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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.