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For a single resource, assume unit-capacity calls arrive as a Poisson process of rate $a$, holding times have independent exponential distributions with mean $h$, and a call is rejected when all $C$ units are occupied. Let $\nu =ah$. Its occupancy is a birth-death process on $\{0,\dots ,C\}$, with birth rate $a$ below capacity and death rate $k/h$ in state $k$. Detailed balance for a birth-death process givesBy Poisson arrivals see time averages, the Erlang loss formula isThis derivation assumes exponential holding times; the standard insensitivity result extends the same formula to independent holding times of the same mean, but is not needed here.

A loss network with fixed routing has finitely many resources $j$, capacities $C_j$, and call types $r$ with a fixed nonempty set of resources. Type $r$ has Poisson arrivals of rate $a_r$, independent holding times of mean $h_r$, and offered load ${\nu }_r=a_r{h}_r$. It is accepted only if every resource on its route has a free unit, and otherwise is lost. In the displayed equations use the unit-incidence matrix $A_{jr}=1$ for $j\in r$ and zero otherwise. The equations concern this unit-demand model, rather than arbitrary multi-unit requirements.

The Erlang fixed point approximation replaces the correlated resource availabilities by independent ones. A type-$r$ call survives blocking at resources other than $j$ with approximate probability ${\prod }_{i\in r\setminus \{j\}}(1-B_i)$. Treating this filtering as Poisson thinning gives the reduced offered loadApply the single-resource formula to obtain $B_j=E({\rho }_j,C_j)$. Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.

Assume first $C_j\ge 1$. The right side is a continuous map from $[0,1{]}^J$ into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let $m_C(\rho )=\rho [1-E(\rho ,C)]$ be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yieldsThus $E$ increases continuously from zero to one and $m_C$ increases from zero to $C$.

Write $x_j=-\log (1-B_j)\ge 0$, and define ${\rho }_C(x)=E(\cdot ,C{)}^{-1}(1-e^{-x})$ and ${\ell }_C(x)=m_C({\rho }_C(x))$, with ${\ell }_C(0)=0$. Each ${\ell }_C$ is continuous and strictly increasing. A fixed point satisfiesThese are precisely the stationary equations for the convex potential for the Erlang fixed pointThe first term is convex; the second is strictly convex because each ${\ell }_{C_j}$ strictly increases. Hence $F$ is strictly convex. Its gradient cannot vanish at two distinct points: the difference of gradients dotted with the difference of points is strictly positive, with a positive contribution from every changed coordinate in the integral term. This also covers boundary coordinates $x_j=0$, using the right derivative there. Every fixed point has $B_j<1$, since its offered load is finite, so the change of variables is valid. There is exactly one fixed point. If a capacity is zero, its blocking probability is one; delete it and all routes using it before applying the argument to the remaining positive capacities. Unused resources have blocking probability zero.

A birth-death chain on ${\mathbb{N}}_0$ has generator entries $q_{n,n+1}={\lambda }_n$, $q_{n,n-1}={\mu }_n$ for $n\ge 1$, ${\mu }_0=0$, $q_{nn}=-({\lambda }_n+{\mu }_n)$, and zero elsewhere. Assume the usual conservative nonexplosive chain. The detailed balance for a birth-death process are ${\pi }_n{\lambda }_n={\pi }_{n+1}{\mu }_{n+1}$.

If they hold, each incoming term cancels an outgoing term, giving $\pi Q=0$ and thus invariance. Conversely the invariance equation at zero gives ${\pi }_0{\lambda }_0={\pi }_1{\mu }_1$. At state $n$, write the stationarity equation as equality of the neighbouring probability currentsInduction from the zero boundary current gives detailed balance for a birth-death process for every edge. Thus invariant measures are exactly the detailed-balance measures in this standard birth–death setting. The argument is the generator characterization; a stationary probability law additionally needs finite normalization. For potentially explosive processes, formal generator balance alone would require additional boundary qualifications.

Put $K=N+s$, the total capacity. The customer count is an M-M-s-K queue, hence a birth-death process on $\{0,1,\dots ,K\}$. Its nonzero transition rates areThe first rate is zero at $K$ because an arrival finding the shop full is rejected. The second rate is the number of busy servers times the individual service rate.

Write $\rho =\lambda /\mu$. The detailed balance for a birth-death process recursion isThereforewhereThis is the stationary distribution of an M-M-s-K queue.

The state space is finite, and every state communicates with every other through successive births and deaths because $\lambda ,\mu >0$. The chain is therefore an irreducible Markov chain, so its invariant distribution is unique. Finally the adjacent-state identitiesare detailed balance, while all nonadjacent rates vanish in both directions. Part a now shows that the queue is reversible in equilibrium.

The queue length is an M-M-1 queue and hence a birth-death process with birth rate $\lambda$ and death rate $\mu$ away from zero. The detailed-balance equations areWriting $\rho =\lambda /\mu$, they give ${\pi }_n={\pi }_0{\rho }^n$. Since $\rho <1$, this measure is summable and normalization yieldsThe existence of this invariant probability distribution for the irreducible nonexplosive chain proves positive recurrence.

For $i\ge 1$, the total holding rate is $(\lambda +1)2^i$, and the jump chain moves upward and downward with probabilitieswhile it moves from zero to one with probability one. Thus it is a reflected nearest-neighbour random walk. Recurrence is unaffected by the holding times, so it is recurrent exactly when $p\le q$. In the given range,For $\lambda >1$ the jump chain is transient.

The detailed-balance equations arewhere the formula also covers the boundary $i=0$. Hence every candidate invariant measure hasAt $\lambda =1$ this is summable and the chain is nonexplosive, givingFor $\lambda =2$ it is not summable. When $1<\lambda <2$, it is a summable formal solution of $\pi Q=0$, but the chain explodes, so it is not an invariant probability for the minimal process; this is precisely the caveat in invariant distribution of an explosive chain. Consequently

It remains to classify explosion. At $\lambda =1$, the recurrent jump chain visits zero infinitely often. The holding time at zero has rate one, so nonexplosion from recurrent visits to a slow state shows that the accumulated time is infinite almost surely.

For $\lambda >1$, let $V_i$ be the total number of visits of the transient jump chain to $i$. Using the stated uniform visit bound and writing $q_i=(\lambda +1)2^i$ for $i\ge 1$,because ${\sum }_i{2}^{-i}<\infty$. Conditional on the jump path, this is the expected sum of all holding times, so the total lifetime is finite almost surely. This is explosion of an upward-biased walk with geometrically increasing rates. Therefore

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M-M-1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.

Put $\rho =\lambda /\mu$. The unique stationary distribution of an $M/M/s/K$ queue iswhere $Z$ is the sum of the displayed weights. Adjacent terms satisfy detailed balance for a birth-death process, so the queue is reversible in equilibrium.