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Theory of Constraints Handbook - James Cox Iii [140]

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The assembly departments should operate to a priority list that specifies what units should be assembled in what quantities and in what sequence.

The question often asked is, “What about at the CCR or bottleneck?” Is a detailed schedule specifying sequence and quantity of production necessary and should this be carefully monitored and controlled? If the CCR or bottleneck has sequence-dependent setups, then the setup time depends on what is currently on the resource and what product is next. A simple case is an operation that applies color. Going from a light color to a dark color requires minimal cleanup, but going from black to white will require extensive and thorough cleaning. Therefore, it is important to produce to a defined sequence and this list will have to be provided to the CCR. If this is not the case, and the process times are a small fraction of the total production lead time, then sequence even at the CCR is not that critical and no additional step beyond controlling material release is necessary. Figure 8-7 shows the schedule control points where sequence and a time frame for actions are important in controlling the flow through a factory.

Finally, there is the shipping or completion point of the batch. It is the most important schedule control point in that every batch has to meet the date when it is scheduled to be completed. Failure of a batch to meet this date or even the anticipation that a batch might fail to meet this date will trigger corrective actions as described in the section on BM.

Managing Flow with DBR—An Example


We illustrate the DBR system with a simple example10 in this section. The plant represented by the PFD in Fig. 8-8 shows a relatively simple plant with five different types of resources—each pattern in the diagram is a different type of resource (labeled R1, R2, R3, R4, R5). The number in each box in the flow diagram is the time to process one unit at that step. For example, the first step (A-1 on the bottom left of the grid used in Fig. 8-8) is performed by the R2 resource and takes 4 min for each unit. Similarly, the step corresponding to the grid point B-3 is an assembly operation performed by the R5 resource and takes 8 min per unit to assemble. To assemble a unit at B3, a unit must be completed at both A1 and C1. That assembly can then be used at the A-5 operation to make Product A or by C-5 to make Product D. The number of units of a specific resource type is indicated on the left-hand side of Fig. 8-8 and shows that there is only one resource of type R1. We also see that there are two resources of type R2. Similarly, we have two resources of types R3 and R4 and one resource of type R5. The setup time for each resource is indicated next to the resource—R1-type resource has a setup time of 15 min, R2-type resources have a setup time of 120 min, and so on. The number just below a node in the flow centric represents the number of units that are available for the resource (the WIP) at this point in time—there are 15 units available for operation E-5 performed by resource R1, etc. The demand for the three finished products is indicated at the top of the diagram and is the weekly demand for the product. For the current week, the demand for each product is as follows: 40 units of Product A, 50 units for Product D, and 40 units for Product F.

FIGURE 8-7 Schedule control points in a plant with assembly and divergence.

Based on this product structure, the demand, the material in process, and the process information for each product, we can compute the load for each type of resource. For this, we calculate the number of units that need to be processed at a given step and then multiply that by the time to process a unit. For example, at Step A-1, which feeds both Product A and Product D, the total number of units to be processed is 65 (40 units for A + 50 units for D – 25 units of WIP at B-3). The total time required of the R2 resource for this production is (Number of units to be produced) × (Time to produce one unit) or 65 × 4 min = 260 min. In a similar manner, we compute

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