Computation Section
Subunit Inventory Analysis
 Systems - Push Network




A worksheet with the push tree structure shown is below. The example has the same structure as the figure at the left. Our illustration is a doctor's office that performs a specialized procedure. Patients enter the process at operation 1 at a rate of 3 per hour. After a registration procedure that takes on average 0.3 hours, half the patients to to operation 2 where they see a 0.5 hour movie that explains the procedure. Three patients see the movie together. After the movie, the patients go to the payment area where a secretary schedules the actual procedure for a later date.

The other half of the patients have already seen the movie, so they go from the entering operation to the treatment operation at 3. This operation treats one patient at a time and the treatment requires 0.5 hours. In 90% of the cases, the treatment is successful. If the treatment is successful, the patient goes to the payment area. If it is not successful, the patient goes to a diagnosis room where the doctor tries to identify the problem. Again this process takes 0.5 hours.

Only 60% of the patients passing through diagnosis return to the treatment operation. The others leave the system, perhaps to return on another day. The diagnosis office also receives one patient an hour that does not pass through the other steps. Patients thus receiving diagnosis are returning from a prior visit or are referred by other doctors.

In the model shown below, we treat the entering, diagnosis and pay operations as queues to reflect the random nature of arrivals to these operations. The movie is a batch operation with a lot size of 3. All other lot sizes are 1 throughout the system. Since there is no variability in the treatment operation, we treat this as a process. We do not place a value on the products, in this case patients, nor do we deal with setup costs or setup times. It is the nature of the problem that we are most interested in the number of patients in the system and the time they spend in the office.

Discussion about the network is below the worksheet illustration.


Starting in row 21, we enter data that describes the external flows and network structure. In row 23 we see the flow of 3 per hour entering the system at Enter and one per hour entering Diagnose. The flows through the operations are computed using matrix algebra and are shown in row 24. These flows are transferred to row 4 for the analysis.

The network structure is defined in the Transfer Out matrix, P. A nonzero entry (i,j) indicates the proportion of the flow passing through operation i that is passed to operation j. The first row shows the division of the entry flow to the movie and treatment operations. All the movie participants go directly to the payment operation. 90% of those passing through treatment go to payment, while the remaining 10% go to the diagnosis operation. 60% of the diagnosis patients return to treatment while the others leave the system. Note that the entries in a row of P need not sum to 1. In a case such as this where the entries represent mutually exclusive alternatives, proportions that do not sum to 1 imply that the missing patients leave the system.

The Augmented Matrix is constructed by the add-in. Its inverse is used in the computation of the flows.

For the network structure the input and output lot sizes are not automatically related. Although it is still logically correct that the input lot size of an operation should equal the output lot size of its immediate predecessors, the network option allows an operation to have any number of successors and predecessors. Since the lot sizes would be difficult to automatically relate, we leave the coordination of lot sizes to the modeler.

Although it is easiest to think of the entries of the transfer out matrix as probabilities or proportions, the entries of the matrix can have any non-negative entries as long as the augmented matrix is not singular.

Reviewing the results for the example, we see that the total WIP is 9.16 patients. The average time a patient spends in the office is 2.29 hours. A culprit in this system is the movie. Although the movie only lasts 0.5 hours, the average time spent in this operation is almost three hours. The problem is the time it takes to gather the 3 patients for the movie and then to split them up for final processing. Since patients arrive for the movie every 45 minutes, it would be much better to use a batch size of 1.

It is interesting to vary other parameters of the system. For example, what is the effect of changing the maximum utilization of the queues from 80% to 90%? This reduces the need for two employees, but the patient time in the office more than doubles.

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tree roots

Operations Management / Industrial Engineering
by Paul A. Jensen
Copyright 2004 - All rights reserved