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 $\lambda = 2/3$.

  • $\begingroup$ Since the λ is 2/3, so while calculating the P(queue empty) = 1- λ, the probability should be 1 - 2/3 = 1/3 $\endgroup$ – Srishti Bindal Feb 5 at 19:27
  • $\begingroup$ 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 :) $\endgroup$ – Sam T Feb 6 at 9:32

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