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For offered loads on consecutive routes $a,b,c,d$ of the reversible proportionally fair allocation on a four-cycle, the normalizing sum of the stationary distribution isThis formula applies in the stability region for the reversible four-cycle flow network. Grouping populations on opposite routes gives ${\sum }_{s,t}\left(\genfrac{}{}{0ex}{}{s+t}{s}\right)h_s(a,c)h_t(b,d)$, where $h_s(a,c)={\sum }_{i=0}^s{a}^i{c}^{s-i}$. For distinct opposite loads, substituting $h_s=(a^{s+1}-c^{s+1})/(a-c)$ and summing the four geometric generating series proves the formula; continuity covers coincident loads. If all four loads equal $z<1/2$, it reduces to $(1-2z+2z^2)/(1-2z{)}^3$.

Only active routes, those with $n_r>0$, enter the proportional fairness inequality and the weighted logarithmic utility. Their rates must be positive; inactive routes have no per-flow rate to determine and may be assigned zero. This convention removes otherwise meaningless terms $0\log 0$ and $0/0$.

For positive active rates, concavity of the logarithm gives, for any positive feasible competitor $y$,Thus the proportional fairness inequality proves that $x$ maximizes the log-utility objective. A competitor with an active zero rate has utility $-\infty$ and cannot improve it. Conversely, if $x$ maximizes utility, the right directional derivative along $x+\epsilon (y-x)$ at $\epsilon =0$ is nonpositive, giving exactly that inequality. Positive resource capacities and finitely many active routes allow a feasible allocation with every active rate positive, so an optimizer cannot have an active zero rate. Strict concavity gives uniqueness of the active rates. Hence proportional fairness is equivalent to the stated log-utility maximization.

Index the four adjacent-pair routes in cyclic order and put $z_i=n_i{x}_i$, the aggregate service on route $i$. The resource constraints areThey involve every odd-even route pair, so they are equivalent toFor given maxima, increasing each active odd service to $p$ and each active even service to $q$ remains feasible and improves utility. Apart from the constant $-{\sum }_i{n}_i\log n_i$, the objective therefore reduces toWhen both groups are present the constraint is tight, and differentiation gives $P/p=Q/(1-p)$. If one group is empty, the other receives unit aggregate service on each of its active routes. Thus the opposite-route reduction for a proportionally fair four-cycle yields, with $T=P+Q>0$,Inactive rates can be set to zero, including every rate in the empty state. In particular $n_{\{1,2\}}{x}_{\{1,2\}}=(n_{\{1,2\}}+n_{\{3,4\}})/{\sum }_r{n}_r$ whenever that route is active. This proves the required formula even when other routes are empty.

For the flow-level network model, an active document's residual size retains an exponential distribution with parameter ${\mu }_i$. Receiving service at rate $x_i(n)$ gives completion intensity ${\mu }_i{x}_i(n)$, so $n_i$ independent active documents give total completion intensity ${\mu }_i{n}_i{x}_i(n)$. Consequently the population continuous-time Markov chain hasAll death intensities are zero at $T=0$. Their total is bounded by ${\sum }_i{\mu }_i$, and the total arrival rate is constant, proving nonexplosion.

Put ${\rho }_i={\nu }_i/{\mu }_i$. The required stationary weight uses a binomial coefficient, as printed in the PDF; the converted TeX incorrectly rendered it as an ordinary fraction. DefineFor an odd route, the neighboring binomial coefficient ratio iswhile its death intensity at the neighboring state is ${\mu }_i(P+1)/(T+1)$. Their product is ${\nu }_i$, so $w(n)q(n,n+e_i)=w(n+e_i)q(n+e_i,n)$. For an even route, replace $P$ by $Q$ in the same calculation. This proves detailed balance and the stationary law of a proportionally fair four-cycle:A finite $B$ is essential for this expression to be a probability distribution. We determine its exact convergence condition rather than asserting stationarity for arbitrary arrival rates. LetGrouping the states by odd and even population givesWith $A=\max (a,c)$ and $E=\max (b,d)$, the bounds $A^p\le h_p(a,c)\le (p+1)A^p$ and $E^q\le h_q(b,d)\le (q+1)E^q$ show that convergence holds exactly when $A+E<1$. For necessity, the lower bound sums to ${\sum }_{k\ge 0}(A+E{)}^k$, which diverges at or above one. For sufficiency, choose slightly larger $A^{\prime },E^{\prime }$ with $A^{\prime }+E^{\prime }<1$; the polynomial factors are absorbed into constants times $(A^{\prime }{)}^p(E^{\prime }{)}^q$, giving a convergent geometric generating series. Thus the stability region for the reversible four-cycle flow network isEquivalently, the offered load on each resource, ${\rho }_1+{\rho }_4$, ${\rho }_1+{\rho }_2$, ${\rho }_2+{\rho }_3$, or ${\rho }_3+{\rho }_4$, is strictly below its unit capacity. For positive arrival rates this finite normalizing constant gives the unique stationary distribution of the irreducible process. At or beyond the boundary, the displayed invariant weights are not summable and do not define a stationary probability law.

If an explicit normalizing constant is desired, $h_p(a,c)=(a^{p+1}-c^{p+1})/(a-c)$ for $a\ne c$, and ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. Hence for $a\ne c$, $b\ne d$ in the stable region,Equal-parameter cases are obtained by continuity or directly from the positive series. In particular the empty-traffic limit is $B=1$.