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A linear flow network has a through route using all resources and one local route for each individual resource. Under proportional fairness with unit resource capacities, the per-flow through rate is $1/N$, where $N$ is the total number of flows. Its exponential-document model has a balance function of a flow-level network given by $\left(\genfrac{}{}{0ex}{}{N}{n_0}\right)$.

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 $B=1$.

Only routes with $n_r>0$ need a per-flow rate: absent flows contribute no utility and have no completion events. Interpret the proportional-change sum and the logarithmic objective on these active routes. With positive capacities and finitely many nonempty routes, a strictly positive feasible active-rate vector exists. Define its weighted logarithmic utility by $U(x)={\sum }_{r:n_r>0}{n}_r\log x_r$, with value $-\infty$ if an active rate is zero.

For positive feasible $x,y$, the elementary concavity inequality $\log u\le u-1$ givesThe last sum is nonpositive by proportional fairness. Feasible competitors with a zero active rate have utility $-\infty$, so they cannot improve it either. Thus a proportionally fair allocation maximizes weighted logarithmic utility. Conversely, if $x$ maximizes this utility, the feasible line segment $x+\theta (y-x)$ has nonpositive right derivative at $\theta =0$, giving precisely ${\sum }_r{n}_r(y_r-x_r)/x_r\le 0$. This also covers a competitor with zero coordinates, since $x+\theta (y-x)$ stays positive for sufficiently small $\theta$. The strictly concave function $U$ makes the active-rate optimizer unique.

For the linear flow network, put $M={\sum }_{i=1}^I{n}_i$ and $N=n_0+M$. If $n_i>0$ and resource $i$ is not saturated, its local per-flow rate could be increased without affecting any other resource. This would strictly increase weighted logarithmic utility. HenceIf $n_0>0$ and $M>0$, substitute $x_i=(1-n_0{x}_0)/n_i$ on all active local routes into the utility. The terms depending on $x_0$ areIts derivative vanishes when $1/x_0=M/(1-n_0{x}_0)$, so $x_0=1/N$. Its second derivative is negative, giving the unique maximizer. If $M=0<n_0$, utility increases up to the capacity endpoint $x_0=1/n_0=1/N$. If $n_0=0$, each active local route uses its entire local resource. Altogether the proportionally fair allocation on a linear flow network isAbsent-route rates may be set to zero, and departures are zero at the empty state $N=0$. Notice that these are per-flow rates: the aggregate through service is $n_0/N$, while each active local route gets aggregate service $M/N$.

In the document-transfer flow-level network model, let $e_r$ be the unit vector for route $r$. Independent Poisson processes of flow arrivals give transitions $n\to n+e_r$ at rate ${\nu }_r$. An active document on route $r$ receives service at rate $x_r(n)$ and has an exponential distribution of size with parameter ${\mu }_r$. By the memoryless property, its instantaneous completion rate is ${\mu }_r{x}_r(n)$. Summing over the active documents gives the continuous-time Markov chain intensitiesThus through-route departures have rate ${\mu }_0{n}_0/N$, and active local departures have rate ${\mu }_{i}M/N$. All other off-diagonal intensities vanish. The diagonal intensity is the negative of the sum of the outgoing intensities. Each route's aggregate service is at most one, so the total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$; the process is a nonexplosive continuous-time Markov chain.

Set ${\rho }_r={\nu }_r/{\mu }_r$, assume ${\mu }_r>0$, and define the balance function of a flow-level network byThe adjacent-state ratios of this binomial coefficient areThey equal the aggregate service rates just obtained. Therefore, with $w(n)=H(n){\prod }_r{\rho }_r^{n_r}$,These are the detailed balance for a continuous-time Markov chain equations on every adjacent pair. Summing incoming and outgoing pairs proves stationarity whenever the weights can be normalized. Hence the stationary law of a linear flow network isThe coefficient here is a binomial coefficient, not merely the total number of flows. Both it and the denominator in $x_0=1/N$ are visible in the original PDF but missing from the converted TeX.

We can also determine exactly when the normalizing constant is finite. For $0\le {\rho }_0<1$, summing first over $n_0$ with local counts fixed and applying the negative binomial series givesThe remaining sums are independent geometric series. Thusprovided ${\rho }_0+{\rho }_i<1$ for every $i$. These inequalities are also necessary: if ${\rho }_0\ge 1$, the states with every local count zero already give a divergent sum; otherwise any local ratio at least one makes its geometric series diverge. Thus the stationary formula requires strict offered load below capacity at every resource. With positive arrival rates, the process is irreducible, and these normalizable detailed balance weights give its unique stationary distribution. Outside these inequalities, the displayed weights are not a probability distribution. Zero arrival rates can be handled on the reachable class by setting the corresponding population to zero.

For the increasing-supply resource-price dynamics, assume a finite network, positive weights $w_r$, nonempty routes, nonnegative prices, and initially positive prices on every used resource. Resource $j$ announces price ${\mu }_j$, and a route sees total price $s_r={\sum }_{j\in r}{\mu }_j$. Maximizing $w_r\log x-s_{r}x$ over $x>0$ gives $x_r=w_r/s_r$: this is the route's weighted proportional fairness demand. The supply $q_j({\mu }_j)$ increases with its price, and the multiplicative resource-price dynamics raise a positive price when total demand exceeds supply and lower it otherwise.

The boundary equilibria of multiplicative resource prices show why the positivity qualification is necessary. With two resources, one route using both, $w=1$, ${\kappa }_1={\kappa }_2=1$ and $q_j(u)=u$, the prices $(0,1)$ and $(1,0)$ are distinct equilibria with well-defined route price one. The positive equilibrium is $(1/\sqrt{2},1/\sqrt{2})$. A zero coordinate stays zero because it multiplies its own derivative. Thus the printed claim about all trajectories is false if these boundary initial conditions are admitted. We now prove the intended convergence for positive initial prices; unused resources may start at zero.

For positive-price convergence for increasing resource supplies, define $Q_j(u)={\int }_0^u{q}_j(\eta )d\eta$. Strict increase of $q_j$ makes $Q_j$ strictly convex, even without differentiability of $q_j$. On the convex domain $\mu \ge 0$ with every $s_r>0$, the supplied function $V={\sum }_r{w}_r\log s_r-{\sum }_j{Q}_j({\mu }_j)$ is strictly concave, since each logarithmic term is concave and the negative integral terms give strictness in every changed coordinate. Its gradient isIt is also coercive toward large prices: for fixed $a>0$, $Q_j(u)\ge q_j(a)(u-a)$ when $u\ge a$, while the positive logarithmic terms grow only logarithmically in the largest coordinate. If any route price tends to zero with the other prices bounded, $V\to -\infty$. Consequently every nonempty superlevel set is compact and bounded away from zero route prices. $V$ attains a unique maximum ${\mu }^{\ast }$.

For a used resource, the derivative at a zero coordinate is ${\sum }_{r:j\in r}{w}_r/s_r>0$, since $q_j(0)=0$. Such a coordinate cannot vanish at the maximum. An unused resource has its unique maximizing coordinate at zero. Thus ${\mu }^{\ast }$ is positive on used resources and satisfies demand equal to supply there. It is the unique equilibrium in that positive domain.

Along any trajectory,The initial superlevel set traps the trajectory and supplies a global upper bound $M$ on every price, and a lower bound on every route price. On any finite time interval $g_j$ is bounded, so ${\mu }_j(t)={\mu }_j(0)\exp ({\kappa }_j{\int }_0^t{g}_j(\mu (u))du)$ stays positive if initially positive. The bounded trajectory therefore exists for all positive times.

More strongly, each used resource has a uniform positive demand boundChoose ${\delta }_j>0$ with $q_j({\delta }_j)<c_j/2$. While $0<{\mu }_j\le {\delta }_j$, its derivative is positive, so ${\mu }_j(t)\ge \min ({\mu }_j(0),{\delta }_j)>0$. Unused prices satisfy ${\dot{\mu }}_j=-{\kappa }_j{\mu }_j{q}_j({\mu }_j)$ and converge to zero.

The nonnegative function $dV/dt$ has finite integral, since $V$ is increasing and bounded above. It is uniformly continuous along the trajectory: on the compact trapped set, the vector field is bounded, the trajectory is Lipschitz, and ${\sum }_j{\kappa }_j{\mu }_j{g}_j^2$ is a continuous, hence uniformly continuous function of $\mu$. The criterion uniformly continuous integrable functions vanish at infinity applies here: otherwise separated intervals around positive peaks would give an infinite integral. Therefore $dV/dt\to 0$. Since used prices stay bounded away from zero, $g_j(\mu (t))\to 0$ on every used resource. Every limit point is thus the unique maximizer ${\mu }^{\ast }$, with zero prices on unused resources. Compactness now implies $\boxed{\mu (t)\to {\mu }^{\ast }}$. Only continuity and strict increase of the supply functions were used; differentiability or an unbounded supply was not required. The boundary counterexample remains a necessary qualification of the literal source assertion.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

A feasible per-flow allocation satisfies $x_r>0$ for each active route and ${\sum }_r{A}_{jr}{n}_r{x}_r\le C_j$. It has proportional fairness if every feasible competitor obeysBy the first-order inequality for the strictly concave logarithm, this is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible allocations. Inactive-route rates are immaterial; set them to zero as a convention.

Write $a=n_{\{1\}}$, $b=n_{\{2\}}$, $c=n_{\{1,2\}}$, and $N=a+b+c$. The capacity constraints are $a{x}_1+c{x}_0\le 1$ and $b{x}_2+c{x}_0\le 1$. For $c>0$, put $q=c{x}_0$. Each active local route uses its remaining capacity $1-q$. Omitting constants, the utility isIts maximum has $q=c/N$; when $a+b=0$, the maximum is the endpoint $q=1$. Hence for active routesThese formulas also give $x_1=1/a$ and $x_2=1/b$ when $c=0$. When $N=0$ there is no allocated service. Equivalently, the aggregate route rates for $N>0$ areThis is the proportionally fair allocation on a two-resource linear network for a two-resource linear flow network.

For independent Poisson processes of documents and independent document sizes with exponential distribution of rate ${\mu }_r$, the memorylessness of the exponential distribution makes the population a continuous-time Markov chain. Its transition intensities areThe departure intensity sums the $n_r$ independent completion hazards ${\mu }_r{x}_r$. There is no departure on an inactive route. Set ${\rho }_r={\nu }_r/{\mu }_r$, with ${\mu }_r>0$.

To prove the proposed stationary distribution, define the balance function of a flow-level networkFor an active local route, $\Psi (n-e_r)/\Psi (n)=(a+b)/N$, while for the through route this ratio is $c/N$. Thus ${\Lambda }_r(n)=\Psi (n-e_r)/\Psi (n)$. The weightssatisfy detailed balance on every arrival-departure pair:Each total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$, so the chain is a nonexplosive Markov chain; normalized detailed-balance weights therefore give its stationary distribution, the stationary law of a two-resource linear flow network.

Use the negative binomial series to sum first over $c$:The remaining two geometric series giveAll summands are nonnegative. Consequently the sum is finite exactly under the stability conditions for a two-resource linear flow networkThese say that the offered load on each unit-capacity resource is strictly below one. Equality is insufficient: at least one geometric series then diverges. If all arrival rates are positive, the chain is irreducible and the normalizable law is its unique equilibrium law.

Finally put $q_i={\rho }_i/(1-{\rho }_0)$, $i=1,2$. Summing over $c$ yieldsThus the two local-route counts are independent, each with geometric distribution on the nonnegative integers. The conditional through-route count has negative binomial distributionFor positive traffic rates this depends on the local counts, so all three route counts are not independent. More quantitatively, $\mathrm{Cov}(a,c)={\rho }_0\mathrm{Var}(a)/(1-{\rho }_0)>0$, and likewise for $b$. If zero arrival rates are allowed, the exceptional independent cases are ${\rho }_0=0$, or ${\rho }_1={\rho }_2=0$, when the relevant counts are deterministic zero. This distinction is local-count independence with through-flow dependence.

Only routes with $n_r>0$ contribute to proportional fairness or to the weighted logarithmic utility; omit inactive routes from expressions with a rate in the denominator. Active routes have positive rates whenever a positive feasible allocation exists. For a feasible competitor $y$, the elementary logarithm inequality $\log u\le u-1$ givesA zero competitor rate on an active route has utility $-\infty$ and also cannot improve the objective. Therefore a proportionally fair allocation maximizes the logarithmic objective. Conversely, the feasible set is a convex set, and differentiating the objective along $x+t(y-x)$ at $t=0+$ shows that an optimizer satisfies the proportional fairness inequality. Strict concavity gives uniqueness on active coordinates; rates assigned to nonexistent flows are immaterial.

In the linear flow network, put $z_0=n_0{x}_0$ and $z_i=n_i{x}_i$ for active routes. A local route $i$ uses only resource $i$. If $n_i>0$ and $z_0+z_i<1$, increasing $x_i$ increases the objective without violating any other constraint. Thus each active local route saturates its resource:Write $L={\sum }_{i=1}^I{n}_i$ and $N=n_0+L$. For $n_0>0$ and $L>0$, substitute $z_i=1-z_0$ into the objective. Apart from constants its dependence on $z_0$ isIts derivative vanishes when $n_0/z_0=L/(1-z_0)$, and its second derivative is negative. Hence the per-flow through-route rate isIf $L=0<n_0$, the optimal through-route aggregate rate is one, so $x_0=1/n_0=1/N$ still holds. If $n_0=0<L$, each active local route has $x_i=1/n_i$, also given by the displayed local formula. When $N=0$ there are no flows and no departures. Inactive coordinates need not be assigned these undefined per-flow formulas.

Independent arrivals and exponential document sizes make this a flow-level network model. A document of type $r$ has residual-size hazard ${\mu }_r$ per unit of data transmitted; when its transmission speed is $x_r(n)$, its completion hazard per unit time is ${\mu }_r{x}_r(n)$. Thus the continuous-time Markov chain of flow counts has transition intensitiesFor $N>0$ these departure intensities are ${\mu }_0{n}_0/N$ for the through route and ${\mu }_{i}L/N$ for an active local route. Total arrival intensity is constant and total departure intensity is bounded by ${\sum }_r{\mu }_r$, ensuring a nonexplosive Markov chain.

Set ${\rho }_r={\nu }_r/{\mu }_r$, the offered traffic, and define the balance function of a flow-level networkThe binomial coefficients satisfyThese identities also cover states with only one kind of active route. They show that the weight $a(n)=\Phi (n){\prod }_{r=0}^I{\rho }_r^{n_r}$ satisfiesThus detailed balance for a continuous-time Markov chain proves that this is a reversible Markov chain and reduces the stationary distribution calculation to normalization.

Sum over $n_0$ first. The negative binomial series gives, for fixed local counts and $L={\sum }_i{n}_i$,All summands are nonnegative, so the Tonelli theorem permits exchanging the sums. The partition sum is thereforeIt is finite exactly when ${\rho }_0+{\rho }_i<1$ for every $i$, for $I\ge 1$. These are the resource-load conditions: resource $i$ receives the through-route offered traffic plus its local offered traffic. The normalized stationary distribution isWith positive arrival rates this is the unique stationary distribution of the irreducible feasible flow-count chain. Zero arrival rates restrict its closed class to the corresponding zero coordinates.

Summing out $n_0$ also shows that the local counts are independent, with geometric distributions on $0,1,\dots$:Hence their mean flow counts areThe binomial balance factor is essential: independent geometric counts for all routes would not give these departure rates or the correct resource-load normalization.

On unit-capacity links $1,\dots ,I$, route zero uses every link and route $i$ uses only link $i$. Let $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. Under proportional fairness, every active local route has $n_0{x}_0+n_i{x}_i=1$. Maximizing $n_0\log w_0+M\log (1-w_0)$ for $w_0=n_0{x}_0$ gives $w_0=n_0/N$ and the displayed per-flow rates on active routes. At the empty state departures vanish, and rates of absent flows may be set to zero.

For local counts $a,b$, through count $c$, and $N=a+b+c>0$, proportional fairness gives aggregate through service $c/N$. Every active local route receives aggregate service $(a+b)/N$, divided equally among its active flows. Inactive routes receive zero aggregate service. This maximizes weighted logarithmic utility subject to both unit-capacity constraints.

For the proportionally fair allocation on a linear flow network, independent Poisson flow arrivals at rates ${\nu }_r$ and exponential document sizes of rates ${\mu }_r$ give a continuous-time Markov chain with departures ${\mu }_r{n}_r{x}_r(n)$. Put ${\rho }_r={\nu }_r/{\mu }_r$. Under ${\rho }_0+{\rho }_i<1$ for each local route, the normalizing constant is $Z=(1-{\rho }_0{)}^{I-1}/{\prod }_i(1-{\rho }_0-{\rho }_i)$. The binomial coefficient weight has adjacent-state ratios equal to the aggregate service rates, proving detailed balance. For more general network topologies, proportional fairness need not give this reversible product form.

A two-resource linear flow network has one local route on each of two resources and one through route using both. With unit capacities and proportional fairness, its exponential-document flow-level network model has the stationary law of a two-resource linear flow network. The local counts are independent in equilibrium despite their dependence on the shared through-route count.

The utility ${\sum }_r{n}_r\log x_r$ sums logarithmic utilities over individual flows. Its first-order condition gives proportional fairness. Its negative is a strictly convex function on active flow rates, giving uniqueness of those rates on a convex set of feasible allocations.