Answer to Question #89449 in Statistics and Probability for sunil

Question

a postal clerk can service a customer in 3 minutes, the service time being exponentially distributed. The inter-arrival time of customers is also exponentially distributed with an average of 12 minutes during the early morning and an average of 5 minutes during the afternoon peak period. Determine the average queue length and the expected waiting time in the queue during the two periods

Answer 1

Average queue length is average number of customers in system $L.$

We have an M/M/1 system. We also have:

$$\lambda=3, \mu_1=12, \mu_2=5$$ $$\rho={\lambda \over \mu}$$

Hence

$$\rho_1={3 \over 12}={1 \over 4}, \ \rho_2={3 \over 5}$$

Little’s rule provide the following results:

$$L=\lambda W, \ L_q=\lambda W_q$$

$$W=W_q+{1 \over \mu}$$

For the M/M/1 queue, we can prove that

$$L_q={\rho^2 \over 1-\rho}$$ $$L_{q1}={({1 \over 4})^2 \over 1-{1 \over 4}}={1 \over 12}$$

$$L_{q2}={({3 \over 5})^2 \over 1-{3 \over 5}}={9 \over 10}$$

$$W_{q1}={L_{q1} \over \lambda}={1 \over 12(3)}={1 \over 36}$$

$$W_{q2}={L_{q2} \over \lambda}={9 \over 10(3)}={3 \over 10}$$

$$W_1=W_{q1}+{1 \over \mu_1}={1 \over 36}+{1 \over 12}={1 \over 9}$$

$$W_2=W_{q2}+{1 \over \mu_2}={9 \over 10}+{1 \over 5}={11 \over 10}$$

$$L_1=\lambda W_1=3({1 \over 9})={1 \over 3}$$

$$L_2=\lambda W_2=3({11 \over 10})={33 \over 10}$$

The average queue length is ${1 \over 3}$ during early morning and ${33 \over 10}$ during afternoon peak period.

The expected waiting time in the queue is ${1 \over 36} min$ during early morning and ${3 \over 10} min$ during afternoon peak period.