Question

On average, 4 individuals (detainee or visitor) per hour use the public telephone in the sheriff’s detention area, and this use has a Poisson distribution. The length of a phone call varies according to a negative exponential distribution, with a mean of 5 minutes.

When an arrival can expect to wait 3 minutes or longer for the phone, a second phone is justified.

If the probability of having to wait, by either detainee or visitor, at all exceeds 0.6, an additional phone is justified. How much must the arrival rate per hour increase to justify a second phone?

Answer 1

In this case the arrival rate = 4 and the service rate = 60/5 = 12 per hour

If the waiting time Wq is greater than 3 minutes = 3/20 = 0.15 hours, a second phone is justified. The waiting time Wq is given as below

It is required that waiting time be greater than 0.15 hours to justify the second phone. Hence we have the below

21.6 - 1.8

21.6 2.8

7.714

It is also given that if the probability of having to wait, by either detainee or visitor, at all exceeds 0.6. Probability of waiting Pw is given as below

It is required that probability of having to wait, by either detainee or visitor, to exceeds 0.6. to justify the second phone. Hence we have the below

7.2

Now combining the above 2 we can see that if the arrival rate becomes 7.2 per hour the second phone is justified. Hence the the arrival rate per hour must increase to 7.2 to justify a second phone