LaRosa Machine Shop (LMS) is studying where to locate its toolbin facility on the shop floor. The locations of the fiveproduction stations appear in figure shown below.
| Location | |
| Station | X | Y | Demand |
| Fabrication | 1.0Â Â Â | 4.0Â Â Â | 12Â Â Â Â Â |
| Paint | 1.0Â Â Â | 2.0Â Â Â | 24Â Â Â Â Â |
| Subassembly 1 | 2.5Â Â Â | 2.0Â Â Â | 13Â Â Â Â Â |
| Subassembly 2 | 3.0Â Â Â | 5.0Â Â Â | 7Â Â Â Â Â |
| Assembly | 4.0Â Â Â | 4.0Â Â Â | 22Â Â Â Â Â |
In an attempt to be fair to the workers in each of theproduction stations, management has decided to try to find theposition of the tool bin that would minimize the sum of thedistances from the tool bin to the five production stations.We define the following decision variables:
X = horizontal location of the tool bin
Y = vertical location of the tool bin
We may measure the straight line distance from a station to thetool bin located at (X,Y) by using Euclidean(straight-line) distance. For example, the distance fromfabrication located at the coordinates (1,4) to the tool binlocated at the coordinates (X,Y) is given by .
| (a) | Suppose we know the average number of daily trips made to thetool bin from each production station. The average number of tripsper day are 12 for fabrication, 24 for Paint, 13 for Subassembly 1,7 for Subassembly 2 and 22 for Assembly. It seems like we wouldwant the tool bin closer to those stations with high averagenumbers of trips. Develop a new unconstrained model that minimizesthe sum of the demand-weighted distance defined as the product ofthe demand (measured in number of trips) and the distance to thestation. |
| Min:Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â |
| |
| (b) | Solve the model you developed in part (a). |
| If required, round your answer to six decimal places. Do notround intermediate calculation. |
| X =Â Â |
| Y =Â Â |
| |
| (c) | The solution to the un-weighted model is X = 2.230 andY = 3.349. Comment on the differences between theunweighted distance solution given and the demand-weighted solutionfound in part (b). |
