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A stable equilibrium M-M-s queue has a Poisson process of departures with the same rate as its input. Its present population is also independent of past departures, a component of quasireversibility. The birth-death process satisfies detailed balance; reversed upward jumps have a constant arrival rate, so they are generated by an independent Poisson process.

Put $\mu ={\mu }_1+{\mu }_2$. In this heterogeneous two-server queue, the total population alone is not a Markov process: when only one customer is present, its departure rate depends on which server is occupied. Use states $0,a,b,2,3,\dots$, where $a$ and $b$ distinguish the two single-customer possibilities, and $n\ge 2$ records the total population. The memoryless property of the exponential distributions gives these nonzero transition rates:and $q_{n,n+1}=\nu$ for $n\ge 2$, $q_{n,n-1}=\mu$ for $n\ge 3$. In particular, a departure from state $2$ leaves the other server occupied.

The detailed balance equations suggestFor example, both ${\pi }_a\nu ={\pi }_2{\mu }_2$ and ${\pi }_b\nu ={\pi }_2{\mu }_1$ hold; these are the two transitions that would be lost by using only population as the state. All other adjacent-state detailed balance equations hold as well. Since $\nu /\mu <1$, the geometric series converges, and normalization givesThus the stationary distribution exists, and the requested probability isWith equal server rates ${\mu }_1={\mu }_2=m$, this reduces to ${\nu }^2/[m(2m+\nu )]$, as for an M-M-s queue with two servers.

To prove the output claim, put the continuous-time Markov chain in its stationary distribution. The preceding detailed balance equations make it a reversible Markov chain: time reversal of a continuous-time Markov chain has the same transition rates. A forward departure becomes a reversed arrival. In every reversed state, the total rate of such arrivals is $\nu$: from $0$ the rates $\nu /2$ add, from $a$ or $b$ the upward rate is $\nu$, and from every $n\ge 2$ it is also $\nu$. The reversed process can therefore be constructed with an independent rate-$\nu$ Poisson process of arrival epochs, choosing the idle server by a fair coin when appropriate. Consequently reversed arrival counts on disjoint time intervals are independent Poisson random variables with the corresponding means. Reversing those intervals proves that the stationary departure stream is a rate-$\nu$ Poisson process, rather than merely a stream with mean rate $\nu$.

Stationarity matters here. If the queue starts empty, a departure by time $h$ requires both an arrival and a completed service time, so its probability is $O(h^2)$; a rate-$\nu$ Poisson process would have probability $\nu h+O(h^2)$. Thus the output assertion is the equilibrium assertion in the context of the preceding stationary calculation.

First consider an M-M-K queue with $K$ servers, arrival rate $\lambda$, and service rate $\beta$ at each server. Assume $\lambda <K\beta$ so that a stationary distribution exists. Its population is a birth-death process with rates $q_{n,n+1}=\lambda$ and $q_{n,n-1}=\beta \min (n,K)$. The equilibrium probabilities satisfy detailed balanceThus the population process is a reversible Markov chain. In its time reversal of a continuous-time Markov chain, forward departures become upward jumps whose rate is the constant $\lambda$, independent of the state. They can be generated by a rate-$\lambda$ Poisson process independent of the initial reversed state. Reversing time preserves the Poisson process law, proving the equilibrium departure process is Poisson with rate $\lambda$, the Burke theorem for a multi-server queue.

The same argument gives the departure component of quasireversibility needed for the tandem calculation: the present upstream population is independent of its entire past departure process. Indeed, these departures become future arrivals in the reversed chain, independent of its initial population. First check this on finite time windows, then extend to the full past. A Poisson output alone, without this independence, would not justify multiplying queue-empty probabilities.

Now let $a=\nu \mu$, where $\mu$ denotes the mean service time at passport control, so the service rate at each of its two servers is $1/\mu$. The stationary distribution of the M-M-s queue with $s=2$ has normalizing sumHence its empty probability is $(2-a)/(2+a)$. Its output feeds a stable M/D/1 queue with rate $\nu$ and deterministic service time $\tau$. In a stationary work-conserving single-server queue, the busy fraction is the rate of completed work, $\nu \tau$, so the empty probability is $1-\nu \tau$.

The downstream stationary workload is a causal function of past upstream departures. For example, it is the limit of workloads constructed from arrivals in increasingly long finite past intervals; stability makes this stationary construction well defined. The backward independence just proved therefore makes the two present queue populations independent. Consequently

With walking time $\sigma$, the downstream arrival stream is the upstream departure stream shifted by $\sigma$. Its present workload depends only on departures up to time $-\sigma$. The number walking at time zero is the number of departures in $(-\sigma ,0]$, which has Poisson distribution of mean $\nu \sigma$ and is independent of the earlier departures. The upstream present population is independent of both parts of its past departure stream. Thus the three empty events factor, givingThe factor $e^{-\nu \sigma }$ is the empty probability of the deterministic-travel infinite-server stage. Both printed stability inequalities ensure the stationary calculations apply.

The number present is an M-M-s queue: births occur at rate $\lambda$, deaths at rate $\mu \min (n,s)$. Both rates are bounded, so the chain is nonexplosive. For $\lambda ,\mu >0$ and an integer $s\ge 1$, the detailed-balance weights with ${\pi }_0=1$ areTheir sum is finite precisely when $\lambda <s\mu$, giving positive recurrence and a stationary probability distribution after normalization.

For completeness, the birth–death recurrence criterion is ${\sum }_{n\ge 0}({\lambda }_n{\pi }_n{)}^{-1}=\infty$. It follows by solving the harmonic hitting-probability recurrence: successive differences are proportional to $({\lambda }_n{\pi }_n{)}^{-1}$, and divergence makes the probability of escaping to infinity before reaching zero vanish. The geometric tail of the weights makes this sum diverge when $\lambda \le s\mu$ and converge when $\lambda >s\mu$. Thus $\lambda =s\mu$ gives null recurrence, since the chain is recurrent but has no finite invariant normalization, and $\lambda >s\mu$ gives transience. The critical case is not positive recurrent merely because arrival and maximum service rates balance.

In equilibrium, a queue is quasireversible when its present state is independent of subsequent arrival times and preceding departure times, for each customer class. For an M-M-s queue, detailed balance and time reversal of a continuous-time Markov chain show that past departures become future arrivals generated independently of the present state. This backward independence is useful in tandem calculations where downstream occupancy is determined by earlier departures.