# Queuing Theory m/m/1 system

1. A fast-food restaurant has one drive-through window. An average of 40 customers per hour arrives at the window. It takes an average of 1 minute to serve a customer. Assume that interarrival and service times are exponential.

Q.What is the probability that a customer will have to wait before being served?

I suggest having a read of the book Stochastic Networks by Frank Kelly and Elena Yudovina. In particular, you can find a section titled "$$M/M/1$$ Queues" on page 22 of the book. There the invariant distribution is determined: writing $$\lambda = 40/60 = 2/3 < 1$$ for the arrival rate (with departure rate $$1$$), it is given by $$\pi(j) = (1 - \lambda) \cdot \lambda^j \quad\text{for}\quad j = 0,1,2,... \, .$$ This is valid if and only if $$\lambda < 1$$. (For $$\lambda > 1$$ one can show that no invariant distribution exists: there is a unique invariant measure, up to scaling, but this measure is not summable.)
Now, once we know the invariant distribution, we're done: "customer wait" equals "queue non-empty". So, in equilibrium, $$P(\text{customer does not have to wait}) = P(\text{queue empty}) = \pi(0) = 1 - \lambda,$$ and hence your probability in question is $$\lambda = 2/3$$.
• I think there's a little bit of confusion with the terminology. Here "empty" means that the person won't have to wait. Your OP asks "what is the probability that a customer will have to wait". "Will wait" means "non-empty", and $P(\text{non-empty}) = 1 - P(\text{empty})$. Does that clear it up for you? If not, feel free to ask for clarification :) – Sam T Feb 6 at 9:32