Discussion
A canteen has a single cash counter. A customer has to buy coupons from the cash counter before getting the food items. Customers arrive at the counter according to a Poisson process with rate of 2 per 5 minutes. It takes 1.5 minutes on average to buy coupons. What is the long-run probability that the cash counter is free ?
Answer: C
The probability that the counter is free is \(1 - \dfrac{1.5}{2.5}\).
Treat the counter as an M/M/1 (or M/G/1) queue. The arrival rate is \(\lambda = 2/5 = 0.4\) per minute and the mean service time is 1.5 minutes, so the traffic intensity is
\[ \rho = \lambda \times 1.5 = 0.6 = \frac{1.5}{2.5} \]
In the long run the server is busy a fraction \(\rho\) of the time, so it is free with probability \(1 - \rho = 0.4\). 2.5 minutes is the mean time between arrivals. \(1 - 1.5/5\) wrongly uses 5 minutes as the inter-arrival time, \(1 - 1.5/2\) uses the arrival count, and 0 would mean the counter is always busy.
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