Ren Qing-dao-er-ji

School of Computer Science and Technology

School of Science

Xidian University

Yuping Wang, Xiaojing Si

School of Computer Science and Technology

Xidian University

2010 International Conference on Computational Intelligence and Security

Each job has m operations that must be processed at m machines.

The operations of a given job have to processed in a given order.

The objective is to determine the schedule which minimizes the makespan -The time required to complete all the jobs.

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Job Operation routing (processing time )

1 1(3) 2(3) 3(3)

2 1(2) 3(3) 2(4)

3 2(3) 1(2) 3(1)

Genetic algorithms have been tried to solve the job-shop scheduling.

However, the simple genetic algorithm is with a slow convergent speed and is easy to convergeprematurely.

But, the crossover and mutation operators ◦ not sufficiently made use of the characteristics of the

problem structure.

Hence, in this paper, To sufficiently use the information of the problem structure, a new crossover and mutation operators based on the characteristics of the job shop problem were designed

The proposed genetic operators are explained using disjunctive graph theory model

Given an instance of JSSP, it is associate with a disjunctive graph G = (V, A, E)

with V being the set of nodes (operations )

A the set of conjunctive directed arcs

E the set of disjunctive undirected arcs (edges)

V = {0,1,.., N, N +1} , where {0} and {N +1} are special nodes which identify the start and completion of the overall jobs

A = {(i, j) : operation i is an immediate predecessor of operation j in the chain of job }

E = { (i, j): operation i and operation j are processed on the same machine , i, j ∈V }.

For each vertex i∈V , a weight di is associated, and di is the duration of the operation i .

d is 0 for node 0 and N+1

If length of a path is defined as the sum of the weights of the vertices in the path, solving the job shop scheduling problem corresponds to finding an acyclic orientation of G so that the length of the longest path between 0 and N +1 (critical path) is minimized.

In this representation, the chromosome consists of n*m genes.

i.e each job will appear m times exactly.

E.x (3-job and 3 machine problem ) a chromosome is given as [2 1 3 1 2 1 2 3 3]. ◦ So, 1 represents the job 1, 2 represents the job 2 and 3

represents the job 3.

◦ Because each job consists of three operations, it occurs exactly three times in the chromosome.

[2 1 3 1 2 1 2 3 3]

The fitness function is the function of the objectives function and defined as

And the selection probability is

It is driven be the belief that the good gene characteristics preservation and the feasibility are the most important criteria to design crossover operation in JSSP.

In this paper, a new crossover operator based on the characteristic of the JSSP itself was designed. The offspring generated can keep the good characteristics of the problem structure and satisfy the feasibility.

Suppose , there are two parents: parent 1 and parent 2

Parent 1.

Parent 2.

Divide the machine numbers into two complementary sets, such as {1, 3} and {2}.

Combine the operation orders of machines {1, 3} in the parent 1 and the operation orders of machine {2} in the parent 2 to form child 1.

Similarly, Combine the operation orders of machine {2} in the parent 1 and the operation orders of machine {1, 3} in the parent 2 to form child 2.

Child 1.

Child2.

Given an individual chromosome, mutation generates the child by the following procedure:

Step 1. Calculate/specify the critical path of this individual.

Step 2. Permuting two successive operations v and w assigned to the same machine with probability of pm and for which the arc (v, w) is on a critical path in that individual.

For example: the graph of the parent 1is and thecritical path of the parent 1 is 0-1-8-9-10.

Then we know that the operations 1 and 8 areassigned to the same machine 1.

Permuting two successive operations 1 and 8assigned to the same machine with probability ofpm and get the child 1 as shown below.

Experimental results ◦ Population size 100

◦ Cross over probability 0.7

◦ Mutation probability 0.1

◦ 10 independent runs for each test

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