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Suppose I have a queue with $\lambda$ and $\mu$. I can calculate the probability that there are 2 objects in the queue trivially, but how can I compute, for example, the probability that it takes an object less than $n$ units of time to be processed?

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    Just to clarify; are you considering the long term probabilities (ie when the system is in equilibrium)?2011-05-08
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    Yes. I think all the ones I've seen are like that anyway.2011-05-08
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    Since you're dealing with a $M/M/1$ queue you know that the service time is exponentially distributed with parameter $\mu$, which means that the service time $B$ has probability density function $f_{B}(t) = \mu e^{-\mu t}$. So $P(B \leq t)$ can be found. Does this answer your question?2011-05-08
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    @cardinal, I mixed up some terminology. Sorry about that.2011-05-08
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    @Stijn, no problem. I removed my comment since you fixed yours in time. Cheers.2011-05-08
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    @Stijin wait, how do I find P with the pdf?2011-05-08
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    I'll just repost my comment as an answer then, if you don't mind :x2011-05-08

2 Answers 2

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As you know from the comments, the processing (service) time of an M/M/1 queue is exponentially distributed, hence the probability is

$P(T_{serve} < t_n) = \int_0^{t_n} \mu \mathrm e^{-\mu t} \mathrm d t = 1-\mathrm e^{-\mu t_n}$

If you want the total waiting time, you will have to add to that the queueing time. See also Little's law.

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    What is the t_n?2011-05-08
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    The $t_n$ is in reference to the $n$ units of time in your original post, and $\mu$ is the parameter of the exponential distribution.2011-05-08
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Since you're dealing with a $M/M/1$ queue you know that the service time is exponentially distributed with parameter $\mu$, which means that the service time $B$ has probability density function $f_{B}(t)=\mu e^{-\mu t}$. So $P(B < t) = \int_0^{t} f_{B}(\tau)d \tau$.