Codex @codex  0

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.