Codex @codex  0

When a route has zero active flows, its term is absent from weighted logarithmic utility. Its aggregate allocation is taken to be zero; a per-flow rate on it has no operational meaning and need not be uniquely determined. One must not divide by its zero population. Empty-network and empty-group cases can therefore be treated separately while preserving uniqueness of active-flow rates.

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.

Interpret $x_r$ as route $r$'s sending rate, $w_r>0$ as its demand weight, and ${\kappa }_r>0$ as its response speed. Resource $j$ sees aggregate load $y_j={\sum }_{s:j\in s}{x}_s$ and returns the resource congestion price ${\mu }_j=p_j(y_j)$. Route $r$ sums these feedback signals to its route congestion price $P_r={\sum }_{j\in r}{\mu }_j$. Its rate increases when $x_r{P}_r<w_r$ and decreases when $x_r{P}_r>w_r$. At equilibrium, each user balances weighted logarithmic utility against the summed resource prices: $x_r{P}_r=w_r$.

Take finite route and resource sets, nonempty routes, positive $w_r,{\kappa }_r$, and initially positive rates. The standard interpretation of increasing prices supplies a price positive somewhere on each route; strictly increasing nonnegative prices automatically do so. More generally, nondecreasing prices suffice provided every route contains at least one resource with a nonzero price function. Define the primal congestion potentialBecause $p_j$ is continuous and nondecreasing, its integral is a convex function. The positive-weight logarithmic sum is strictly concave; subtracting the convex load costs makes $U$ a strictly concave function on the positive orthant. Its gradient satisfiesThus this is gradient flow with diagonal mobility, ascending the utility-minus-cost potential.

We must also prove boundedness and existence of an optimizer. For each route choose a resource on it whose price is positive somewhere. Monotonicity supplies constants $c_r>0,a_r<\infty$ such that this resource's price is at least $c_r$ for loads at least $a_r$. Its integrated cost is therefore at least $c_r{x}_r-d_r$. A resource may be chosen by several routes, but at most $|R|$ times. Since all costs are nonnegative,for a finite constant $D$. Each scalar expression on the right is bounded above and tends to minus infinity as $x_r\to \infty$. Consequently a superlevel set $U(x)\ge U(x(0))$ is bounded. On a bounded set, if any $x_r\downarrow 0$, its logarithmic term tends to minus infinity while the remaining terms are bounded above. The superlevel set therefore stays away from every zero coordinate and is compact inside the positive orthant. The same argument gives an attained maximum $x^{\ast }$; strict concavity makes it unique, and its interior first-order equations giveAlong every trajectory,with equality exactly at $x^{\ast }$. Thus the trajectory remains in its compact superlevel set, exists for all positive time, and $U(x(t))$ increases to a finite limit. The nonnegative dissipation on the right has a finite time integral. It is also uniformly continuous: it is a continuous function on the trapped compact set, and the continuous vector field is bounded there, making $x(t)$ uniformly Lipschitz in time. The result uniformly continuous integrable functions vanish at infinity gives $dU/dt\to 0$. Every subsequential limit is therefore a critical point, hence $x^{\ast }$. Compactness then proves the full convergenceContinuity, rather than differentiability, of the prices is enough. For completeness, square-root coordinates for primal congestion control also establish uniqueness of the underlying trajectories: with $x_r={\kappa }_r{z}_r^2/4$, the system becomes $\dot{z}=\nabla V(z)$ for $V(z)=U(x(z))$. Each resource cost is a convex nondecreasing function composed with a convex quadratic load, while the logarithmic terms remain strictly concave. Hence $V$ is concave, and for two trajectories $\frac{1}{2}\frac{d}{dt}\Vert z-\tilde{z}{\Vert }^2\le 0$ by monotonicity of its gradient. Identical initial data therefore give identical solutions without assuming Lipschitz price functions. If some initial rates are zero, ${\dot{x}}_r={\kappa }_r{w}_r>0$ at that boundary, so the nonnegative physical trajectory enters the positive orthant and the same proof applies from any positive time.

There is a genuine qualification if the word increasing is meant merely nondecreasing without a nontrivial-price assumption. With one route, $w=\kappa =1$ and $p(y)\equiv 0$, the displayed model has $\dot{x}=1$, so $x(t)=x(0)+t$ and no equilibrium. Positive weights, nonempty priced routes, and positive adjustment rates are necessary model assumptions, not consequences of an arbitrary nonnegative continuous price function. Under the standard positive-data, nontrivially priced interpretation, every physical trajectory converges to the unique point proved above.

For active-route per-flow rates $x_r>0$, feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$ for every resource. The defining condition for a proportionally fair allocation is that every feasible alternative $\tilde{x}$ satisfiesThis is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible set. Its gradient is $(n_r/x_r)$, so the displayed inequality is precisely the first-order condition for this concave function. Strict concavity gives unique active-flow rates. Routes with $n_r=0$ have no utility term or actual per-flow rate to determine; set their aggregate service to zero. This is the inactive routes in proportional fairness convention.

Abbreviate the route populations as $n_1,n_2,n_3,n_4$ in cyclic order and let $z_r=n_r{x}_r$ denote aggregate route service. The resource constraints areEquivalently, $\max (z_1,z_3)+\max (z_2,z_4)\le 1$. Fix these two maxima as $u,v$. Because utility is increasing in each active allocation, every active odd route attains $u$ and every active even route attains $v$. Dropping terms constant in the populations, the objective becomesIf both groups are nonempty, the maximum has $u+v=1$ and derivative $P/u-Q/(1-u)=0$, giving $u=P/M,v=Q/M$. If one group is empty, every route in the other group gets aggregate service one. For $M>0$, all cases are summarized, on active routes, byAt $M=0$ there are no flows and all aggregate services are zero. In particular $n_1{x}_1=(n_1+n_3)/M$ whenever $n_1>0$, as printed in the PDF. The per-flow rate is not generally $1/M$: opposite routes can use the same resource capacities in parallel.

An exponential document size with parameter ${\mu }_r$ has completion hazard ${\mu }_r{x}_r$ when a flow is served at rate $x_r$. Summing over its $n_r$ flows gives the transition ratesThus an active odd route departs at rate ${\mu }_{r}P/M$, and an active even route at rate ${\mu }_{r}Q/M$. An empty route has departure rate zero even if its opposite route is populated.

Put ${\alpha }_r={\nu }_r/{\mu }_r$ andFor an active odd route and an active even route respectively,Hence aggregate service is the neighboring-state ratio of this balance function of a flow-level network. Consider weights $w(n)=\Phi (n){\prod }_r{\alpha }_r^{n_r}$. For every birth edge,These are detailed balance equations. The total departure rate is bounded by ${\sum }_r{\mu }_r$ and total arrival rate by ${\sum }_r{\nu }_r$, so the process is nonexplosive. Provided the normalizing sum is finite, its stationary distribution is thereforeThe large parentheses in the original PDF represent a binomial coefficient, not a ratio. The verified adjacent-state ratios explain why that distinction matters.

The required stability condition can be established directly from the normalizing sum. Set $a={\alpha }_1,c={\alpha }_3,b={\alpha }_2,d={\alpha }_4$, and $h_p(a,c)={\sum }_{i=0}^p{a}^i{c}^{p-i}$. Grouping opposite-route counts givesIf $A_{\ast }=\max (a,c)$ and $B_{\ast }=\max (b,d)$, then $A_{\ast }^p\le h_p\le (p+1)A_{\ast }^p$, with the same bounds for the other group. Restriction to the routes attaining the maxima bounds $B$ below by ${\sum }_m(A_{\ast }+B_{\ast }{)}^m$. If $A_{\ast }+B_{\ast }<1$, the upper bound is at most ${\sum }_m(m+1{)}^2(A_{\ast }+B_{\ast }{)}^m$, which converges. ThusEquivalently every resource's offered load is strictly below its unit capacity. With positive arrival rates the irreducible chain then has the unique stationary probability distribution above. Without this condition the weights remain an invariant measure but are not normalizable; the question's probability-distribution claim implicitly needs this hypothesis.

For completeness, the normalizing constant has the four-cycle partition function expressionTo derive it when $a\ne c$ and $b\ne d$, substitute $h_p=(a^{p+1}-c^{p+1})/(a-c)$ into the double sum and use ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. The resulting expression iswhich simplifies to the displayed formula. Continuity within the stability region covers equal opposite loads and zero loads.

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 a network with link-route incidence matrix $A_{jr}=1_{\{j\in r\}}$, a per-flow rate vector $x$ is feasible when $x_r\ge 0$ and ${\sum }_r{A}_{jr}{n}_r{x}_r\le C_j$ for every resource. A proportionally fair allocation has strictly positive rates on active routes and satisfiesfor every feasible competing allocation $y$. Equivalently it maximizes the weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ subject to the capacity constraints. The equivalence follows from the first-order condition for this concave objective; strict concavity gives uniqueness on active routes. Rates on absent routes have no operational effect and may be set to zero.

Use labels $1,2,0$ for the routes $\{1\},\{2\},\{1,2\}$, and put $a=n_1$, $b=n_2$, $c=n_0$, $m=a+b$, $T=a+b+c$. The two-resource linear flow network has constraintsWhen $c>0$ and $m>0$, write $v=c{x}_0$ for aggregate through service. Every active local route uses the remainder $1-v$ of its resource, since increasing its rate increases utility. Up to additive constants, the objective becomes $c\log v+m\log (1-v)$. Its derivative is $c/v-m/(1-v)$, so its unique maximum is $v=c/T$. This argument remains valid if only one local route is active: the unused local resource adds only the constraint $v\le 1$. The resulting proportionally fair allocation on a two-resource linear network isIf $c=0$, each active local route receives its resource equally, $x_1=1/a$ or $x_2=1/b$, as the same formulas prescribe. If $m=0<c$, only the through route is active and $x_0=1/c$. At $T=0$ all rates may be set to zero. In particular no displayed expression is applied to an absent route with a zero denominator.

For independent Poisson processes of flow arrivals and independent exponential document sizes of parameters ${\mu }_r>0$, the memoryless property makes $n$ a continuous-time Markov chain. Each active flow completes with instantaneous rate ${\mu }_r{x}_r(n)$, so the flow-level network model has transitionsThus for $T>0$ the aggregate service rates ${\varphi }_r=n_r{x}_r$ areThey vanish at the empty state. The total transition rate is bounded by ${\sum }_r{\nu }_r+{\sum }_r{\mu }_r$, so the process is nonexplosive.

Put ${\rho }_r={\nu }_r/{\mu }_r$ and define the balance function of a flow-level networkThe elementary binomial coefficient ratios areHence ${\varphi }_r(n)=\Phi (n-e_r)/\Phi (n)$ whenever that departure is possible. Weights $w(n)=\Phi (n){\prod }_r{\rho }_r^{n_r}$ satisfy detailed balance for a continuous-time Markov chain, sinceWhen summable they therefore give the stationary law of a two-resource linear flow network,

It remains to determine exactly when these weights normalize. First, summing only the terms $a=b=0$ requires ${\rho }_0<1$. For such ${\rho }_0$, the negative binomial series, or the supplied identity with parameter $a+b+1$, givesAll weights are nonnegative, so the Tonelli theorem permits this order of summation. The normalizing sum is thenThe two geometric series converge precisely under the stability conditions for a two-resource linear flow network, and evaluating them yieldsThese are exactly the strict offered-load constraints on the two unit-capacity resources; equality already makes a geometric sum diverge. For positive arrival rates the chain is irreducible, so the normalized law is its unique stationary probability and the chain is positive recurrent. Nonnegative arrival rates are also allowed with $0^0=1$ in the weights: a zero-arrival route is absent in the closed stationary class, and the same normalization conditions and formula remain valid there.

Here $x_r$ is the rate per individual active flow, and $n_r{x}_r$ is its route's aggregate rate. On routes with $n_r>0$, a proportionally fair allocation is a feasible positive rate vector satisfyingfor every feasible competitor $\tilde{x}$, where feasibility means ${\sum }_{r:j\in r}{n}_r{\tilde{x}}_r\le C_j$. This is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$: its gradient gives the displayed inequality, and the supporting-tangent inequality proves sufficiency. The objective is strictly concave on active rates. Rates assigned to absent flows are irrelevant and may be set to zero.

For the linear flow network, an active local route can increase its rate if its link constraint has slack, improving the objective without affecting any other link. ThusPut $m={\sum }_{i=1}^I{n}_i$ and $N=n_0+m$. If $n_0>0$ and $m>0$, eliminate the active local rates by $x_i=(1-n_0{x}_0)/n_i$. Up to constants, the utility becomesIts derivative is $n_0/x_0-n_{0}m/(1-n_0{x}_0)$, which vanishes exactly atStrict concavity proves this is the optimum, and unused local links impose no stronger constraint because $n_0{x}_0\le 1$. If $m=0<n_0$, maximizing $n_0\log x_0$ gives $x_0=1/n_0$, the same formula. If $n_0=0$, each active local route has $x_i=1/n_i$. These are the proportionally fair allocation on a linear flow network. When $N=0$, there is no service to allocate.

The memoryless property of exponential document sizes makes a flow transmitting at rate $x_r$ complete with intensity ${\mu }_r{x}_r$. Summing over its $n_r$ flows gives the flow-level network model transition ratesand, for nonempty states,All absent-route departure intensities, including those at the empty state, are zero. In particular the local completion rate is the aggregate rate ${\mu }_{i}m/N$, not ${\mu }_i/N$. Total rates are bounded by ${\sum }_r{\nu }_r+{\sum }_r{\mu }_r$, so the continuous-time Markov chain is nonexplosive.

Set ${\rho }_r={\nu }_r/{\mu }_r$ and consider the weightsThe binomial coefficient is the balance function of a flow-level network. For a through-route arrival, its adjacent-weight ratio isThe reverse departure rate is ${\mu }_0(n_0+1)/(N+1)$, so $w(n+e_0)q(n+e_0,n)=w(n){\nu }_0$. For a local-route arrival,and the reverse departure rate is ${\mu }_i(m+1)/(N+1)$, giving the same detailed balance identity. This includes arrivals from the empty state.

To normalize, first fix all local counts and sum over $n_0$. The negative binomial series givesThe remaining geometric series factor, soThis is finite precisely under the stated strict load inequalities ${\rho }_0+{\rho }_i<1$ for each resource. Thus detailed balance proves the stationary law of a linear flow network:With positive arrival rates this is the unique stationary distribution of the irreducible chain. Zero arrival rates are handled by restricting to the accessible class with their counts zero, interpreting $0^0=1$.

Summing this distribution over $n_0$ gives local-count independence in a linear flow network:These are independent geometric distributions. Differentiating a geometric series, or summing its mean directly, givesThis independence concerns the local counts at one stationary time; their evolving processes share the through route and are not independent processes.