Abstract
In the flowshop scheduling problem, n jobs are to be processed on m machines. The order of the machines is fixed. We assume that a machine processes one job at a time and a job is processed on one machine at a time without preemption. Let t p ( i , j ) denote the processing time of job j on machine i and t c ( i , j ) denote the completion time of job j on machine i . Let Jj denote the j -th job and M i the i -th machine. The completion times of the jobs are obtained as follows. For i = 1, 2,..., m and j = 1, 2,..., n , t c ( M 1 , J 1 ) = t p ( M 1 , J 1 ); t c ( M i , J 1 ) = t c ( M i− 1 , J 1 ) + t p ( M i , J 1 ); t c ( M 1 , J j ) = t c ( M 1 , J j− 1 ) + t p ( M 1 , J j ); t c ( M i , J j ) = m ax{t c ( M i− 1 , J j ), t c ( M i , J j− 1 )} + t p ( M i , J j ). Makespan is defined as the completion time of the last job, t c ( M m , J n ). The goal is to obtain the n -job sequence that minimizes the makespan. The search space consists of n ! possible job sequences. The problem is NP-hard.
Original language | American English |
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Journal | Applications and Science in Soft Computing |
DOIs | |
State | Published - Jan 1 2004 |
Disciplines
- Algebra
- Mathematics