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
1
Expert's answer
2019-05-15T08:41:30-0400

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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