Question

Consider an M/M/1 queue in which the expected number of customers in the system is 4, and the expected waiting time in the system is 80 minutes. What is the probability that a customer’s service time is less than 40 minutes?

Answer 1

Answer:

Step-by-step explanation:

Given

Expected no of customer in the system is 4 i.e. length of queue is $L=4$

Expected waiting time in the system $W=80\ min$

Length of queue is given by

$L=\frac{\rho }{1-\rho }$

$4=\frac{\rho }{1-\rho }$

thus $\rho =\frac{4}{5}$

where $\rho =\frac{mean\ arrival\ rate}{mean\ service\ rate}$

L is also given by

$L=\lambda \times W$

therefore

$\lambda =\frac{4}{80}=\frac{1}{20}$

where $\lambda$=mean arrival rate

$\mu$=mean service rate

Probability that customer service rate is less than 40 minutes is

$P=1-\rho e^{-\mu (1-\rho )\cdot t}$

$P=1-0.8\times e^{-\frac{1}{16}\times 0.2\times 40}$

$P=1-0.4852$

$P=0.5147$