I was reading about the M/M/1 Queue and that we assume new customers arrive according to a Poisson distribution, and each customer takes an amount of time to service that is drawn from an exponential distribution?

I understand what it means to have an arrival rate based on Poisson distribution - if an average of 10 come per hour, you'll have 9 a few times, 6 very infrequently, 11 pretty often etc - but what does it mean that service is exponential?

Does this mean that as you serve more customers, the average time it takes to service them rises exponentially? Wouldn't the system crash soon after a few customers?

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    $\begingroup$ Google for 'exponential distribution'. $\endgroup$ – reinierpost Jun 30 '14 at 8:13
  • $\begingroup$ I understand what Exponential Distribution is, but not in this context $\endgroup$ – CodyBugstein Jun 30 '14 at 11:09
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    $\begingroup$ Then I don't understand what it is you don't understand. Can you make the question more specific? $\endgroup$ – reinierpost Jun 30 '14 at 13:57
  • $\begingroup$ @reinierpost Not sure how to make it more specific... I'm asking, what does it mean that service time is exponential? Relative to what? If I tell you grades in a high school math class followed an exponential distribution, that might mean that a low [but gradually rising] number of students scored under 85 and then a rapidly increasing amount scored between 85 and 100 (with 100 being the most common grade). What does it mean in terms of M/M/1? Does it mean that as I serve more customers, the time it takes to serve them shoots up like an exponential curve? $\endgroup$ – CodyBugstein Jun 30 '14 at 14:11
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    $\begingroup$ @Imray The exponential distribution is not the same thing as exponential growth. If somebody says "Google something", you should always Google it before asserting that you don't need to because you already understand that thing... $\endgroup$ – David Richerby Jun 30 '14 at 14:31

In an exponential distribution, the probability is DECREASING exponentially -- hence little risk that service time will grow. On the contrary, the mean service time is fixed, at the inverse of the rate. This distribution is actually very rare in practice, but is chosen because it is Markovian: the probability to terminate servicing a client does not depend on the past, and in particular on how much time you already spent servicing that client. That yields very nice mathematical properties.

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    $\begingroup$ Indeed. The distribution is sometimes called the "negative exponential distribution" to emphasize that it's exponential decrease, not increase. $\endgroup$ – David Richerby Jun 30 '14 at 14:33
  • $\begingroup$ the probability is DECREASING exponentially... the probability of what? $\endgroup$ – CodyBugstein Jun 30 '14 at 20:07
  • $\begingroup$ The probability that it will take time $t$ to serve a customer is proportional to $e^{-\lambda t}$ for some $\lambda$. See Wikipedia. $\endgroup$ – Peter Shor Jun 30 '14 at 22:40

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