Question

In a single pump petrol station, vehicles arrive at the rate of 20 customers per
hour and petrol filling takes 2 minutes on an average. Assume the arrival rate is
Poisson probability distribution and service rate is exponentially distributed,
determine
(a) What is the probability that no vehicles are in the petrol station?
(b) What is the probability that 1 customer is filling and no one is waiting in the queue?
(c) What is the probability that 1 customer is filling and 2 customers are waiting in the
queue?
(d) What is the probability that more than 2 customers are waiting?

Answer 1

To solve this problem, we can use the M/M/1 queuing model, where "M" stands for Poisson arrivals and exponentially distributed service times with a single server.

Given:
Arrival rate (λ) = 20 customers per hour
Service rate (μ) = 30 customers per hour (60 minutes divided by 2 minutes per customer)

(a) Probability that no vehicles are in the petrol station:
Since the system is in steady state, we can use the formula for the utilization factor (ρ) in an M/M/1 system:
ρ = λ / μ
ρ = 20 / 30 = 2/3

The probability that the server is busy (P0):
P0 = 1 - ρ = 1 - 2/3 = 1/3

(b) Probability that 1 customer is filling and no one is waiting in the queu:
The probability that the system is in state 1 (P1):
P1 = ρ(1-ρ)

(c) Probability that 1 customer is filling and 2 customers are waiting in the queue:
The probability that the system is in state 3 (P3):
P3 = (ρ^3 / 2!) * (1-ρ)

(d) Probability that more than 2 customers are waiting:
The probability that the system is in states 3, 4, 5, ... ∞ (P>2):
P>2 = 1 - (P0 + P1 + P2) = 1 - (1/3 + ρ(1-ρ) + (ρ^2 / 2!) * (1-ρ))

By plugging the values of ρ and calculating, you can find the exact probabilities for each situation.