Answer to Question #127238 in Statistics and Probability for Deep

Question

6. Arrival rate of telephone calls at a telephone booth are according to Poisson distribution with an average time of 9 minutes between two consecutive arrivals. The length of telephone call is assumed to be exponential distributed with mean 3 minutes;

A. Determine the probability that a person arriving the booth will have to wait.

B. Find the average queue length that is form time to time.

C. Telephone company will install a second booth when convinced that an arrival would expect to have to wait at least 4 minutes for phone. Find the increase in flow of arrivals which will justify a second booth.  

D. What is the probability that an arrival will have to wait for more than 10 minutes before the phone is free?

E. What is the probability that he will have to wait for more than 10 minutes before the phone is available and the call is also complete?

F. Find the fraction of the day that phone will be in use.

( M/M/1: ∞/FCFS Model)

Answer 1

Arrival rate, λ=1/9 per minute

service rate, µ=1/3 per minute

a. p=λ/µ=0.33

b. L=µ(µ-λ)=(1/3)/((1/3)-(1/9))=1.5

c. average waiting time in the queue

W=λ/(µ(µ-λ))

The waiting time toinstalla second booth is 4minutes

4=λ/((1/3)((1/3)-λ))

λ=4/21=0.19 arrivals/minute

increase in flow of arrivals=0.19-1/9=0.079 per minute

d. P(W>10)= $\lambda/\mu\int^\infin_{10}(\mu-\lambda)e^{-(\mu-\lambda)^t}dt=1/3e^{-10(1/3-1/9)}=1/30=0.033$

e. P=$\int^\infin_{10}(\mu-\lambda)e^{-(\mu-\lambda)^t}dt=\mu/\lambda(1/30)=0.1$

f. The expected fraction of day that the phone will be in use is equal to $\lambda/\mu=(1/9)/(1/3)=0.33$