To solve this problem, we need to use the M/M/1 queuing model, which assumes a single server (cashier) and exponential distribution for both inter-arrival times and service times.
First, let's calculate the arrival rate (λ) and service rate (μ).
Given:
- Average arrival rate (λ) = 10 customers / 30 minutes
Since λ is the average rate over a time interval, we can convert it to the rate per unit time:
λ = 10 customers / 30 minutes = (10/30) customers per minute = 1/3 customers per minute
- Average service rate (μ) = 1 customer / (2.5 minutes)
To find the service rate per minute, we take the reciprocal of the average service time:
μ = 1 customer / 2.5 minutes = 2/5 customers per minute
Now, let's calculate the traffic intensity (ρ), which is the ratio of arrival rate to service rate:
ρ = λ / μ = (1/3) / (2/5) = 5/6
Since ρ > 1 for a queue length to exceed six, the queue length can never exceed six.
Next, let's find the expected time spent by a customer in the system (T) using the Little's Law formula:
T = L / λ
where L is the average number of customers in the system (queue + being served).
The expected number of customers in the system is given by:
L = ρ / (1 - ρ)
Substituting the value of ρ, we get:
L = (5/6) / (1 - 5/6) = 5
Finally, substituting the value of L and λ into the Little's Law formula:
T = L / λ = 5 / (1/3) = 15 minutes
Therefore, the probability that the queue length exceeds six is 0, and the expected time spent by a customer in the system (queue + being served) is 15 minutes.
In a supermarket, the average arrival rate of customers is 10 every 30 minutes, following Poisson process. The average time taken by a cashier to list and calculate the customer’s purchase is two and a half minutes following exponential distribution. What is the probability that the queue length exceeds six? What is the expected time spent by a customer in the system?
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