Question #89449

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

Expert's answer

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

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

λ=3,μ1=12,μ2=5\lambda=3, \mu_1=12, \mu_2=5ρ=λμ\rho={\lambda \over \mu}

Hence


ρ1=312=14, ρ2=35\rho_1={3 \over 12}={1 \over 4}, \ \rho_2={3 \over 5}

Little’s rule provide the following results:


L=λW, Lq=λWqL=\lambda W, \ L_q=\lambda W_q

W=Wq+1μW=W_q+{1 \over \mu}

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


Lq=ρ21ρL_q={\rho^2 \over 1-\rho}Lq1=(14)2114=112L_{q1}={({1 \over 4})^2 \over 1-{1 \over 4}}={1 \over 12}

Lq2=(35)2135=910L_{q2}={({3 \over 5})^2 \over 1-{3 \over 5}}={9 \over 10}

Wq1=Lq1λ=112(3)=136W_{q1}={L_{q1} \over \lambda}={1 \over 12(3)}={1 \over 36}

Wq2=Lq2λ=910(3)=310W_{q2}={L_{q2} \over \lambda}={9 \over 10(3)}={3 \over 10}

W1=Wq1+1μ1=136+112=19W_1=W_{q1}+{1 \over \mu_1}={1 \over 36}+{1 \over 12}={1 \over 9}

W2=Wq2+1μ2=910+15=1110W_2=W_{q2}+{1 \over \mu_2}={9 \over 10}+{1 \over 5}={11 \over 10}

L1=λW1=3(19)=13L_1=\lambda W_1=3({1 \over 9})={1 \over 3}

L2=λW2=3(1110)=3310L_2=\lambda W_2=3({11 \over 10})={33 \over 10}

The average queue length is 13{1 \over 3} during early morning and 3310{33 \over 10} during afternoon peak period.

The expected waiting time in the queue is 136min{1 \over 36} min during early morning and 310min{3 \over 10} min during afternoon peak period.



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