Question

Consider an espresso stand with a single barista. Customers arrive at the stand at the rate of 28 per hour according to a Poisson distribution. Service times are exponentially distributed with a service rate of 35 customers per hour. The probability that the server is busy is:

Answer 1

The probability that the server is busy is P=0.56.

Explanation:

We have a quieing theory problem with M/M/1

In queing theory, the probability of the server being busy can be expressed as:

`P=(\lambda)/(\lambda+\mu)`

being μ: the time between services and λ: the time between customers arrival.

Then we can calculate:

`P=(\lambda)/(\lambda+\mu)=((1)/(28) )/((1)/(28)+(1)/(35)  ) =(0.036)/(0.036+0.029)=0.56`

The probability that the server is busy is P=0.56.