Extended gating cyclic service model

In this subsection we shall consider the situation in which we have extended gating service with no possibilities for overtime. During the production interval of a type we can work on any order of this type, but not on orders of another type. If somewhere during the production interval there are no (more) orders of the type, we must wait until the interval is finished or until new orders arrive. Jobs that are not finished by the end of the production interval have to wait until the next production opportunity. In the next production interval the work on these orders will be continued, so no extra work outside the normal working hours will be done. The set-up will be done just before the production interval. We are especially interested in the average delivery times for the different types. Therefore we must make some assumptions about the demand and service time distribution and about the sequence set S, which may enable us to calculate the delivery times. 

For every type of product the orders are supposed to arrive according to a Poisson process with parameter X,-, and the service time for an order of type i is, for 

We want to avoid that the queue length for some of the types becomes infinitely long. Therefore it is necessary that for every type the available capacity pa cycle is large enough to produce the average demand during one cycle. This can be described by the following restriction for the capacity set 

Let L be the average (long run) number of ordos in the queue. For an order arriving in the arbitrary interval [0,7), the delivery time can consist of three parts  

Although these three parts are not independent, we may consider them separately in order to determine the expected delivery time of an arbitrary order arriving in the [0,7)-interval. 

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We will compare this production model with the extended gating cyclic service model that has been described in Subsection 3.2.1. In a production cycle we have a fixed time Ti available for the production of type i including one set-up. Sometimes this time may not be used entirely, but at other times the time will not be enough to produce all orders, leading to orders that have to wait until the next cycle. By means of iteration, we can determine the optimal values for and the... 

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