Question

Assume that the arrival of airplanes at a one-runway airport is a Poisson distribution with a mean rate of 8 planes per hour. The landing time (service time, not service rate which would be planes/min) is an exponential distribution with a mean of 5 minutes per plane. What is the mean number of planes in a holding (planes circling around waiting to land) pattern (compute exact number)? a. 1 b. 1.33 c. 2 d. 1.67

Answer 1

Answer:

Option b that is 1.33 is the right choice.

Step-by-step explanation:

Given:

Mean rate of arrival $(\lambda)$ = 8 planes/hr

Service time = $5$ minute/plane

Mean service rate  $(\mu)$ = $\frac{60}{5}$ = $12$ planes/hr

Applying the concept Poisson-distributed arrival and service rates (exponential inter-arrival and service times)(M/M/1) process:

We have to find mean number of planes waiting to land that is mean number of customers in the queue .

Mean number of customers in queue $(L_q)$ .

⇒ $L_q=\frac{\rho \lambda}{\mu-\lambda}$

   Considering, $\rho=\frac{\lambda}{\mu}$ , $\rho$ is also mean number of customers in service.

⇒ $L_q=\frac{ \lambda^2}{\mu(\mu-\lambda)}$

⇒ Plugging the values.

⇒ $L_q=\frac{8^2}{12(12-8)}$

⇒ $L_q=\frac{64}{48}$

⇒ $L_q=1.33$

So,

Mean number of planes in holding and waiting to land = 1.33