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A loss network models calls that require several resources simultaneously and are rejected if any required capacity is unavailable; rejected calls do not queue. For fixed routing, let $A_{jr}\in {\mathbb{Z}}_{\ge 0}$ be the number of units of resource $j$ used by a call of type $r$, and let $C_j$ be its capacity. Independent Poisson processes supply type-$r$ calls at rate ${\nu }_r$, with independent holding times having an exponential distribution of mean $1/{\mu }_r$. Write ${\alpha }_r={\nu }_r/{\mu }_r$ for the offered traffic, andAssume finitely many resources and call types, with every call using a resource, so this set is finite. The occupancy continuous-time Markov chain has ratesIts stationary distribution isIndeed, for each feasible upward transition,which is detailed balance for a continuous-time Markov chain. Thus this is a reversible Markov chain. With positive arrival rates it is irreducible on $\mathcal{N}(C)$: departures reach the empty state, and any feasible state can be assembled by arrivals. Its stationary distribution is consequently unique.

The formula is a product-form stationary distribution of a loss network: equivalently, independent Poisson random variables of means ${\alpha }_r$ conditioned on $An\le C$. The conditioning couples the occupancies, so the resources are generally not independent. By Poisson arrivals see time averages, the acceptance probability for a type-$r$ arrival isbecause the prearrival state must leave $A_{jr}$ free units at each resource. Set the numerator to zero if its capacity vector has a negative entry. This exact formula is often expensive to evaluate, which motivates the Erlang fixed point approximation.

The same occupancy stationary distribution extends to independent general holding-time distributions with these means by insensitivity of loss networks; the exponential assumption above makes the occupancy process itself a continuous-time Markov chain and permits the direct detailed balance proof.

For a fixed occupancy $n$, let $x_r>0$ be the per-flow rate on every active route, so feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. A proportionally fair allocation is a feasible allocation satisfying, for every other feasible $\tilde{x}$,The sum counts the proportional changes over individual flows. By the first-order condition for a concave function, it is exactly the allocation solvingThis is weighted logarithmic utility optimization, and its negative objective is a strictly convex function, which gives unique rates on active routes. Routes with $n_r=0$ have no per-flow rate to determine.

Index the four routes by $1=\{1,2\}$, $2=\{2,3\}$, $3=\{3,4\}$, $4=\{4,1\}$, and let ${\Lambda }_r=n_r{x}_r$ be each route's total service. WriteThe four capacity constraints say ${\Lambda }_1+{\Lambda }_2\le 1$, ${\Lambda }_2+{\Lambda }_3\le 1$, ${\Lambda }_3+{\Lambda }_4\le 1$, and ${\Lambda }_4+{\Lambda }_1\le 1$. Equivalently,For fixed maxima $u,v$, every active route in the first group optimally takes total service $u$, and every active route in the second takes $v$, because its logarithmic objective is increasing. Up to the constant $-{\sum }_r{n}_r\log n_r$, the objective therefore becomes $P\log u+Q\log v$ with $u+v\le 1$. Its maximizer is $u=P/N$, $v=Q/N$ when $N>0$, with the evident endpoint choices when one group is empty. Hence the proportionally fair allocation on a four-cycle isSet ${\Lambda }_r=0$ on inactive routes. If $N=0$, all service totals are zero. In particular, for $n_1>0$, $n_1{x}_1=P/N$, as required. Two disjoint active routes in the same group can each receive the same total service; service totals need not sum to one over the entire network.

With independent document arrivals given by Poisson processes of rates ${\rho }_r$ and independent unit-mean exponential distributions for document sizes, each of the $n_r$ active documents has completion rate $x_r$. The flow-level network model is consequently a continuous-time Markov chain withAt $n=0$ all downward rates are zero. The total departure rate is at most four, so the bounded total jump rate ensures no explosion.

The original PDF's displayed factor is a binomial coefficient, not the plain quotient produced by the TeX extraction. The candidate stationary distribution for this reversible four-cycle flow model isUse $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$. For $r\in \{1,3\}$ and $n_r>0$,so $\pi (n)q(n,n-e_r)=\pi (n-e_r){\rho }_r$. For $r\in \{2,4\}$ the corresponding ratio is ${\rho }_{r}N/Q$. Thus detailed balance for a continuous-time Markov chain holds for every adjacent pair, proving the displayed stationary distribution whenever $B<\infty$.

For completeness, its existence condition can be made explicit. Put $\alpha =\max ({\rho }_1,{\rho }_3)$ and $\beta =\max ({\rho }_2,{\rho }_4)$. At fixed $P,Q$,Therefore the sum over states with $N=m$ is at most $(m+1{)}^2(\alpha +\beta {)}^m$, using the binomial theorem. This is summable if $\alpha +\beta <1$. Conversely, restricting to the single maximizing route in each group leaves the series ${\sum }_{m\ge 0}(\alpha +\beta {)}^m$, which diverges if $\alpha +\beta \ge 1$. HenceThese are exactly the four strict resource-load inequalities. With positive arrival rates the Markov chain is irreducible, so the normalized stationary distribution is unique.

A continuous-time Markov chain is a reversible Markov chain in equilibrium if, when started in a stationary distribution $\pi$, the process $(X_t:0\le t\le T)$ has the same finite-dimensional distributions as $(X_{T-t}:0\le t\le T)$ for every $T>0$.

For transition rates $q_{ij}$, the detailed balance for a continuous-time Markov chain equations areThey imply invariance directly. For each state $j$,because $q_{jj}=-{\sum }_{k\ne j}{q}_{jk}$. Thus $\pi Q=0$, so $\pi$ is an invariant distribution of a continuous-time Markov chain.

For a loss network with feasible occupancies $An\le C$, its stationary distribution is proportional to ${\prod }_r{\alpha }_r^{n_r}/n_r!$. It is the law of independent Poisson random variables conditioned on the resource constraints. With exponential holding times, detailed balance for a continuous-time Markov chain proves the formula directly.