ISyE 6201: Manufacturing Systems Instructor : Spyros Reveliotis Fall 2006 Solutions for Homework #3



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Ca2 for the inter-arrival times experienced at the first station. Since we are told that batches are released to the line at a constant rate, there is no variability in the inter-arrival times, and therefore, Ca2 is equal to zero (This is typical in repetitive, high throughput manufacturing).
W
e will start with a configuration, that selects the option with the lowest cost at each workstation, and utilizes the minimum number of machines that provides a machine utilization lower than one for the considered throughput.
The result is that we have 3, 3, 6, and 3 machines at station 1, 2, 3, and 4 respectively. From this configuration, we see that resulting total cycle time is higher than the cycle time constraint (less than 8 hrs) with a total cycle time of 21.40 hrs.
The largest portion of cycle time is at station 2, 13.34 hrs, so we will work on this first. By adding one more machine of the same technology option, the total cycle time goes down to 8.25 hrs and the cost goes up to $1,010, with a resulting benefit of 0.26 hr/$-added (You can check by editing the number of machine at station 2 in the spreadsheet in the previous page to be 4 machines). We now try to replace the machines at station 1 by a machine of option 2, which has a smaller variability. This selection is entered in our spreadsheet by setting the corresponding part processing CV to be 1 and the corresponding cost per machine equal to $85. With four machines of option 1 at station 2, the cycle time goes down to 6.75 hrs, i.e., our cycle time requirement is met. At the same time the total cost increases to $1,115.
Alternatively, we consider the possibility of adding one more machine of option 1 at station 1. The results for this configuration are given below:


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