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Represent $N$ telephone lines by a closed migration process with free-line, operator and automated-service colonies. Accepted arrivals have rate $\nu$ when a free line exists, $c$ operators complete at rate $\lambda \min (k,c)$, and $i$ automated calls complete at rate $\mu i$. Define $D_c(k)={\prod }_{h=1}^k\min (h,c)$ andThe product-form stationary distribution of a closed migration process and Poisson arrivals see time averages give the displayed blocking probability. This extends a single-operator model without treating the independent operators as a server with rate $c\lambda$ when fewer than $c$ calls are present.

// Target: probability-and-statistics.bigb

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 $\nu (1-P_{\mathrm{loss}})=\lambda \mathrm{Pr}\{O>0\}=\mu \mathbb{E}[C]$.

Use one particle for each telephone line, including unused lines. Let $f$ be the number of free lines, $h$ the number of callers waiting for or being checked by an operator, and $a$ the number using automated service. Then $f+h+a=N$. These are the three colonies of a closed migration process, with directed cycle $f\to h\to a\to f$ and departure ratesRejected attempted calls leave the state unchanged. They must not be represented as extra customers in the waiting stage.

Equivalently the state is $(h,a)$ with $h+a\le N$. Its transitions areThe directed cycle has equal traffic weights; choosing each weight to be $\nu$ gives the unnormalized product-form stationary distribution of a closed migration processWe can verify global balance for a continuous-time Markov chain directly. Weighted incoming transitions from $(h-1,a)$, $(h+1,a-1)$ and $(h,a+1)$ contribute, respectively,Their sum equals the weight times the outgoing rate. This proves stationarity after normalization, without falsely asserting reversibility of the directed migration cycle.

For $n=h+a$ occupied lines, summing over the automated population givesThe normalizing constant is $G_N={\sum }_{n=0}^{N}H(n)$. The attempted calls form an independent Poisson process, so Poisson arrivals see time averages: the rate of attempted arrivals seeing a full state is $\nu$ times its stationary probability, and division by the total attempted-arrival rate gives the loss proportion. ConsequentlyThe numerator is essential; the PDF contains it even though the converted TeX loses it.

With two independent operators, the total check-completion rate is $\lambda \min (h,2)$. In particular it is only $\lambda$ when one caller is present. DefineThe human-stage weight becomes $(\nu /\lambda {)}^h/D_2(h)$, while the free and automated weights are unchanged. The same global-balance calculation applies using these departure rates. ThusThis is the blocking probability in a finite-line sequential-service network with two operators. It remains valid for $N=0$, when every attempted call is lost; for $N=1$ it coincides with the single-operator answer.

Assume $\nu ,\lambda ,\mu >0$ and regard each line as an individual circulating through three colonies. Then $n$ is a closed migration process on $\{n\in {\mathbb{Z}}_{\ge 0}^3:n_1+n_2+n_3=N\}$. With $e_j$ the coordinate vectors, its nonzero transition intensities areThe memoryless property of the exponential distributions and the independent Poisson process make these counts a continuous-time Markov chain. In particular, the accepted-arrival rate is $\nu$, not $\nu n_1$: free lines are availability tokens, not independent callers. An attempted arrival when $n_1=0$ is lost and causes no state change, so it does not enter the off-diagonal generator.

For the cyclic routing $1\to 2\to 3\to 1$, take equal traffic weights at all colonies. The product-form stationary distribution of a closed migration process gives weightsHere is a direct global balance for a continuous-time Markov chain verification. Set ${\varphi }_1(k)=\nu 1_{\{k>0\}}$, ${\varphi }_2(k)=\lambda 1_{\{k>0\}}$, ${\varphi }_3(k)=\mu k$, with cyclic subscripts. The predecessor for a $j\to j+1$ transition is $m=n+e_j-e_{j+1}$, allowed when $n_{j+1}>0$. The product weights obeySumming these incoming weights over the three edges gives $w(n){\sum }_j{\varphi }_j(n_j)$, precisely the outgoing weight. Thus $\pi (n)=w(n)/G_N$ is stationary. The finite state space is irreducible, so this stationary distribution is unique.

Let $k=n_2+n_3$ be the occupied-line count. Summing over its automated-service count $i=n_3$ givesso $G_N={\sum }_{k=0}^{N}H(k)$. The external Poisson process sees the stationary state probabilities by Poisson arrivals see time averages. A call is lost exactly when $n_1=0$, equivalently $k=N$. Therefore

For two independent identical operators, the middle transition rate becomes $\lambda \min (n_2,2)$: one call receives one operator's service, while two or more calls keep both operators busy. Define $D_2(0)=1$ and $D_2(m)={\prod }_{h=1}^m\min (h,2)=2^{m-1}$ for $m\ge 1$. The same global balance for a continuous-time Markov chain calculation applies toConsequently the blocking probability in a finite-line sequential-service network isThe subtraction corrects the $n_2=0$ term, for which $D_2(0)=1$ rather than $2^{-1}$. In particular $H_2(0)=1$, and when $N\le 1$ replacing one operator by two does not change blocking.

For an irreducible closed migration process with finite population $N$, positive departure rates and routing matrix $P$, choose positive traffic weights satisfying $\theta =\theta P$. On ${\sum }_j{n}_j=N$, its stationary distribution has the displayed form. For a predecessor $m=n+e_j-e_k$, its weighted transition into $n$ is $\pi (n)({\theta }_j/{\theta }_k)P_{jk}{\varphi }_k(n_k)$. Summing over $j$ gives $\pi (n){\varphi }_k(n_k)$ by the traffic equation, and summing over $k$ proves global balance for a continuous-time Markov chain. The distribution need not satisfy detailed balance.

// Target: probability-and-statistics.bigb

In a one-operator closed migration process representing a finite-line switchboard, let $K$ count calls waiting for or receiving the operator's connection service and let $I$ count connected calls. At a stationary distribution, the expected drift of $K$ is $\nu \mathrm{Pr}\{K+I<N\}-\lambda \mathrm{Pr}\{K>0\}=0$, and that of $I$ is $\lambda \mathrm{Pr}\{K>0\}-\mu \mathbb{E}[I]=0$. Poisson arrivals see time averages identifies the full-state probability with $P_{\mathrm{loss}}$. Thus the accepted arrival rate, operator throughput and completed-call rate coincide. This is a useful independent check on the product-form stationary distribution of a closed migration process.