Opposite-route reduction for a proportionally fair four-cycle ($\max (z_1,z_3)+\max (z_2,z_4)\le 1$)

Write $z_r=n_r{x}_r$ for aggregate route service. The four resource constraints are all sums of one odd and one even route service, so they are equivalent to the displayed maximum constraint. For fixed maxima, increasing each active service to its group's maximum improves weighted logarithmic utility. Maximizing $P\log p+Q\log q$ with $p+q\le 1$ gives $p=P/(P+Q)$ and $q=Q/(P+Q)$, where $P=n_1+n_3$, $Q=n_2+n_4$. Inactive routes are omitted; the same formula covers states where one group is empty.