Question

Customers enter the waiting line at 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 is the average number of customers waiting in line behind the person being served

Answer 1

Answer:

The average number of customers waiting in line behind the person being served is 0.7619.

Step-by-step explanation:

In queueing problem, an M/M/1 queue model is an arrangement with a single queue, where arrivals at the queue are approximated by a Poisson distribution with mean λ and the service times follows an exponential distribution with mean μ.  

The mean number of arrivals in this system is given by:

$\text{Average number of customers}=\frac{\rho}{1-\rho}$

Here the variable ρ is defined as:

$\rho=\frac{\lambda}{\mu}$

From the average number of arrivals in the system, we can calculate the average number of  arrivals in the queue by subtracting the average number of arrivals in service.

That is,

$\text{Average number of customers in the queue}=\frac{\rho}{1-\rho}-\rho =\frac{\rho^{2}}{1-\rho}$

The information provided is:

The average number of arrivals of customers is, λ = 4.

The average service rate of a single server is, μ = 7.

Compute the average number of customers waiting in line behind the person being served as follows:

$\text{Average number of customers in the queue}=\frac{\rho}{1-\rho}-\rho =\frac{\rho^{2}}{1-\rho}$

                                                                   $=\frac{(4/7)^{2}}{1-(4/7)}\\\\=\frac{(4/7)^{2}}{(7-4)/7}\\\\ =\frac{16}{49}\times \frac{7}{3}\\\\=0.7619$

Thus, the average number of customers waiting in line behind the person being served is 0.7619.