By J. J. Verzijl

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Period 4 3 14 22 4 1 2 3 1 job 1 job 2 job 3 job 4 job 5 8 Departrnent order 2 job 1 job 2 4 3 2 Departrnent order 3 job 1 job 2 job 3 6 4 12 1 1 2 4 Departrnent order 4 job 1 job 2 job 3 job 4 12 2 6 4 2 I 5 Departrnent order 5 job 1 job 2 5 2 2 Departrnent order 6 job 1 18 3 plan unit = 1 h, planned period = 8 h The identifieation of the department orders appears on line 1. The number of planned periods for eaeh department order are added. The theoretieal minimum throughput time of departinent order 1 is eight planned periods, for departmnet order 4 five planned periods, ete.

If the FAN column contains no number 3, then the departmen torder can start independently of the rest of the orders. In this example, however, there is a department order with a FAN 3, so we must proceed to the next one. The subsequent one is a SAN 6 and in column 3 there is also a FAN equal to 6. Once again we proceed to the next Une, and again we find a SAN number which can be found in the next column. Proceeding in this fashion we eventually find that on line 6 there is a SAN 0, while in column 3 no FAN 0 can be found.

We call these time units, 'pre-allocated time', and we schedule them by first unloading the capacity resource involved before any job can start unloading. 3 Sales performance and proportioning In utilising its planned machines the factory must of course be flexible. However, there are limits beyond which flexibility becomes unreasonable. In particular, a discrepancy can exist between the performance of the commercial department and the flexibility of the factory, such that under-utilisation cannot be avoided.

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