1. on average 2 customers per hour use the public telephone in the sheriff\'s de
ID: 389607 • Letter: 1
Question
1. on average 2 customers per hour use the public telephone in the sheriff's detention area, and this use has a Poisson distribution. The length of the phone call follows exponential distribution with a mean of 6 minutes. The sheriff will install a second telephone booth Continue to postwhen an arrival can expect to wait 3 minutes or longer for the phone. What is the arrival rate that justifies a second telephone booth?
2. in a waiting line system, there is only one server. Customer arrivals follow Poisson distribution with mean inter-arrival time 5 minutes. Service time follow exponential distribution and a server can serve 15 customers per hour on average. What is the average waiting time?
Explanation / Answer
Interval between arrival of 2 customers = 30 mins
Given = 1/30 = 0.033 person per minute.
= 1/6 = 0.166 person per minute.
Probability that a person arriving at the booth will have to wait,
P (w > 0) = 1 – P0
= 1 – (1 - / ) = /
= 0.033/0.166 = 0.198
The installation of second booth will be justified if the arrival rate is more than the waiting time. Expected waiting time in the queue will be,
Wq = l/ m (m-l)
Where, E(w) = 3 and = (say ) for second booth.
= 0.16
Hence the increase in arrival rate is, 0.16-0.033 = 0.127 arrivals per minute.
Average number of units in the system is given by,
Ls= /1-
=0.198/1-0.198
= 0.25 customers
Probability of waiting for 3 minutes or more is given by
0.84 percent of the arrivals on an average will have to wait for 3 minutes or more before they can use the phone.
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