing importance, there is better hope that the local optimum so found will be the global one, or close to it. he makespan of a feasible schedule is determined by the longest path in G(D) from to. In job shop problems fixing the value of variable means fixing the sequence in which jobs are to be processed on a given machine. Minimise makespan: find a selection of disjunctive arcs that minimises the length of the longest path (the critical path). Iteration M set of m machines M M machines for which sequence of jobs has already been determined in previous iterations Analysis of machines still to be scheduled i {MM } Define a singlemachine problem 1 r j L max for machine i i,j Jobs... j... r ij d ij longest path from to (i,j) C max (M ) longest path from (i,j) to + pij 9 Bottleneck selection A machine k with the largest maximum lateness is a bottleneck. L max ( k) = max ( Lmax ( i)) i { M M } 1. chedule machine k according to the sequence which minimises the corresponding L max (singlemachine problem). 2. Insert all the corresponding disjunctive arcs in the graph.. Insert machine k in M. C max (M k) C max (M ) + L max (k) 1 Resequencing of all machines scheduled earlier Aim: to reduce the makespan Do for each machine l {M  k} delete the disjunctive arcs associated with the machine l formulate a single machine problem for the machine l and find the sequence that minimises L max (l) Insert the corresponding disjunctive arcs. 11 hifting Bottleneck Algorithm tep 1. et the initial conditions M = set of scheduled machines. Graph G is the graph with all the conjunctive arcs and no disjunctive arcs. et C max ( ) equal to the longest path in graph G. tep 2. Analysis of the machines still to be scheduled olve the simple problem for each machine still to be scheduled: formulate a single machine problem with all operations subject to release dates and due dates. tep. Bottleneck selection he machine with the highest cost is designated the bottleneck. Insert all the corresponding disjunctive arcs in graph G. Insert machine which is the bottleneck in M. 12 Automated cheduling, chool of Computer cience and I, University of Nottingham 2
3 tep. Resequencing of all the machines scheduled earlier Find the sequence that minimised the cost and insert the corresponding disjunctive arcs in graph G. tep. topping condition If all the machines are scheduled (M = M) then OP else go to tep 2. Example. Jobs Machine Procesing imes equence 1 1, 2, p 11 =1, p 21 =, p 1 = 2 2, 1,, p 22 =, p 12 =, p 2 =, p 2 = 1, 2, p 1 =, p 2 =, p = 1,1,2,2 1, 2,, 1 1 Iteration 1. M = C max ( )=22 p1j 1 r1j d1j , L max (1)= 1,1,2,2 1, 2,, L max (1, 2, ) = max {, 2, } = L max (1,, 2) = max {, 2, } = L max (2, 1, ) = max {, 11,...} >11 L max (2,, 1) = max {,, 1} = 1 L max (, 1, 2) = max {,, } = L max (, 2, 1) = max {,, 1} = 1 1 1,1,2,2 p2j r2j 1 d2j ,,1 L max (2)= 1, 2,, L max (1, 2, ) = max {, 1, 1} = 1 L max (1,, 2) = max {,, 2} = 2 L max (2, 1, ) = max {,, } = L max (2,, 1) = max {, , } = L max (, 1, 2) = max {, 1, 19} = 19 L max (, 2, 1) = max {,, } = 1 1,1 1,1,2,2,2,2 1, 2,, 1, 2,, Machine pj rj 1 1 dj L max (1, 2) = max {, } = L max (2, 1) = max {, } = Machine Jobs 2 pj rj dj 1 22 L max (2, ) = max {, } = L max (, 2) = max {, } = L max ()= L max ()= 1 1 Automated cheduling, chool of Computer cience and I, University of Nottingham
4 L max (2)= 2,,1 L max (1)=, L max ()= L max ()= Machines 1 and 2 are bottlenecks. 1,1 1,2,2 1, 2,, is chosen as a bottleneck! C max ({1}) = C max ( ) + L max (1) = 22 + = Iteration 2. M = {1}, C max ({1})=2 1,1 1 Machine pj rj 1 1 dj 2 2 or L max ()=1,2,2 1, 2,, Machine L max ()= Machines 2 and are bottlenecks. p 2j r 2j 1 1 d 2j 2 1 2, L max (2)=1 is chosen as a bottleneck! ,1 1,2,2 1, 2,, C max ({}) = C max ({1}) + L max (2) = = 2 Can we decrease C max ({})? Will resequencing machine 1 give any improvement? 1,1 1,2,2 1, p1j 1 r1j d1j original sequence:, 2,, C max ({}) = 2 L max (1, 2, ) = max {, , 1} = L max (1,, 2) = max {, , } = L max (2, 1, ) = max {, 11,} =11 L max (2,, 1) = max {, , 1} = 1 L max (, 1, 2) = max {1,, } = L max (, 2, 1) = max {1, 2, 1} = 1 2 Resequencing machine 1 does not give any improvement. 2 Automated cheduling, chool of Computer cience and I, University of Nottingham
5 Iteration. M = {} C max ({})=2 Machine pj rj 1 1 dj 2 2 or L max () = 1,1 1,2,2 1, 2,, Machine L max () = 1 2 No machine is a bottleneck! Machine Machine t Automated cheduling, chool of Computer cience and I, University of Nottingham
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