Answer to Question #237059 in Operations Research for opr

Question

Arrivals at a telephone booth are considered to be Poisson with an average time of 10 minutes between one arrival and the next. The length of a phone call is assumed to be distributed exponentially with mean of 3 minutes.

i) What is the probability that a person arriving at the booth will have to wait?

ii) What is the average length of the queues that form time to time?

The telephone department will install a second booth when convinced that an arrival would expect to have to wait at least three minutes for the phone. By how much must the flow of arrivals be increased in order to justify a second booth?

Answer 1

Given λ= 1/10 = 0.10 person per minute.

μ = 1/3 = 0.33 person per minute.

(i) Probability that a person arriving at the booth will have to wait,

P (w > 0) = 1 – P₀= 1 – (1 - λ / μ) = λ / μ= 0.10/0.33 = 0.3

(ii) 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,

W_q = l/ m (m-l)

Where, E(w) = 3 and λ = λ (say ) for second booth. Simplifying we get

λ = 0.16

Hence the increase in arrival rate is, 0.16-0.10 = 0.06 arrivals per minute.

(iii) Average number of units in the system is given by,

L_s= ρ/1- ρ= 0.3/1-0.3= 0.43 customers

(iv) Probability of waiting for 10 minutes or more is given by

$P(W\ge10)=\int^{\alpha}_{10}\frac{\lambda}{\mu}(\mu-\lambda)e^{-(\mu-\lambda)}dt=$

$=\int^{\alpha}_{10}0.3\cdot 0.23e^{-0.23}dt=0.069(\frac{e^{-0.23t}}{-0.23})|^{\alpha}_{10}=0.03$

This shows that 3 percent of the arrivals on an average will have to wait for 10 minutes or more before they can use the phone.