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 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

Correct option is C

Concept:

The value of  E(X)=0×P₀+1×P₁+2×P₂+3×P₃

Solution:

we have,

A single server (the petrol dispensing unit),

Customers arrive according to a Poisson process with rate λ=1 per minute,

Service times are exponentially distributed with mean 12 minutes (service rate μ=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  λ=1, service rate μ=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 λ=1 ,

Service rate μ=2.

Let Pn be the steady-state probability that there are n customers in the system. We need to find

P₀P₁,P₂,P₃ the probabilities for 0, 1, 2, and 3 customers in the system.

Using the birth-death process relations, we get

​Thus,​

Now, since the total probability must sum to 1,

P₀+ P₁+ P₂+ P₃= 1

Substituting the values of P₁,P₂,P₃ in terms of P₀:

​Factoring out P₀,

​Hence correct option is (c).​