# Expected Waiting Time in a Queue with exponential distribution

I am trying to solve the following problem in my exercise sheet towards preparation of competitive exam:

A post office has 2 clerks. $$A$$ enters the post office while 2 other customers, $$B$$ and $$C$$, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has the Expo($$\lambda$$) distribution.

What is the expected time spent by $$A$$ in the post office?

My take:

Total time that A spends in the office is the sum of waiting time + the time taken to serve A.

$$E(\text{Total Time}) = E(\text{Waiting Time}) + E(\text{Service Time})$$

$$E(\text{Service Time}) = \frac{1}{\lambda}$$

Now I computed that the Probability that A is served at any of the counter is $$1/2$$. Therefore,

$$E(\text{Waiting Time}) = 1/2 \times E(\text{Service Time of B}) + 1/2 \times E(\text{Service Time of C})$$

$$=\frac{1}{\lambda}$$

Which gives me $$E(\text{Total Time}) = \frac{2}{\lambda}$$ But the answer given is $$\frac{3}{2\lambda}$$.

Please point out the mistake in what I am thinking or if you think there's a better way to do it.

Thanks.

## 1 Answer

The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $$B$$ and $$C$$. This is because $$A$$ takes the first available opening, so whoever finishes first, $$A$$ chooses that clerk. That is to say, if $$W_A$$ is the waiting time of $$A$$ and $$S_B$$, $$S_C$$ are the service times of $$B$$ and $$C$$ respectively, then $$W_A = \min(S_B, S_C),$$ and it is not generally true that $$\operatorname{E}[W_A] = \operatorname{E}[S_B]/2$$.

Rather, you must compute the minimum order statistic of $$B$$ and $$C$$'s service times. Since these are iid exponentially distributed with rate $$\lambda$$, it follows that $$\Pr[W_A > t] = \Pr[\min(S_B, S_C) > t] = \Pr[(S_B > t) \cap (S_C > t)] \overset{\text{ind}}{=} \Pr[S_B > t] \Pr[S_C > t] \overset{\text{id}}{=} (e^{-\lambda t})^2.$$ That is to say, the probability that $$A$$ waits more than $$t$$ to begin service is the product of the probabilities that both $$B$$ and $$C$$ take more than $$t$$ to be finished. And this makes perfect intuitive sense--because if $$A$$ must wait more than $$t$$, then that means both $$B$$ and $$C$$ are still occupied, which means their service times are also more than $$t$$.

Now that you know $$\Pr[W_A > t] = e^{-2\lambda t}$$, what can you say about $$\operatorname{E}[W_A]$$?

• Many thanks to you :) – Vizag May 3 at 22:33
• You say "it is not generally true that $\operatorname{E}[W_A] = \operatorname{E}[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time. – Henry May 3 at 22:47
• @Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct. – heropup May 4 at 1:00