Question

A graphics reproduction firm has four units of equipment that are automatic but occasionally become inoperative...

A graphics reproduction firm has four units of equipment that are automatic but occasionally become inoperative because of the need for supplies, maintenance, or repair. Each unit requires service roughly 1.0 minutes each hour, or, more precisely, each unit of equipment runs an average of 75 minutes before needing service. Service times vary widely, ranging from a simple service (such as pressing a restart switch or repositioning paper) to more involved equipment disassembly. The average service time, however, is ten minutes.

Equipment downtime results in a loss of $60 per hour. The one equipment attendant is paid $18 per hour.

Using finite queuing analysis, answer the following questions. In your calculations, use the value of "X" in Exhibit 10.10  that is closest to the value you compute.

a. What is the average number of units in line? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

b. What is the average number of units still in operation? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

c. What is the average number of units being serviced? (Round your answer to 2 decimal places.)

d. What is the total service cost of the system per hour, including downtime and labor costs? (Round your answer to 2 decimal places.)

Business   Operations-Management   
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Answer #1

Average service requirement rate = 1/75 / min or 60/75 per hour =0.8

Average service time = 6 per hour

rho = 60/75 x1/6 = 2/15

1. Avg units waiting = lambda2 / mu ( mu- lambda)

= 0.64 / 6 ( 6-0.8) = 0.02

2. avg no. of units in system = lambda / mu-lambda = 0.8 / 6-0.8 = 0.8 /5.2 = 0.1538

Avg units being served = 0.1538-0.02 = 0.1518

Avg units in operation = 4-0.1518 = 3.8482

total time ( waiting + service) = 1/mu-lambda = 1/ 6-0.8 = 0.1923 hours

Avg time waiting = lambda / mu ( mu-lambda) = 0.8 /6 x5.2 = 0.025

Service time = .1923-0.025 = 0.1666 hour

Cost of downtime = 60 x0.1923 = 11.538

Cost of service = 18x0.1666 =2.99

Total cost per case = 11.538+2.99 = 14.53

Total cost per hour = 60/75 x14.53 = 11.62

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