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In an $M/M/1/2$ queue what is the probability that a packet arrives and finds one packet in the queue, given that we have an arrival rate of $\lambda$ and departure rate $\mu$.

I have thought of a number of possible solutions. One of them being simply calculating the probability of having two packets in the system as $(1-\rho)\rho^2$. Or the probability that a packet arrives and the previous one hasn't left yet as $\frac{\lambda}{\mu + \lambda}$.

What would be the correct way of thinking about it?

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  • $\begingroup$ What do you mean by "the probability that a packet arrives and finds one packet in the queue"? Over a certain period of time? Or do you mean: For any given packet, what is the probability that when it arrives it finds one packet in the queue? $\endgroup$
    – joriki
    Commented Dec 12, 2019 at 0:04

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Assuming you are referring to the steady state of the system, we have the balance equations \begin{align} \lambda\pi_0 &= \mu\pi_1\\ \lambda\pi_1 &= \mu\pi_2, \end{align} from which $\pi_1 = \frac\lambda\mu \pi_0$ and $\pi_2 = \left(\frac\lambda\mu\right)^2\pi_0$. Let $\rho = \frac\lambda\mu$. From $\pi_0+\pi_1+\pi_2=1$ we have $$ \pi_0(1 + \rho + \rho^2) = 1 \implies \pi_0 = \frac1{1+\rho+\rho^2}. $$ It readily follows that $\pi_1 = \frac\rho{1+\rho+\rho^2}$ and $\pi_2 = \frac{\rho^2}{1+\rho+\rho^2}$. The probability that an arriving packet finds one packet in the system is exactly $\pi_1$.

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  • $\begingroup$ Shouldn't it be $\lambda\pi_1=2\mu\pi_2$? When there are two customers, the departure rate applies to each of them. $\endgroup$
    – joriki
    Commented Dec 12, 2019 at 0:39
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    $\begingroup$ There is only one server though, so the departure rate would only apply to the customer being served. $\endgroup$
    – Math1000
    Commented Dec 12, 2019 at 0:41

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