Theorem 4.1 — optimal capacity allocation φ_j = a_j + √(a_j f_j)/∑√(a_k f_k) · (F − ∑ a_k f_k)/f_j

A communication network has $J \ge 1$ channels. Channel $j$ has average arrival rate $a_j > 0$ and, if given capacity $\phi_j > a_j$, holds on average $a_j/(\phi_j - a_j)$ customers. Capacity on channel $j$ costs $f_j > 0$ per unit, and the capacities must satisfy the cost constraint (4.2)

where the budget satisfies $F > \sum_k a_k f_k$. Among all capacity vectors $\phi$ with $\phi_j > a_j$ for every $j$ and satisfying (4.2), the mean number of customers in the network

is minimized by the allocation

Precisely: $\phi^*$ is feasible, it attains the minimum over the feasible set, and every other feasible $\phi$ gives a strictly larger mean number of customers.

Each channel first receives the capacity $a_j$ needed to carry its traffic; the excess budget $F - \sum_k a_k f_k$ is then shared in proportion to $\sqrt{a_j f_j}$. As the book observes, minimizing the mean number of customers in the network is equivalent to minimizing the average time a customer spends in it.

Formalization Note The book leaves implicit that $a_j > 0$, $f_j > 0$, $J \ge 1$ and $F > \sum_k a_k f_k$; without the last the feasible set is empty. The stability condition $\phi_j > a_j$ is part of the feasible set. The uniqueness clause (strict inequality for every other feasible point) slightly strengthens the book's "The optimal allocation is"; it holds because the objective is strictly convex.