Answer to Question #89722 in Statistics and Probability for AMAR

Question

a) For Rail booking there are two reservation counters for customers, who arrive in Poisson fashion at an average rate of 10 per hour. The service time for booking clerks at both the counters are exponentially distributed with mean time 5 minutes. These counters remain open for 12 hours per day.
i) Find the hours of the day for which all the clerks are busy.
ii) Find the probability that both the clerks are idle, one is idle.
iii) Find the expected proportion of idle time for clerks.
iv) Find the expected waiting time of customers in the

Answer 1

i) Find the hours of the day for which all the clerks are busy. 

Two counters can serve:

$${2\ customers \over 5\ min}={10\ customers \over 25\ min}$$

So clerks are busy 25 minutes in one hour.

Then clerks are busy per day:

$$25\cdot 12=300 \min=5\ h$$

The clerks are busy 5 hours per day.

ii) Find the probability that both the clerks are idle, one is idle. 

When n=0, the system is idle.

Pn = probability of exactly n customers in the system.

$$\rho={\lambda \over 2\mu}$$ $$\lambda=10, \mu=12, \rho={10 \over 2(12)}={5 \over 12}$$

$$P_0={1-\rho \over 1+\rho}={1-{5 \over 12} \over 1+{5 \over 12}}={7 \over 17}\approx0.411765$$

$$P_1={\lambda \over \mu}P_0={2\rho(1-\rho) \over 1+\rho}$$

$$P_1={10 \over 12}\cdot{7 \over 17}={35 \over 102}\approx0.343137$$

The probability that both the clerks are idle is ${7 \over 17}\approx0.411765.$

The probability that one is idle is ${35 \over 102}\approx0.343137.$

iii) Find the expected proportion of idle time for clerks. 

$$1-\rho=1-{\lambda \over 2\mu}$$

$$1-\rho=1-{10\over 2(12)}={7 \over 12}\approx0.583333$$

iv) Find the expected waiting time of customers in the system $W$ and expected waiting time of customers in queue $W_q$

For the M/M/2 queue,

$$\rho={\lambda \over 2\mu}$$

$$P_0={1-\rho \over 1+\rho}$$

$$L_q={P_0(\lambda/ \mu)^2\rho \over 2!(1-\rho)^2}={2\rho^3 \over 1-\rho^2}$$

$$W_q={L_q \over \lambda}={2\rho^3 \over \lambda(1-\rho^2)}={\rho^2 \over \mu(1-\rho^2)}$$

$$W=W_q+{1 \over \mu}={1 \over \mu(1-\rho^2)}$$

$$\lambda=10, \mu=12, \rho={10 \over 2(12)}={5 \over 12}$$

$$W_q={({5 \over 12})^2 \over 12(1-({5 \over 12})^2)}={25 \over 1428}\approx0.017507$$

$$W={25 \over 1428}+{1 \over 12}={12 \over 119}\approx0.100840$$