Question

 

The College of Business and Public Administration at Benton University has a copy machine on each floor for faculty use. Heavy use of the five copy machines causes frequent failures. Maintenance records show that a machine fails every 2.5 days (or A = 0.40 failure/day). The college has a maintenance contract with the authorized dealer of the copy machines. Because the copy machines fail so frequently, the dealer has assigned one person to the college to repair them. The person can repair an average of 2.5 machines per day. Using the finite-source model, answer the following questions:

A) What is the average utilization of the maintenance per person

B) On average, how many copy machines are being repaired or waiting to be repaired

C) What is the average time spent by a copy machine in the repair system (waiting and being repaired)?

Answer 1

It has that every 2.5 days a machine fails. This is that 1 / 2.5 = 0.4 machine fails per day.

On the other hand they say that the person repairs 2.6 machines per day.

Data:

N = 7

λ = 0.4 machine / day

\ miu = 2.6 machine / day

a. What is the average utilization of the maintenance person?

rho = / ={0.4}/ {2.6} = 0.1539

The average utilization of the maintenance person is 0.1539

Answer:-

b. On average, how many copy machines are being repaired or waiting to be repaired?

Lq=N− λ λ+μ (1−p 0 ) (number of machines in queue)

Probability that there are no machines in the repair system

p 0 =[∑ (N−n)! N! ( μ λ ) n ]   −1 

p​0​​=(4.6027)​−1​​=0.2174

Lq=N−​λ*​​λ+μ​​(1−p​0​​)

Lq=7−​0.4​​*0.4+2.6​​(1−0.2174)=1.1305 machine

On average there are 1.1305 copier machines waiting to be repaired

Answer:-

W=W​q​​+​μ​​1​​ Time spent by a copy machine in the repair system​

W​q​​=​(N−L)λ​​L​q​​​​ Queue waiting time

L=L​q​​+(1−p​0​​) number of machines in the repair system

L=L​q​​+(1+p​0​​)=1.1305+(1−0.2174)=1.1305+0.7826=1.9131 machine (number of machines in the repair system​ )

W​q​​=​(N−L)λ​​L​q​​​​=W​q​​=​(7−1.9131)0.4​​1.1305​​=0.5556 day (Queue waiting time)

W=0.5556+​2.6​​1​​=0.9402 day (Time spent by a copy machine in the repair system​ )

The average time spent by a copy machine in the repair system​ is 0.9402 day