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A closed migration process is a continuous-time Markov chain with state $n=(n_1,\dots ,n_m)\in {\mathbb{Z}}_{+}^m$, conserved population ${\sum }_i{n}_i=N$, and transition ratesHere ${\varphi }_i(0)=0$, ${\varphi }_i(k)>0$ for $k>0$, and $P=(p_{ij})$ is a fixed stochastic routing matrix. Take $p_{ii}=0$ for convenience; self-routing can instead be omitted from the actual jumps. Assume irreducible routing. Let positive traffic weights satisfy ${\theta }_j={\sum }_i{\theta }_i{p}_{ij}$. Define $d_i(0)=1$, $d_i(k)={\prod }_{h=1}^k{\varphi }_i(h)$. The product-form stationary distribution of a closed migration process isTo prove it, an incoming transition $i\to j$ to $n$ comes from $m=n+e_i-e_j$, and is possible only if $n_j>0$. The weight ratio givesSumming over $i$ and using the traffic equations gives $\pi (n){\varphi }_j(n_j)$. Summing then over $j$ equals $\pi (n){\sum }_j{\varphi }_j(n_j)$, the outgoing probability flux, proving global balance for a continuous-time Markov chain. The state space is finite and irreducible, so this normalized stationary distribution is unique. Detailed balance is not required; a directed routing cycle, for example, is not reversible.

For the switchboard, make each of the $N$ lines one circulating individual. Use colonies $F,O,C$ for free lines, calls waiting for or receiving the operator's connection service, and already connected calls. The transitions form the cycle $F\to O\to C\to F$, with departure ratesThe first rate is $\nu$, not $f\nu$: incoming calls form one external Poisson process, not one per free line. An arrival when $f=0$ changes no state and is counted as lost. Exponential connection and conversation times make this the required continuous-time Markov chain; only one call receives the operator's service at a time, whereas every connected call independently completes at rate $\mu$.

The cyclic traffic weights can all be chosen equal to one. Multiplying all unnormalized weights by the same ${\nu }^N$ gives, for $f+k+i=N$,If $n=k+i$ is the number of occupied lines, sum over the $n+1$ possibilities for the connected-call count:Thus the normalizer is ${\sum }_{n=0}^{N}H(n)$ and the full-state probability is $H(N)/{\sum }_{n=0}^{N}H(n)$. By Poisson arrivals see time averages, this is also the long-run fraction of attempted calls rejected:One can also see the arrival-rate argument directly: the lost-call counting process has conditional intensity $\nu 1_{\{F=0\}}$, so its equilibrium rate is $\nu \pi (F=0)$; dividing by the total attempted-call rate $\nu$ gives the same fraction. As an independent check, stationary flow conservation in a finite-line switchboard gives $\nu (1-P_{\mathrm{loss}})=\lambda \mathrm{Pr}\{O>0\}=\mu \mathbb{E}[C]$.