Process Flow Analysis Economics and Resources

The economic analysis for the system is shown below. Data in white fields are provided by the user, while numbers in yellow fields are Excel formulas. Assuming sales at their maximum values, the system generates a net income of \$7350. Goldratt would call this the throughput of the system. It is the revenue from sales less the cost of raw materials. The operating expense of \$4500 should be deducted from this figure to obtain the profit for the system.

The unit-raw material costs for each product are computed and shown in column J. Column K holds the unit-operating profit for the products. Goldratt would call these numbers the unit-throughput values for the products.

This solution may or may not be feasible with respect to the resources available for production. The resources are represented by the times available for the machines which are shown in the table below. The add-in creates the list of machines (resources) used from the four products. For each machine, data specifies the number, time available, percent available and maximum utilization. The results of columns K through M are computed from this data. Column N shows the amount of each resource used for the sales quantities entered in the table above. Column O shows that only machine B exceeds 100% utilization, so this machine must be the bottleneck. We also note that machine R is a potential source of problems since its utilization is 95%.

We could stop at this point with the bottleneck identified, but we continue to illustrate the next step of analysis.

The foregoing analysis has neglected a number of features associated with the situation. An important factor for all real systems is statistical variability. Although the data presents operating times and setup times as single numbers, these times may vary from unit to unit and from lot to lot. Machine time can also be lost due to failures. Although machine B is barely a bottleneck with its utilization exceeding capacity by only 1%, statistical variability assures that the machine can probably not be operated at full capacity. One way to plan for this is to set the maximum utilization in column K to some number less than 100%. In the table below, we set the maximum utilization to 90%. Now the resources used by machines B and R exceed the resource available indicating that there may be two bottlenecks.

With bottlenecks in the system, we cannot manufacture the entire weekly requirements. Rather, we must select a product mix that maximizes total throughput (net income on this worksheet). The add-in provides access to a linear programming model to find the optimum product mix. To create an LP model click the Build LP/IP model button on the worksheet.
 The LP model describing the product mix problem is shown below. All parameters of the model are determined automatically from the process and economic data. The model has already been solved in the figure below. Review of the solution shows that the full production of finished good C is not met by the optimum solution. Column D shows the amounts of the resources used by the solution. Clearly machine B is a bottleneck since all of its capacity is used. Machine R is not a bottleneck because the solution does not use all of its available capacity.

Returning to the Project worksheet we find that the sales per week have been translated from the LP solution. There is a reduction in net income from the unconstrained case. The resource utilizations are all less than 90% as required for this solution. As expected the resource utilization for machine B is exactly at 90% while all the others are less.

The Project worksheet provides a queuing analysis for the machines. The queuing model assumes that units are processed independently with Poisson arrival and service processes.

Although one might argue about the validity of the queuing analysis assumptions, these results do show the affects of statistical variability. The bottleneck machine has the greatest queuing delay, while the near bottleneck, machine R, also shows considerable delay. The delays at the less utilized machines are much smaller.

Operations Management / Industrial Engineering
Internet
by Paul A. Jensen