A drive-through car-wash service has one station for washing cars. Cars arrive according to a Poisson process with rate $\lambda = 10$ per hour. If the service time is approximately a fixed value of $4$ minutes, what is the expected wait in queue?

My attempt: On average, one person arrives per $6$ minutes, while the service rate is fixed at $4$ minutes, we infer that on average, nobody arrives would need to wait for service. Thus the expected waiting time in queue is $0$.

My question: Is my thought above correct? Note that we are not given the information about the distribution of the service time.

  • $\begingroup$ No it is not correct. The minimum waiting time is zero. The expected (average) waiting time must necessarily be longer unless literally nobody ever has to wait. You are given information about the distribution of the service time. You are told that it's 'approximately a fixed value of four minutes' which is to say that you are to treat the service time as exactly four minutes for every customer (i.e. the service time is deterministic. its distribution is an atom at 4 minutes). $\endgroup$ – spaceisdarkgreen Mar 12 '17 at 23:15
  • $\begingroup$ @spaceisdarkgreen: thank you so much for your thought. Is the correct answer = 1/15 hours = 4 mins then? The formula for $W_q$ is here: en.wikipedia.org/wiki/M/D/1_queue $\endgroup$ – user177196 Mar 12 '17 at 23:23
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    $\begingroup$ yes that looks right $\endgroup$ – spaceisdarkgreen Mar 13 '17 at 1:00
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    $\begingroup$ I get $8/7$ for the mu parameter and $49/16$ for $k$ (that way the mean is mu*k = 7/2), but maybe we're using different parametrization. Not sure if it's a bad sign that you don't get an integer $k$. It's certainly a well-defined distribution (it's more commonly called a Gamma distribution). When $k$ is not an integer, you can't express it as the sum of $k$ exponentials. Don't know enough to know if that's a problem here. Would depend if the model of the service time is as the sum of exponentials (i.e. service completes after a poisson process clicks $k$ times). $\endgroup$ – spaceisdarkgreen Mar 14 '17 at 1:24
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    $\begingroup$ I used the same parametrization as in the side bar here en.wikipedia.org/wiki/Erlang_distribution . with mu = 1/lambda. I'm gonna punt on the second point. If you know it's the sum of exponentials, then yeah, $k=3$ is really close and that's probably it (maybe $3.5$ or $4$ is an approximate value). But I don't have any expertise here. $\endgroup$ – spaceisdarkgreen Mar 16 '17 at 0:13

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