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A company needs to locate three departments (X, Y, and Z) in the three areas (I, II, and III) of a new facility. They want to minimize interdepartmental transportation costs, which are expected to be $0.50 per load meter moved. An analyst has prepared the following flow and distance matrices: Distances (meters) From/To I II III I - 10 20 II - 10 III - Flows (loads per week) From/To X Y Z X - 0 80 Y 30 - 150 Z 100 130 - If the company were to locate departments X, Y, and Z in areas I, II, and III, respectively, what would be the total distance (in meters) loads would be moved each week?

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Parrain

Based on the distance between locations, the total distance moved in the week will be 6,700 meters.

Distance between the departments

Assuming that Departments X, Y and Z are in areas I, II and III respectively. The following will be true:

  • Department X is located 10 meters from Y but 20 meters from Z.
  • Department Y is located 10 meters from Z.

Total distance traveled

This can be found by multiplying the loads carried across departments by the distance between the locations of these departments.

= ( (Loads from X to Y x distance of X and Y) + (Loads from X to Z x Distance of X and Z)) + ( (Loads from Y to X x distance of X and Y) + (Loads from Y to Z x Distance of Y and Z)) + ( (Loads from Z to Y x distance of Z and X) + (Loads from Z to Y x Distance of Y and Z))

= ( (0 x 10) + 80 x 20)) + ( (30 x 10) + (150 x 10) ) + ( (100 x 20) + 130 x 10) )

= 1,600 + 1,800 + 3,300

= 6,700 meters

In conclusion, the total distance would be 6,700 meters.

Find out more on minimizing costs at https://brainly.com/question/1380316.

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