= Solution
An <open migration process> has finitely many colonies and state $n\in\mathbb N^J$. External arrivals to colony $j$ form independent <Poisson processes> with rates $\nu_j$. When $n_j=k>0$, individuals leave that colony at total rate $g_j(k)>0$, with $g_j(0)=0$. A departing individual moves to $k$ with probability $p_{jk}$ or exits with probability $p_{j0}=1-\sum_kp_{jk}$. The routing choices and all arrival and service randomness are independent. The routing <matrix> $P=(p_{jk})$ is substochastic and transient: every individual eventually exits, equivalently its <spectral radius> is less than one. Self-routing is allowed, but creates no state change.
The <traffic equations of an open migration process>, using row vectors, are
$$
\alpha_j=\nu_j+\sum_k\alpha_kp_{kj},\qquad
\alpha=\nu(I-P)^{-1}.
$$
The <inverse matrix> exists because $\sum_{m\geq0}P^m$ converges; its $(k,j)$ entry counts expected visits to $j$ by an individual entering at $k$. Thus $\alpha_j$ includes both external arrivals and internal migrations. Restrict first to the accessible colonies with $\alpha_j>0$.
Set
$$
b_j(0)=1,\qquad b_j(k)=\frac{\alpha_j^k}{\prod_{h=1}^kg_j(h)},\qquad Z_j=\sum_{k\geq0}b_j(k).
$$
If each $Z_j<\infty$, the candidate <product-form stationary distribution of an open migration process> is
$$
\boxed{\pi(n)=\prod_j\frac{b_j(n_j)}{Z_j}.}
$$
Here is a <global balance for a continuous-time Markov chain> proof, which does not incorrectly assume that the routing is reversible. The weight ratio is $\pi(n+e_j)/\pi(n)=\alpha_j/g_j(n_j+1)$. After dividing the incoming balance equation at $n$ by $\pi(n)$, external arrivals contribute $\sum_{j:n_j>0}\nu_jg_j(n_j)/\alpha_j$, exits contribute $\sum_j\alpha_jp_{j0}$, and migrations from $k$ to $j\ne k$ contribute $\sum_{j:n_j>0}\sum_{k\ne j}g_j(n_j)\alpha_kp_{kj}/\alpha_j$. Combining the first and third contributions using the <traffic equations of an open migration process> yields $\sum_jg_j(n_j)(1-p_{jj})$. Summing those same <traffic equations of an open migration process> gives $\sum_j\alpha_jp_{j0}=\sum_j\nu_j$. The incoming rate is therefore exactly the outgoing rate
$$
\sum_j\nu_j+\sum_jg_j(n_j)(1-p_{jj}).
$$
This proves stationarity. With finite arrival rates, finite service rates in each state and transient routing, the process is nonexplosive: only finitely many individuals enter in a finite interval, and each makes finitely many routing transitions before exit, almost surely. On the accessible <irreducible Markov chain>, a finite normalizer gives the unique <stationary distribution> and <positive recurrence>. Colonies with zero effective arrivals are empty in this equilibrium. For a nonempty accessible colony, a divergent $Z_j$ prevents normalization of the equilibrium weights.
For the single-server specialization, $g_j(k)=\phi_j$ for $k>0$. Each colony has the <stationary distribution of an M-M-1 queue>:
$$
\Pr\{n_j=k\}=\left(1-\frac{\alpha_j}{\phi_j}\right)\left(\frac{\alpha_j}{\phi_j}\right)^k,\qquad
\mathbb E n_j=\frac{\alpha_j}{\phi_j-\alpha_j},
$$
requiring $\phi_j>\alpha_j$ on every colony with positive traffic. The service rates do not change the <traffic equations of an open migration process>.
For positive $\alpha_j$, write $B=F-\sum_j\alpha_j$ and $s_j=\phi_j-\alpha_j$. A stable allocation requires $B>0$ and $s_j>0$, with $\sum_js_j=B$. The <Cauchy-Schwarz inequality> gives
$$
\left(\sum_j\sqrt{\alpha_j}\right)^2
=\left(\sum_j\sqrt{\frac{\alpha_j}{s_j}}\sqrt{s_j}\right)^2
\leq B\sum_j\frac{\alpha_j}{s_j}.
$$
Equality holds precisely when $s_j$ is proportional to $\sqrt{\alpha_j}$. Hence the unique optimal <service-capacity allocation in an open migration process> and its mean population are
$$
\boxed{\phi_j=\alpha_j+\frac{(F-\sum_k\alpha_k)\sqrt{\alpha_j}}{\sum_k\sqrt{\alpha_k}},\qquad
\min\mathbb E\sum_jn_j=\frac{(\sum_j\sqrt{\alpha_j})^2}{F-\sum_j\alpha_j}.}
$$
There is no stable allocation when $F\leq\sum_j\alpha_j$ and some traffic is positive. If some $\alpha_j=0$, those empty colonies contribute no population; allocate the spare capacity to the positive-traffic colonies by the same rule. Allowing zero capacity at an unused colony attains this optimum. If strictly positive $\phi_j$ are required even there, the displayed value is only an infimum, approached as their capacities tend to zero. If every $\alpha_j=0$, the equilibrium is empty and any admissible allocation minimizes its mean population at zero.
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