Question

Customers enter the waiting line to pay for food as they leave a cafeteria on a first-come, first-served basis. The arrival rate follows a Poisson distribution, while service times follow an exponential distribution. If the average number of arrivals is four per minute and the average service rate of a single server is seven per minute, what proportion of the time is the serverbusy?A) 0.43B) 0.57C) 0.75D) 0.25E) 0.33

Answer 1

Answer:

The answer is B) 0.57.

Step-by-step explanation:

In this problem we have to apply queueing theory.

It is a single server queueing problem.

The arrival rate is  $\lambda=4$ and the service rate is $\mu=7$.

The proportion of time that the server is busy is now as the "server utilization"and can be calculated as:

$p=\frac{\lambda}{c\mu} =\frac{4}{1*7}=0.57$

where c is the number of server (in this case, one server).