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Arrivals occur at rate $\lambda$ according to a Poisson process the service time have an exponential distribution with parameter $1/\mu$ in an M/M/1 queue, where $\mu$ is the mean service rate where $\mu\geq \lambda$ (stability condition).

According to Burke's theorem the departure process of an M/M/1 queue is a Poisson process with rate parameter $\lambda$ if the arrival follows a Poisson process with rate parameter $\lambda$.

Now, lets assume arrival rate is 2 jobs/sec ($\lambda=2$) and service rate is 3 jobs/sec ($\mu=3$). Then the departure rate should be 3 jobs/sec under the condition that all jobs leave the server after getting service. But according to the Burke's theorem 2 jobs/sec should be leaving the server. What I'm missing here?

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    $\begingroup$ I'm not sure about this, but if it is at a steady state, then shouldn't the arrivals be the same as the departures? Otherwise the queue wouldn't be steady. $\endgroup$
    – OFRBG
    Jul 25, 2015 at 21:46
  • $\begingroup$ We know that for stability the condition $\mu>\lambda$ is sufficient. $\endgroup$
    – Justin
    Jul 25, 2015 at 21:53
  • $\begingroup$ @precision If I may... please reread OFRBG's comment more carefully. $\endgroup$
    – Did
    Jul 25, 2015 at 22:01
  • $\begingroup$ No special explanations are needed here. If x job/sec arrive to the single server queueing system with one infinite capacity queue, not more that x job/sec can depart from it. Why in the M/M/1 queue it is exactly x job/sec, that depart from the system on average? It follows from the formulas. This is all because of exponential distributions. $\endgroup$
    – rrv
    Nov 30, 2021 at 7:33

3 Answers 3

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The OFRBG comment is the thing (we know the long term average departure rate must be less than or equal to the long term average arrival rate). But here is some more intuition:

Consider an $M/M/1$ queue with $0 < \lambda < \mu$. Define $\rho = \lambda /\mu$. The steady state distribution is $p_k = (1-\rho)\rho^k$ for $k \in \{0, 1, 2, \ldots\}$, where $p_k$ is the steady state probability of being in state $k$. Thus, $p_0=Pr[\mbox{idle}] = 1-\rho$. Now suppose we arrive at a time $t$ when the system is in steady state. Let $T$ be the remaining time to the next departure. Let's calculate $E[T]$ by conditioning on whether or not the system is busy at time $t$:

\begin{align} E[T] &= E[T|\mbox{busy}]\rho + E[T|\mbox{idle}](1-\rho)\\ &= (1/\mu)\rho + (1/\lambda + 1/\mu)(1-\rho) \\ &= 1/\lambda \end{align}

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  • $\begingroup$ Thanks a lot. I can see it now. $\endgroup$
    – Justin
    Jul 25, 2015 at 22:03
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The departure rate must equal the arrival rate at equilibrium. Although, the service rate is larger than the arrival rate, the server is not busy all of the time and cannot serve customers who aren't there. (Do you know how to find the proportion of time the server is idle in terms of $\lambda$ and $\mu$?)

Notice that Burke's theorem makes a statement about the ARRIVAL rate, not the service rate.

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The departure rate is $3$ jobs/sec when the server works. If the server is not working the departure rate is $0$ jobs/sec. On average the departure rate is $2$ jobs/sec.

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