Example: Determine comparable units and production rates

The examples in this topic show how comparable units are determined for production rate calculations when the entities on the line can produce different items for a work order.

These examples do not apply to work orders created in MES Web Portal. All of the jobs that are running for work orders created in MES Web Portal will produce the same item at each entity.

The line for these examples is set up as follows:

The entities in the line have following default rates and standard items.

The entity production UOMs are as follows:

UOM conversions are defined in MES Client. The general procedure for creating the UOM conversions for a line is:

1. Create all of the required UOMs.

2. Identify the UOM that will be used for the item that is defined as the standard item for the line.

3. Create conversion factors for each UOM so that all items can be specified in the UOM of the standard item.

For instructions about creating UOMs and defining UOM conversion factors, see the Units of Measure chapter in the MES Client User Guide or help.

For the example line, the UOM conversion factors listed in the following table would be used. Note that the conversions shown in italics do not need to be defined for the sake of these examples, but are shown to illustrate how some of the rest of the conversions are obtained.

Assume that there is one work order A1 running on the entire line when no entities are disabled, with the following jobs, items, and production rate information.

If the line's production UOM is batches per minute, where the batch size is 1, but there is no standard item for the line so that the item made in the job running on the first entity is used, the rates at each entity will be as follows:

Since the sum of the fillers at the second position is 192, and the sum of the two cappers at the third position is 200, the slowest position on the line is the second. Thus, assuming all entities can be part of the bottleneck, the line's bottleneck comprises the two fillers, and the line's performance rate is 192 bottles of A per minute.

As work order A1 progresses, a point will be reached where its first jobs are done. If the jobs for work order A1 are completed on the washer, both fillers, and capper 1, and there is no other work order's jobs started yet on any of these entities, the rates at these entities will be set to their default values. Since no job is running at the first entity and the default item produced by it is bottles of C, all the rates will be expressed in terms of bottles of C.

Since the sum of the fillers at the second position is 64, and the sum of the two cappers at the third position is 65, the slowest position on the line is the fifth. Thus, assuming all entities can be part of the bottleneck, the line's bottleneck is the palletizer, and the line's performance rate is 53.25 bottles of C per minute.

Continuing the example, as work order A1 progressed, another work order, B1, is assigned to the line and is started. If the jobs for work order A1 reach the same point as above (that is, completed on the washer, both fillers, and capper 1), and the jobs for work order B1 are as follows:

Then once the jobs for the washer and filler 1 are started, the operations for filler2 and capper1 from work order B1 will be used to determine the rates at those entities even though they are not yet running, because of the jobs being run at the upstream entities (the washer and filler1 respectively). So since the item produced by the first entity is now bottles of B, all the rates will be expressed in terms of bottles of B.

Since the sum of the fillers at the second position is 120, and the sum of the two cappers at the third position is 100, the slowest position on the line is the third. Thus assuming all entities can be part of the bottleneck, the line's bottleneck comprises the two cappers, and the line’s performance rate is 100 bottles of B per minute.