= Solution
A <loss network> admits a call only when all the resources it requires have available capacity. Calls which fail admission disappear rather than waiting. Under <fixed routing>, class $r$ has a fixed nonzero integer resource-requirement vector $a_r$, <Poisson process> rate $\nu_r$, and independent <exponential distributions> with parameter $\mu_r$. Let $A$ have columns $a_r$ and let $C$ be the resource-capacity vector. The feasible call counts are
$$
\mathcal S=\{n\in\mathbb Z_+^R:An\leq C\}.
$$
Write $\alpha_r=\nu_r/\mu_r$ for the <offered load>. On $\mathcal S$, the Markov <transition rates> are $q(n,n+e_r)=\nu_r$ if the new state is feasible, and $q(n,n-e_r)=\mu_rn_r$.
The <product-form stationary distribution of a loss network> is
$$
\boxed{\pi(n)=Z(C)^{-1}\prod_r\frac{\alpha_r^{n_r}}{n_r!},\qquad
Z(C)=\sum_{n\in\mathcal S}\prod_r\frac{\alpha_r^{n_r}}{n_r!}.}
$$
Indeed, whenever $n+e_r$ is feasible,
$$
\pi(n)\nu_r=\pi(n+e_r)\mu_r(n_r+1).
$$
These pairwise equalities give <detailed balance>, including the truncated boundary: there is no transition to an infeasible state in either direction. The finite reachable chain is irreducible under positive arrival and holding rates, so this is its unique <stationary distribution>. It can also be viewed as independent Poisson offered populations conditioned on all <capacity constraints>. Conditioning usually makes the actual resource occupancies dependent.
An arriving call of class $r$ sees this <stationary distribution> by <Poisson arrivals see time averages>. Its acceptance probability is
$$
P_r=\sum_{n\in\mathcal S:\,A(n+e_r)\leq C}\pi(n)
=\frac{Z(C-a_r)}{Z(C)},
$$
where a <partition function> with a negative capacity is zero. This is the <blocking partition ratio for a loss network>. Thus its blocking probability is $1-P_r$, admitted arrival rate is $\nu_rP_r$, and mean number present is $\alpha_rP_r$, by balancing admitted births with departures. With a single unit-demand resource, the formula reduces to the <Erlang loss formula>
$$
E(a,C)=\frac{a^C/C!}{\sum_{k=0}^Ca^k/k!}.
$$
The stationary count law is also insensitive to the shape of independent holding-time distributions in the standard fixed-routing model when their means are unchanged. With nonexponential holding times, counts alone need not be Markov; residual holding times are included in the state for this insensitivity assertion. State-dependent admission controls and <alternative routing> do not automatically preserve the same simple product form.
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