= Solution
A <loss network> models calls that need several resources simultaneously and are rejected, rather than queued, when insufficient capacity remains. Take finite resource and route sets. Let $A_{jr}\geq0$ be the integer amount of resource $j$ required by a route-$r$ call, $C_j$ its capacity, and
$$
\mathcal S=\{n\in\mathbb Z_+^R:An\leq C\}.
$$
Under <fixed routing>, each arriving call has a predetermined resource requirement. Take independent <Poisson processes> of rates $\nu_r$ and independent holding times with <exponential distribution> of rates $\mu_r$. A feasible arrival changes $n$ to $n+e_r$ at rate $\nu_r$; a departure changes it to $n-e_r$ at rate $\mu_rn_r$. Put $\rho_r=\nu_r/\mu_r$ and assume each route uses a finite positive-capacity resource, so the state space is finite.
The <product-form stationary distribution of a loss network> is
$$
\boxed{\pi(n)=\frac1{G(C)}\prod_r\frac{\rho_r^{n_r}}{n_r!},\qquad
G(C)=\sum_{n\in\mathcal S}\prod_r\frac{\rho_r^{n_r}}{n_r!}}.
$$
For any feasible adjacent pair,
$$
\pi(n)\nu_r=\pi(n+e_r)\mu_r(n_r+1).
$$
These <detailed balance equations> prove stationarity and make the process a <reversible Markov chain>. Equivalently, <Independent random variables> with <Poisson distributions> of means $\rho_r$ are conditioned on satisfying the joint capacity constraints. The conditioning makes resource occupancies dependent even though the unconstrained counts are independent.
By <Poisson arrivals see time averages>, a route-$r$ arrival sees acceptance <probability> $G(C-A_r)/G(C)$, interpreting the numerator as zero for a negative capacity. Hence its blocking <probability> is $1-G(C-A_r)/G(C)$ and the <expected value> of its number in service is $\rho_rG(C-A_r)/G(C)$. This connects a stationary occupancy law to observable rejection and carried traffic.
The <insensitivity of loss networks> extends this occupancy formula to independent general holding-time distributions with the same <expected values>, under the usual fixed resource requirements and admission rule. Counts alone then need not be a <Markov chain>; residual holding times belong in a Markov description. The invariant occupancy formula survives. This is useful because detailed call-duration distributions can be difficult to estimate, while their means are much easier to measure.
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