= Solution
For unit resource requirements, treat the resource-blocking events as approximately independent. Let $B_j$ be the approximate blocking probability at resource $j$. Screening class-$r$ traffic by every resource other than $j$ produces the <reduced-load approximation>
$$
a_j=\sum_{r:j\in r}\alpha_r\prod_{k\in r\setminus\{j\}}(1-B_k),
\qquad
\boxed{B_j=E(a_j,C_j).}
$$
The modeled resource is excluded from its own screening product. For general integer requirements, the standard generalized approximation is $a_j=(1-B_j)^{-1}\sum_r A_{jr}\alpha_r\prod_k(1-B_k)^{A_{kr}}$; in the unit-requirement case this reduces to the displayed formula. The approximate route acceptance is $\prod_j(1-B_j)^{A_{jr}}$. The <independence> used here is an approximation to the correlated exact <stationary distribution>, not an extra property of that law.
Here is a proof of fixed-routing existence and uniqueness. Suppose first that every retained capacity $C_j$ is a positive integer. Links of zero capacity permanently reject their routes and can be removed. If $K$ has the <upper-truncated Poisson distribution> with parameter $a$, its mean is
$$
m_C(a)=a[1-E(a,C)],\qquad m_C'(a)=\frac{\operatorname{Var}_a(K)}a>0\quad(a>0).
$$
The <derivative> follows by differentiating the normalized weights: $dP(K=k)/da=P(K=k)(k-m_C(a))/a$. The probability at the upper endpoint similarly gives
$$
\frac{d}{da}E(a,C)=\frac{E(a,C)}a[C-m_C(a)]>0.
$$
Thus $E(\cdot,C)$ increases from zero to one, while $m_C$ increases from zero to $C$. This is the <monotonicity of the Erlang blocking probability>.
Set $y_j=-\log(1-B_j)$. For $y\geq0$ define
$$
u_j(y)=e^{-y}E(\cdot,C_j)^{-1}(1-e^{-y}).
$$
Here the inverse is with respect to <offered load>, and the value at zero is zero. This function is the carried load evaluated at the inverse blocking coordinate, so it is continuous and strictly increasing, with limit $C_j$ as $y\to\infty$. Consider the <convex potential for the Erlang fixed point>
$$
F(y)=\sum_r\alpha_r\exp\left(-\sum_jA_{jr}y_j\right)
+\sum_j\int_0^{y_j}u_j(z)\,dz.
$$
Each exponential of a <linear function> is convex. Each integral is <strictly convex> because its <derivative> $u_j$ is strictly increasing. Their sum is <strictly convex> on the <nonnegative orthant>. It is also coercive: for large $y_j$, $u_j(y_j)$ is bounded below by a positive constant, so the integral grows without bound; the other terms are nonnegative. Hence $F$ attains a unique minimum.
At a positive coordinate, the <first-order condition> is
$$
u_j(y_j)=\sum_r A_{jr}\alpha_r
\exp\left(-\sum_k A_{kr}y_k\right).
$$
At $y_j=0$, the <derivative> equals minus the right-hand side and is nonpositive. A boundary minimum can therefore occur there only when that load is zero, in which case the <derivative> is zero and the same equation holds. Since $1-B_j=e^{-y_j}$, division by this positive acceptance probability makes the <first-order condition> precisely the reduced-load equations above. Conversely every fixed point is such a minimum. Thus \b[<fixed routing> has exactly one <Erlang fixed point>], including zero blocking at unused resources. The proof does not rely on a potentially nonconvergent simultaneous substitution algorithm.
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