Answer to Question #151322 in Statistics and Probability for K.manivarma

Question

7. A telephone exchange has two long distance operators. The telephone company finds that during the peak load long distance calls arrive in a Poisson fashion at an average rate of 15 per hour. The length of service on these calls is approximately exponentially distributed with mean length 5 Minutes

(What is the probability that a subscriber will have to wait for his long distance call during the peak hours of the day?

Dif the subscribers will wait and are serviced in turn, what is the expected waiting time?

Answer 1

Here we have:

Number of operators $s=2$

Call arrival rate 15 per hour or $\lambda =\frac{15}{60}=\frac{1}{4}$

Mean service time $\mu =\frac{1}{5}$

Traffic intensity $\rho=\frac{\lambda}{s\mu}=\frac{5}{8}$

$$P_0= (\sum_{n=0}^{s-1}\frac{(s\rho)^n}{n!}+\frac{1}{s!}\frac{(s\rho)^s}{1-\rho})^{-1}=(\sum_{n=0}^{1}\frac{(2\cdot\frac{5}{8})^n}{n!}+\frac{1}{2!}\frac{(2\cdot\frac{5}{8})^2}{1-\frac{5}{8}})^{-1}=\frac{3}{13}$$ $$P(n=2)=\frac{1}{s}(\frac{\lambda}{\mu})^s\frac{s\mu}{s\mu-\lambda}P_0=\frac{1}{2}\cdot({\frac{5}{4}})^2\cdot\frac{\frac{2}{5}}{\frac{2}{5}-\frac{1}{4}}\cdot\frac{3}{13}=0.48$$

Hence, the probability that a subscriber will have to wait for his long distance call during the peak hours of the day is 0.48.

The expected waiting time if the subscribers will wait and are serviced in turn is

$$W_q=\frac{1}{(s-1)!}(\frac{\lambda}{\mu})^s\frac{\mu}{(s\mu-\lambda)^2}P_0=\frac{1}{1!}\cdot(\frac{5}{4})^2\cdot\frac{\frac{1}{5}}{(\frac{2}{5}-\frac{1}{4})^2}\cdot\frac{3}{13}=3.2$$