City Cab, Inc., uses two dispatchers to handle requests for service and to dispa

Question

City Cab, Inc., uses two dispatchers to handle requests for service and to dispatch the cabs. The

telephone calls that are made to City Cab use a common telephone number. When both

dispatchers are busy, the caller hears a busy signal; no waiting is allowed. Callers who receive a

busy signal can call back later or call another cab company for service. Assume that the arrival

of calls follows a Poisson distribution, with a mean of 40 calls per hour, and that the call

handling time follows an exponential probability distribution with a mean service time of 2

minutes. Based on this information, answer the following questions.

(a) What percentage of the time are both dispatchers idle?

(b) What percentage of the time are both dispatchers busy?

(c) What is the probability that a caller will receive a busy signal if 2, 3, or 4 dispatchers are

used?

(d) If management wants no more than 12% of the callers to receive a busy signal, how many

dispatchers should be used?

(e) Suppose the service time distribution is not exponential, it follows some other

distribution such as a normal, but the mean service time remains at 2 minutes. Does this

make any difference in the model used and in the results? Explain.

(f) Suppose City Cab added call waiting for the two dispatchers so that at most two calls

could be in the hold queue.

(i) What percentage of the time are both dispatchers idle?

(ii) What percent of the time are both dispatchers busy?

(iii)What is the probability a caller will receive a busy signal?

(iv)What is the average number of customers in the system?

(v) On average, how long does a customer wait before talking to a dispatcher?

Explanation / Answer

What percentage of the time are both dispatchers idle?

Arrival rate = 40 calls per hour

Service rate µ = 60 minutes / service time = 60/2 = 30 calls per hour for one dispatchers

But we have two dispatchers to handle service requests

Therefore service rate µ = 30 *2 = 60 calls per hour for two dispatchers

The percentage of the time both dispatchers are idle = 1- /µ = 1- (40/60) =0.3333 or 33.33%

What percentage of the time are both dispatchers busy?

The percentage of the time both dispatchers are busy = /µ = 40/60 = 0.6667 or 66.67%

What is the probability that a caller will receive a busy signal if 2, 3, or 4 dispatchers are used?

The probability that a caller will receive a busy signal if 2 dispatchers are used

Arrival rate = 40 calls per hour

Service rate µ = 30 *2 = 60 calls per hour for two dispatchers

The probability that a caller will receive a busy signal =/µ = 40/60 = 0.6667 or 66.67%

The probability that a caller will receive a busy signal if 3 dispatchers are used

Arrival rate = 40 calls per hour

Service rate µ = 30 *3 = 90 calls per hour for three dispatchers

The probability that a caller will receive a busy signal =/µ = 40/90 = 0.4444 or 44.44%

The probability that a caller will receive a busy signal if 4 dispatchers are used

Arrival rate = 40 calls per hour

Service rate µ = 30 *4 = 120 calls per hour for four dispatchers

The probability that a caller will receive a busy signal =/µ = 40/120 = 0.3333 or 33.33%

If management wants no more than 12% of the callers to receive a busy signal, how many dispatchers should be used?

Let’s assume that the probability that a caller will receive a busy signal = /µ = 12%

And we know that Arrival rate = 40 calls per hour

Service rate µ = ?

/µ = 40/µ = 12%

µ = 40 /0.12 = 333.33 calls per hour

number of dispatchers should be used for 12% of the callers to receive a busy signal = 333.33 calls per hour/ one dispatchers service rate = 333.33/30 = 11.11

but management wants no more than 12% of the callers to receive a busy signal

There the number of dispatchers should be used is the next whole number = 12 dispatchers

the probability that a caller will receive a busy signal if 12 dispatchers are used = /µ = 40/ (30 * 12) = 40/360 = 11.11% (less than 12%)

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