Question

A simple queuing system has the mean interval time of 8 minutes and a mean service time of 4 minutes .

i) Determine the mean service rate and the mean arrival rate.

ii) Determine the traffic intensity.

iii) Determine the mean time a customer spends in the queue and in the system .

iv) What is the expected number of customers in the queue and in the system.

v) What is the probability of having at most four customers in the system.

Answer 1

Solution:

(i) Mean service rate$=\mu=\dfrac1{4}$

And mean arrival rate$=\lambda=\dfrac1{8}$

(ii) Traffic intensity$=\rho=\dfrac{\lambda}{\mu}=\dfrac{1/8}{1/4}=0.5$

(iii) Number in the queue$=L_q=\dfrac{\rho^2}{1-\rho}=\dfrac{0.5^2}{1-0.5}=0.5$

Wait in the queue$=W_q=L_q/{\lambda}=0.5/(1/8)=4\ mins$

Wait in the system$=W=W_q+1/\mu=4+4=8\ mins$

Number of the system$=L=\lambda W=(1/8)\times 8=1$

Thus, the mean time a customer spends in the queue $=W_q=4\ mins$

And in the system$=W=8\ mins$

(iv) The expected number of customers in the queue $=L_q=0.5$

And in the system $=L=1$

(v)  Probability that there are zero customers or units in the system$=P_0=1-\rho=1-0.5=0.5$

Now, the probability of having at most four customers in the system$=P_0+P_1+P_2+P_3+P_4$

 We know that probability that there are n customers in the system$=P_n=\rho^nP_0$

So, the probability of having at most four customers in the system

$=0.5+0.5(0.5)^1+0.5(0.5)^2+0.5(0.5)^3+0.5(0.5)^4 \\=0.5(1+(0.5)^1+(0.5)^2+(0.5)^3+(0.5)^4) \\=0.96875$