Consider a petrol pump which has a single petrol dispensing unit. Customers arrive there in accordance with a Poisson process having rate λ = 1 minutes. An arriving customer enters the petrol pump only if there are two or less customers in the petrol pump, otherwise he/she leaves the petrol pump without taking the petrol (at any point of time a maximum of three customers are present in the petrol pump). Successive service times of the petrol dispensing unit are independent exponential random variables having mean \(\frac{1}{2}\) minutes. Let X denote the average number of customers in the petrol pump in the long run. Then E(X) is equal to  

Concept:

The value of  \(E(X) = 0 \times P_0 + 1 \times P_1 + 2 \times P_2 + 3 \times P_3\)

Explanation:

we have,

A single server (the petrol dispensing unit),

Customers arrive according to a Poisson process with rate \(\lambda = 1 \) per minute,

Service times are exponentially distributed with mean \(\frac{1}{2}\) minutes (service rate \(\mu = 2 \)),

The petrol pump has a maximum capacity of 3 customers (only 3 customers can be in the

system, including the one being served). If there are already 3 customers, new arrivals are

blocked (they leave without entering the system).

This describes an \(M/M/1/3\) queueing system with arrival rate \(\lambda = 1 \) , service rate \(\mu = 2 \), and

a system capacity of 3 (maximum 3 customers).

Let the state X denote the number of customers in the petrol pump (0, 1, 2, or 3). The system has

a finite capacity, so the maximum number of customers at any given time is 3.

The system is modeled as a birth-death process with the following transition rates:

Arrival rate \(\lambda = 1 \) ,

Service rate \(\mu = 2 \).

Let \(P_n\) be the steady-state probability that there are n customers in the system. We need to find

\(P_0, P_1, P_2, P_3\)  the probabilities for 0, 1, 2, and 3 customers in the system.

Using the birth-death process relations, we get

\(\frac{P_1}{P_0} = \frac{\lambda}{\mu} = \frac{1}{2}, \quad \frac{P_2}{P_1} = \frac{\lambda}{\mu} = \frac{1}{2}, \quad \frac{P_3}{P_2} = \frac{\lambda}{\mu} = \frac{1}{2}\)

Thus,

\(P_1 = \frac{1}{2} P_0, \quad P_2 = \frac{1}{2} P_1 = \frac{1}{4} P_0, \quad P_3 = \frac{1}{2} P_2 = \frac{1}{8} P_0\)

Now, since the total probability must sum to 1,

\(P_0 + P_1 + P_2 + P_3 = 1\)

Substituting the values of \(P_1, P_2, P_3\) in terms of \( P_0 \):

\(P_0 + \frac{1}{2} P_0 + \frac{1}{4} P_0 + \frac{1}{8} P_0 = 1\)

Factoring out \( P_0 \),

\(P_0 \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \right) = 1\)

The sum inside the parentheses is,

\(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{8}{8} + \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{15}{8}\)

Thus,

\(P_0 \times \frac{15}{8} = 1 \quad \Rightarrow \quad P_0 = \frac{8}{15}\)

\(P_1 = \frac{1}{2} P_0 = \frac{1}{2} \times \frac{8}{15} = \frac{4}{15}\)

\(P_2 = \frac{1}{4} P_0 = \frac{1}{4} \times \frac{8}{15} = \frac{2}{15}\)

\(P_3 = \frac{1}{8} P_0 = \frac{1}{8} \times \frac{8}{15} = \frac{1}{15}\)

\(E(X) = 0 \times P_0 + 1 \times P_1 + 2 \times P_2 + 3 \times P_3\)

Substituting the values of \(P_0, P_1, P_2, P_3 \):

\(E(X) = 0 \times \frac{8}{15} + 1 \times \frac{4}{15} + 2 \times \frac{2}{15} + 3 \times \frac{1}{15} \)

\(E(X) = \frac{4}{15} + \frac{4}{15} + \frac{3}{15} = \frac{11}{15}\)

Hence correct option is 3).