Question

A campus deli serves 240 customers over its busy lunch period from 11:00 a.m. to 1:00 p.m. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?

Answer 1

Answer:

the time spent in the process is 5.5 minutos

Step-by-step explanation:

Let $\mu = 240/120 = 2$ customers/minutes and $\lambda$ the customer arrival rate. If on average there are 10 customers in process and $L_q$ is the average number of customers waiting, then $L_s = 10$. We know that the average number of clients in the system is given by:

$L_s = L_q+\frac{\lambda}{\mu}=\frac{\lambda^2}{\mu(\mu-\lambda)}+\frac{\lambda}{\mu}$. So,

$10 = \frac{\lambda^2}{2(2-\lambda)}+\frac{\lambda}{2}\\\\10(2(2-\lambda))=\lambda^2+(2-\lambda)\lambda\\\\40-20\lambda = \lambda^2+2\lambda-\lambda^2\\\\\lambda = \frac{40}{22}=1.8182$ costumers/minutes.

On other hand,

$L_q=\frac{\lambda^2}{\mu(\mu-\lambda)}=\frac{1.8182^2}{2(2-1.8182)}=9.092$ costumers.

$W_q=\frac{L_q}{\lambda}=\frac{9.092}{1.8182}=5.0$ minutes is the time spent in the queue.

Finally you have

$W_s=W_q+\frac{1}{\mu}=5.0+\frac{1}{2}=5.5$ is the time spent in the process.