Let us consider the case of LaRosa Machine Shop (LMS). LMS is studying where to locate its tool bin facility on the shop floor. The locations of the five production stations appear in the figure below. Y Fabrication Min Paint 1 Station Fabrication Paint Subassembly 1 Subassembly 2 (x, y) = X Y Demand 1 Assembly 4 1 2 2.5 2 35 4 11 Subassembly 1 Subassembly 2 Assembly In an attempt to be fair to the workers in each of the production stations, management has decided to try to find the position of the tool bin that would minimize the sum of the distances from the tool bin to the five production stations. We define the following decision variables. 25 12 6 16 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 the tool bin located at (X, Y) by using Euclidean (straight-line) distance. For example, the distance from fabrication located at the coordinates (1, 4) to the tool bin located at the coordinates (X, Y) is given by √(x − 1)² + (Y−4)². (a) Suppose we know the average number of daily trips made to the tool bin from each production station. The average number of trips per day are 11 for fabrication, 25 for Paint, 12 for Subassembly 1, 6 for Subassembly 2, and 16 for Assembly. It seems as though we would want the tool bin closer to those stations with high average numbers of trips. Develop a new unconstrained model that minimizes the sum of the demand-weighted distance defined as the product of the demand (measured in number of trips) and the distance to the station. (b) Solve the model you developed in part (a). (Round your answers to three decimal places.) -(C Comment on the differences between the unweighted distance solution given of X = 2.230 and Y = 3.349 and the demand-weighted solution. The demand-weighted solution shifts the optimal location towards the Paint ✓✓station.
Let us consider the case of LaRosa Machine Shop (LMS). LMS is studying where to locate its tool bin facility on the shop floor. The locations of the five production stations appear in the figure below. Y Fabrication Min Paint 1 Station Fabrication Paint Subassembly 1 Subassembly 2 (x, y) = X Y Demand 1 Assembly 4 1 2 2.5 2 35 4 11 Subassembly 1 Subassembly 2 Assembly In an attempt to be fair to the workers in each of the production stations, management has decided to try to find the position of the tool bin that would minimize the sum of the distances from the tool bin to the five production stations. We define the following decision variables. 25 12 6 16 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 the tool bin located at (X, Y) by using Euclidean (straight-line) distance. For example, the distance from fabrication located at the coordinates (1, 4) to the tool bin located at the coordinates (X, Y) is given by √(x − 1)² + (Y−4)². (a) Suppose we know the average number of daily trips made to the tool bin from each production station. The average number of trips per day are 11 for fabrication, 25 for Paint, 12 for Subassembly 1, 6 for Subassembly 2, and 16 for Assembly. It seems as though we would want the tool bin closer to those stations with high average numbers of trips. Develop a new unconstrained model that minimizes the sum of the demand-weighted distance defined as the product of the demand (measured in number of trips) and the distance to the station. (b) Solve the model you developed in part (a). (Round your answers to three decimal places.) -(C Comment on the differences between the unweighted distance solution given of X = 2.230 and Y = 3.349 and the demand-weighted solution. The demand-weighted solution shifts the optimal location towards the Paint ✓✓station.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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