A new copy machine is being installed in a library. The librarian estimates that

Question

A new copy machine is being installed in a library. The librarian estimates that each user will spend 3 minutes on the average with the machine, and wants the average number of users L at the facility at any moment to be at most three. a) Under these conditions, what is the maximum average number of users per hour that the machine can serve? Assume Poisson arrivals, exponential service times, and steady-state conditions. (b) With the maximum allowed arrival rate from (a), what is the average queue time for a user of the machine? (c) Suppose more space were provided for the machine, so that the facility could accomodate twice as many users (L = 6), on average. What would the results of parts (a) and (b) be in this case?

Explanation / Answer

Solution

This is a problem in M/M/1 Queuing System, i.e., Poisson Arrivals/Exponential Service/Single Service Channel.

Back-up Theory

Let = average number of users coming to the library for using the new copier per hour and

µ = average number of users using the new copier per hour.

Given, each user will spend 3 minutes on an average on the copier, µ = 60/3 = 20.

Part (a)

Let E(n) = average number of users in the system, i.e., number of users using the copier plus number of users waiting.

We have the stipulation that E(n) 3 [Given, the librarian wants the number of users at the facility at any moment of time not more than 3]

And we want the maximum value for satisfying the above stipulation.

Now, for M/M/1 system, E(n) = /(µ - ).

So, we want: /(20 - ) 3 or 4 60 or 15.

Thus, the maximum number of users the machine can serve per hour is 15 ANSWER

Part (b)

Under the above value of , average queue time for a user of the machine = E(w) = /{µ(µ - )}

= 15/(20 x 5) = 3/20 hour = 9 minutes ANSWER

Part (c) to work out (a) and (b) under the stipulation E(n) 6.

Max : /(20 - ) 6 or 17 ANSWER 1

E(w) = 17/(20 x 3) = 17/60 hour = 17 minutes ANSWER 2

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