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An operations manager wants to determine the optimum (the least costly) arrangem

ID: 346801 • Letter: A

Question

An operations manager wants to determine the optimum (the least costly) arrangement of four work centers (A, B, C, and D) in four areas (I, II, III, and IV) on an existing shop floor. The handling/transportation cost is 50 +n cents per load per yard. The work flow matrix, distance matrix, and current process layout for this operation are as follow: Work Flow Matrix (Loads per Day) Distance Matrix (Yards) FromTo A B C D | A - 5 12 20 B 2 - 0 15 | C 120 +n 2 - 1 10 FromTo I II III IV 7 | 12 | 23 - 10 20 10 Z The reverse distances are the same. Current Process Layout " B | " C Y » Example: Handling/Transportation Cost 50¢ + ng = $0.50 + $0.45 = $0.95 per load per yard Work Flow from C to A = 20 +n= 20 + 45 = 65 loads

Explanation / Answer

n = 48

Handlng/transportation cost per load per year = 0.50+48*0.01 = 0.98

A. Total work flow between A and C = 12+(20+48) = 80

B. Total handling/transportation cost per day for the current process layout = 0.98*((5+2)*7 + (12+68)*12 + (20+8)*23 + (0+2)*10 + (15+5)*20 + (10+7)*10) = $ 2,198.14

C. Optimum layout is determined by reorganizing the work centers such that Work centers having the maximum wokflow between them are placed the closest to each other. Work centers A and C have the maximum workflow between and hence they are placed the closest to each other at areas I & II.

The next highest workflow is between A and D, so these are also placed together. The resulting optimum layout is following

Total handling/transportation cost per day for optimum layout = 0.98*((5+2)*20 + (12+68)*7 + (20+8)*10 + (0+2)*23 + (15+5)*10 + (10+7)*12) = $ 1401.40

I-C II-A III-D IV-B
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