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Under fixed routing, each call type always requests the same collection of resources. The link-route incidence matrix records those requirements. This differs from alternative routing, in which a rejected direct request can try another route.

For multiplicative resource-price dynamics with positive capacities, nonempty routes and strictly positive initial prices, full row rank of the link-route incidence matrix makesstrictly concave. There is a unique optimizing price vector, possibly on the boundary, and the relative-entropy Lyapunov function for resource prices proves convergence to it. Row-rank deficiency still gives unique optimal route prices and rates, but resource prices may depend on the initial state. Full column rank alone and unrestricted zero initial prices do not yield the same uniqueness theorem.

Take a finite set of source-sink pairs $S$, with fixed demands $b_s\ge 0$, and finitely many permitted routes $R_s$ for each pair. Every positive demand must have at least one available route. Let ${\nu }_r$ be route flow and ${\rho }_j$ link throughput. The source-sink route incidence matrix and link-route incidence matrix arefor simple routes; if a route traverses a link repeatedly, its entry counts those traversals. Thus $H\nu =b$ conserves each demand and $A\nu =\rho$ aggregates link traffic.

At a Wardrop equilibrium, every route carrying positive flow has minimum delay among routes for the same source-sink pair. Ifthere are numbers ${\tau }_s$ such that $c_r(\rho )\ge {\tau }_s$ for $r\in R_s$, with equality whenever ${\nu }_r>0$. This is a nonatomic routing game: one infinitesimal user's route change does not alter the link costs it sees.

For continuous nondecreasing link delays, define the Beckmann potentialEach term is convex because its derivative is nondecreasing. The feasible route-flow set $\mathcal{F}=\{\nu \ge 0:H\nu =b\}$ is nonempty and compact: $0\le {\nu }_r\le b_s$ on each pair's routes. Hence the continuous function $V(A\nu )$ attains a minimum. Its gradient with respect to route flow isIf $r$ carries positive flow and $r^{\prime }$ serves the same pair, moving a small amount of flow from $r$ to $r^{\prime }$ is feasible. At a minimizer its directional derivative $c_{r^{\prime }}-c_r$ is nonnegative. Thus every used route minimizes delay, proving existence of a Wardrop equilibrium.

Conversely, at such an equilibrium, any feasible competitor $\tilde{\nu }$ satisfiesIndeed, the competitor's cost sum for pair $s$ is at least ${\tau }_s{b}_s$, while the current flow's cost sum equals ${\tau }_s{b}_s$. This first-order variational inequality is sufficient for optimality of a differentiable convex function. ThereforeThe argument proves both directions of the claimed optimization equivalence, including unused routes with costs tied at the minimum.

If each $D_j$ is strictly increasing, each integral is strictly convex, hence so is $V$ as a function of link throughputs. The feasible link-throughput set $A\mathcal{F}$ is convex. Two distinct minimizing vectors would have a feasible midpoint with strictly smaller potential, so the equilibrium link-throughput vector is unique. Strictly convex function in $\rho$ need not give strictly convex function in $\nu$, because $A$ can have a nontrivial kernel.

For an explicit route-flow ambiguity, put two parallel links in each of two consecutive stages, with one unit of demand traversing both stages. Order the four routes by the first and second link choices: $(1,1),(1,2),(2,1),(2,2)$. Set all four link delays to $D_j(z)=z$. Theninduces throughput $1/2$ on each link, and every route has delay one. Every member of this family is an equilibrium, though its route-flow vector changes. This is route-flow nonuniqueness at a Wardrop equilibrium with strictly increasing delays.

For Braess paradox, use vertices $s,u,v,t$ and one unit of $s$-to-$t$ demand. The delays are $D_{su}(z)=z$, $D_{ut}(z)=1$, $D_{sv}(z)=1$, and $D_{vt}(z)=z$. Initially the only routes are $sut$ and $svt$. If the upper route carries $p$, their costs are $1+p$ and $2-p$, so equilibrium has $p=1/2$ and common delay $3/2$.

Add the directed link $u\to v$ with delay zero. Let the upper, lower and new middle routes carry $p,q,c$, where $p+q+c=1$. Their costs becomeIf $p>0$, the middle route is strictly cheaper than the upper route, contradicting equilibrium; likewise $q>0$ is impossible. Thus all flow uses the middle route, with $p=q=0,c=1$, and all three route costs are two. Therefore

The potential for the expanded network isThe old flow remains feasible and has potential $5/4$; the new equilibrium has potential $1$. The enlarged feasible set has indeed lowered the minimized potential. But total travel delay is ${\sum }_j{\rho }_j{D}_j({\rho }_j)$, not this integral potential: it rises from $3/2$ to $2$. This explains the paradox within the optimization framework. The optimal total-delay value cannot worsen when the old flow is still allowed; selfish equilibrium minimizes a different function.

The phenomenon also survives strictly increasing delays on every link. Replace each constant-one delay by $1+\epsilon z$ and the new-link delay by $\epsilon z$. Symmetry gives upper and lower flows $p=\epsilon /(1+3\epsilon )$ after expansion, with middle flow $(1+\epsilon )/(1+3\epsilon )$. The common new delay is $2-\epsilon (1-\epsilon )/(1+3\epsilon )$, while the old delay was $3/2+\epsilon /2$. For $\epsilon =1/10$, these are $251/130>31/20$. Thus the example does not depend on interpreting increasing delays as merely nondecreasing.

Use finite sets of links $J$, source-sink pairs $S$, and admissible routes $R$, with at least one route for each pair. Route $r$ carries aggregate flow $x_r$. The link-route incidence matrix has $A_{jr}=1$ if route $r$ uses link $j$ and zero otherwise; repeated traversals, if allowed, are counted with their multiplicity. The source-sink route incidence matrix has $H_{sr}=1$ if $r$ serves pair $s$. Thus $y=Ax$ gives link throughputs and $f=Hx$ gives source-sink demands. Write $R_s$ for the routes serving $s$ andA Wardrop equilibrium routes positive flow only on minimum-delay routes:This is a nonatomic equilibrium: an individual infinitesimal user's route change does not alter the aggregate link delays.

For fixed nonnegative demand $f$, the feasible route-flow set $\{x\ge 0:Hx=f\}$ is a nonempty product of finite-dimensional simplexes and hence a compact set. The Beckmann potentialis continuous, so it attains a minimum over this feasible set. Each primitive is a strictly convex function, since its derivative $D_j$ is continuous and strictly increasing. Its route-flow derivative is exactly the corresponding route delay:If a minimizer had positive flow on $r$ and a lower-delay alternative $q\in R_s$, transferring a small amount from $r$ to $q$ would give directional derivative $L_q-L_r<0$, contradicting minimality. Hence every minimizer is a Wardrop equilibrium.

Conversely, let $x$ be a Wardrop equilibrium and $z$ any feasible competitor. Convexity givesFor each $s$, subtract ${\lambda }_s$ inside this sum. The demand constraint makes ${\sum }_{r\in R_s}{\lambda }_s(z_r-x_r)=0$, and the remaining terms are nonnegative: used routes have $L_r-{\lambda }_s=0$, while unused routes have $x_r=0$, $z_r\ge 0$ and $L_r-{\lambda }_s\ge 0$. Thus $V(Az)\ge V(Ax)$. We have proved both existence and the optimization characterization.

Strict convexity in the link vector makes $y$ unique: two different minimizing link vectors would have a feasible midpoint of strictly smaller potential. Thus link delays and the minimum route delays at the Wardrop equilibrium are also unique. The route vector need not be unique, because different route decompositions may have the same $Ax$ and $Hx$. For example, two successive stages with two parallel links at each stage have four routes. With unit demand and link delay $D(u)=u$, the vectorsall give throughput $1/2$ on each of the four links and delay one on every route. This exhibits route-flow nonuniqueness at a Wardrop equilibrium. The equilibrium link throughputs are unique; the individual route flows need not be.

For elastic demand, make the usual physical assumptions explicit: delays are nonnegative on $[0,\infty )$, every route is nonempty, and $B_s:[0,\infty )\to [0,\infty )$ is finite, continuous and strictly decreasing. PutThe limit exists by monotonicity; it need not be zero. The inverse demand function $P_s$ is continuous, strictly decreasing, satisfies $P_s(M_s)=0$, and tends to infinity as $v\downarrow {\ell }_s$.

Choose any reference $b_s\in ({\ell }_s,M_s)$ and use elastic-demand utility with an interior reference:Each utility is strictly concave, with derivative $P_s$. A reference inside the range avoids a potentially divergent integral from zero. For instance, $B(\lambda )=(1+\lambda {)}^{-1/2}$ has inverse $v^{-2}-1$, whose integral from zero is infinite, although the displayed reference-point primitive is finite at every interior demand.

Give $G_s$ its limiting value at ${\ell }_s$, which may be $-\infty$, and extend it by $-\infty$ outside $[{\ell }_s,M_s]$. This encodes the physical demand range in the objective. After substituting $y=Ax$, $f=Hx$, minimizeIts effective feasible set is contained in $\{x\ge 0:{\ell }_s\le (Hx{)}_s\le M_s\}$, a nonempty compact set. The objective is lower semicontinuous, including a possible value $+\infty$ at a lower endpoint, and is finite when all demands equal their interior reference values. It therefore has a finite-valued minimizer.

The boundary exclusion for elastic Wardrop demand shows that every minimizing demand is interior. If $f_s={\ell }_s$ and the utility there is finite, adding a small flow $h$ on any route serving $s$ costs at most a bounded constant times $h$, while the utility gain is at least $h{P}_s({\ell }_s+h)$, whose ratio to $h$ tends to infinity. This decreases the objective. If that endpoint utility is $-\infty$, a finite minimizer cannot occur there in the first place. At $f_s=M_s>0$, choose a used route. Its delay is strictly positive, since its nonempty links have positive throughput and strictly increasing nonnegative delays. Removing a small amount of its flow saves a positive first-order cost, whereas the utility loss is $o(h)$ because $P_s(M_s)=0$. This also decreases the objective. Hence ${\ell }_s<f_s<M_s$ for every pair.

The interior-demand route derivatives are nowAt a minimizer they are zero on positive route flows and nonnegative on zero route flows, by increasing or decreasing a single route flow. Since $f_s>0$, some route is used, soThese are precisely the elastic-demand Wardrop equilibrium conditions. Conversely, those conditions give the same nonnegative directional derivative for every competitor; convexity then proves global minimality. The equivalent constrained form is the requested minimization of $V(y)-G(f)$ with $Hx=f$, $Ax=y$, $x\ge 0$. Strict convexity determines both $y$ and $f$ uniquely, while their route decomposition can still be nonunique.

A useful starting model treats a road network as a directed flow network and travelers as a divisible population in a nonatomic congestion game: an individual traveler has negligible effect on link congestion. For each source-sink pair $k$, fix a demand $d_k$ and a finite set ${\mathcal{P}}_k$ of allowed routes. Let $f_p\ge 0$ be the flow on route $p$, with ${\sum }_{p\in {\mathcal{P}}_k}{f}_p=d_k$. The link-route incidence matrix $A$ gives link loads $v=Af$. If link $e$ has delay ${\ell }_e(v_e)$, the route travel time isA Wardrop equilibrium is a feasible route-flow vector for which every used route has minimum travel time among the routes serving the same source-sink pair. Equivalently, there are numbers ${\lambda }_k$ such thatThe condition says that a traveler cannot lower their own delay by changing route. It does not assert that their choice minimizes the total delay experienced by all travelers. At a Wardrop equilibrium, unused routes may have exactly the same delay as used ones.

This static flow network model requires delay to depend on present link load, with each source-sink demand fixed. Models from queueing theory instead describe vehicles or packets as discrete customers whose delay depends on random arrivals, service and accumulated queues; a time-dependent model must also track changes of flow and travel-time propagation. Those models answer different questions from a static Wardrop equilibrium. The static model is nevertheless valuable because its route-choice condition has a precise convex optimization formulation, as developed next.

Take Poisson process arrivals of rate $a$, independent holding times with the exponential distribution of parameter $\mu$, unit resource requirement per call, no waiting room, and integer capacity $C\ge 1$. The occupancy is a birth-death process on $\{0,\dots ,C\}$ with birth rate $a$ below $C$ and death rate $k\mu$ at occupancy $k$. Writing $\nu =a/\mu$ for offered traffic, detailed balance for a continuous-time Markov chain givesAn arrival is lost exactly in state $C$, so Poisson arrivals see time averages proves the Erlang loss formulaThis derivation assumes exponential holding times; the insensitivity of loss networks extends the same formula to independent general holding times with the same mean. For completeness, $E(\nu ,0)=1$; for $C\ge 1$, set $E(0,C)=0$.

A loss network with fixed routing has finitely many resources, capacities $C_j$, and call types indexed by nonempty subsets $r$ of resources. A call of type $r$ uses one unit of every resource in $r$ for its entire holding time and is accepted only if all are available. Let $A_{jr}=1_{\{j\in r\}}$ be the link-route incidence matrix and ${\nu }_r$ the type's offered traffic, arrival rate times mean holding time. Different types have independent Poisson processes and independent holding times.

In the reduced-load approximation, availability at distinct resources is treated as independent, and screening at the resources other than $j$ is modeled as independent Poisson thinning. The load offered to $j$ after that screening isApplying the Erlang loss formula to this isolated-resource load gives the Erlang fixed point approximation, $B_j=E(a_j(B),C_j)$. These equations are an approximation because simultaneous admission induces dependence between resource occupancies, and the screened arrival process need not actually be Poisson.

Here is a proof of existence and uniqueness for positive capacities. Write $e_C(a)=E(a,C)$. Its truncated-Poisson mean is the carried load of an Erlang loss resource,For $a>0$, differentiating its stationary probabilities givesThe variance is positive and $m_C(a)<C$ because every state has positive probability. Moreover $e_C$ increases continuously from zero to one, and $m_C$ increases from zero to $C$. Thus $e_C$ has a continuous inverse on $[0,1)$.

Use acceptance coordinates $y_j=-\log (1-B_j)\ge 0$ and defineHere $u_C$ is strictly increasing, continuous, and tends to $C$ as $y\to \infty$. The convex potential for the Erlang fixed point isEach exponential term is convex, and each integral is a strictly convex function of its coordinate because its derivative is strictly increasing. Hence $F$ is strictly convex on the nonnegative orthant. It is also a coercive function: for sufficiently large $z$, $u_{C_j}(z)\ge C_j/2$, so each integral tends at least linearly to infinity. Therefore $F$ attains a unique minimum $y^{\ast }$.

Its partial derivative isAt an interior minimum this derivative is zero. Since $u_C(y)=e^{-y}{e}_C^{-1}(1-e^{-y})$, this is exactly $B_j=e_{C_j}(a_j(B))$. At a boundary coordinate $y_j=0$, the one-sided minimizing condition is ${\partial }_{j}F\ge 0$, whereas the displayed derivative is $-a_j(B)\le 0$. Thus $a_j(B)=0$ and $B_j=E(0,C_j)=0$ there as well. Conversely every solution of the Erlang fixed point approximation has these derivative conditions and hence minimizes the convex potential. No solution has $B_j=1$ at positive capacity, since every $a_j(B)$ is finite and $E(a,C)<1$ for finite $a$. The coordinates therefore give a bijection between solutions and minima, provingIf a capacity is zero, its blocking probability is forced to one. Routes using it contribute zero reduced load to every other resource. Delete those routes and the zero-capacity resources and apply the proof to the remaining network; unused positive-capacity resources have blocking probability zero. This establishes existence and uniqueness also with zero capacities, allowing $B_j=1$ precisely at those resources.

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.

Let $R_s$ be the routes serving source-sink pair $s$. The source-sink route incidence matrix and link-route incidence matrix areThus $Hx=f$ fixes the demands, and $y=Ax$ gives link throughputs. Assume a finite route set with at least one route for each positive demand. A Wardrop equilibrium is a feasible flow for which every used route has minimum delay among the routes for its source-sink pair. Writingthis means that, for each $s$, there is a number ${\lambda }_s$ with $c_r(x)\ge {\lambda }_s$ for $r\in R_s$, with equality whenever $x_r>0$. No individual user can reduce its delay by switching routes while treating the aggregate throughputs as fixed.

The feasible set $K=\{x\ge 0:Hx=f\}$ is a nonempty compact convex set: within each source-sink pair its route flows lie in a simplex of total $f_s$. The Beckmann potentialis continuous and convex, and has $\partial \Phi /\partial x_r=c_r(x)$. It therefore attains a minimum on $K$. For a differentiable convex function, $x$ minimizes $\Phi$ exactly whenIndeed necessity follows by differentiating along the feasible segment from $x$ to $z$, and sufficiency follows from the supporting-tangent inequality. This condition is equivalent to Wardrop equilibrium. Necessity of the route condition follows by moving a small amount of positive flow from a used route to any other route of the same pair. Conversely, under the route condition,because the terms involving $x_r$ vanish and both flows have the same pair totals. This proves both existence and the claimed convex optimization characterization, with $y=Ax$.

Since each $D_j$ is strictly increasing, its primitive is strictly convex. Thus ${\sum }_j{\int }_0^{y_j}{D}_j(u)du$ is strictly convex in $y$. If two minimizing flows had different link throughputs, their midpoint would have strictly smaller potential. Hence the equilibrium link throughputs are unique.

The route flows need not be unique. Consider one unit of demand through two successive stages, each with two parallel links, labelled $a,b$ in the first stage and $c,d$ in the second. The four routes are $ac,ad,bc,bd$, and all link delays are $D_j(u)=u$. For any $0\le t\le 1/2$, chooseEvery link then carries $1/2$ and every route has delay one, so each vector is a Wardrop equilibrium. They all produce the same link throughputs but give distinct route decompositions.

For the toll construction, put $F={\sum }_s{f}_s>0$. The total delay is $L(x)={\sum }_j{y}_j{D}_j(y_j)$ and the average delay is $L(x)/F$, so they have the same minimizers. Compactness of $K$ gives a global minimizer $x^{\ast }$, with $y^{\ast }=A{x}^{\ast }$. Its differentiable objective satisfies the necessary route-flow inequalityThis inequality is necessary even when total delay is not convex, since it follows from the right derivative along each feasible segment.

Use optimal-flow calibrated congestion tolling:Here $(v{)}_{+}=\max (v,0)$. The toll is nonnegative, continuously differentiable and nondecreasing, because $D_j^{\prime }\ge 0$. The perceived cost ${\hat{D}}_j(u)=D_j(u)+T_j(u)$ is strictly increasing. At $y_j^{\ast }$ it equals $D_j(y_j^{\ast })+y_j^{\ast }{D}_j^{\prime }(y_j^{\ast })$, so the displayed necessary inequality makes $x^{\ast }$ a Wardrop equilibrium for these perceived costs, by the preceding Beckmann potential argument. Their strictly increasing costs give unique equilibrium link throughputs. Therefore every tolled equilibrium has link loads $y^{\ast }$ and minimizes average delay. Constant tolls $T_j(u)=y_j^{\ast }{D}_j^{\prime }(y_j^{\ast })$ also suffice; the added positive-part term makes them genuinely traffic-dependent without moving the optimum. Zero total demand is the trivial empty-flow case.

The usual marginal external cost toll $T_j(u)=u{D}_j^{\prime }(u)$ makes its Beckmann potential equal $L$, but concluding that all its equilibria minimize $L$ requires convexity of $L$. Strict increase of delays alone does not supply this. For example, on two parallel links with $D(u)=1-e^{-u}$ and demand six, the equal split $(3,3)$ is an equilibrium under marginal-cost tolls, while its total delay $6(1-e^{-3})$ exceeds $4(1-e^{-4})+2(1-e^{-2})$, the delay at $(4,2)$. The calibrated construction avoids adding a hypothesis absent from the printed question.

For the link-route incidence matrix $A$, the reduced-load approximation and Erlang loss formula giveIntroduce $y_j=-\log (1-B_j)$ and accepted route flows $x_r={\nu }_r{e}^{-(A^{T}y{)}_r}$. Multiplying the reduced load by the link's own acceptance probability givesThese are precisely the first-order optimality conditions for the convex potential for the Erlang fixed point, sinceEach exponential term is a convex function, and the integral of the strictly increasing function $U(\cdot ,C_j)$ is a strictly convex function. Hence $F$ is a strictly convex function. Moreover $U(z,C_j)\to C_j>0$ as $z\to \infty$, by the saturation of the carried load of an Erlang loss resource. Its integrals tend to infinity at least linearly, so $F$ is a coercive function on the nonnegative orthant and has a unique minimizer.

At $y_j=0$, $U(0,C_j)=0$. If link $j$ is used by any positive-rate route, then ${\partial }_{j}F<0$, excluding a boundary minimizer there. An unused link instead minimizes at $y_j=0$ and has zero derivative. Thus the minimizer satisfies the displayed flow equations for every link. Conversely, those equations imply $\nabla F=0$ and therefore minimize the convex potential for the Erlang fixed point. Finally $U(y,C)=a[1-E(a,C)]$ with $y=-\log (1-E(a,C))$ converts the flow equations back to $B_j=E(a_j,C_j)$. The Erlang fixed point approximation has exactly one solution under fixed routing.

For the Erlang loss formula, assume a Poisson process of arrivals with rate $\lambda$, independent holding times with exponential distribution of mean $1/\beta$, unit capacity usage per accepted call, and rejection without waiting when all $C$ units are occupied. Here $C$ is a positive integer and the offered load is $a=\lambda /\beta$. The call population on $\{0,\dots ,C\}$ is a birth-death process; detailed balance givesBy Poisson arrivals see time averages, the fraction of arrivals rejected is the time probability of full occupancy. ThereforeThe paper's $\nu$ is the offered load, rather than necessarily an arrival rate. Exponential holding times suffice for this derivation; no unproved insensitivity of loss networks assumption is needed.

A loss network with fixed routing has capacities $C_j$, call types $r$, and a link-route incidence matrix $A$ with $A_{jr}=1$ exactly when a type-$r$ call requires link $j$. An accepted call holds one unit on every link of its route for its holding time; if any required link is full, the call is rejected everywhere. Take independent Poisson processes of calls, independent holding times, and route offered loads ${\nu }_r$. For the finite-variable formulation below, take $C_j\ge 1$; a zero-capacity link and all routes using it may be removed first.

The Erlang fixed point approximation treats blocking at different links as independent and thins the traffic offered to link $j$ by acceptance at the other links. Its reduced offered load isThe equality uses $A_{jr}\in \{0,1\}$: only routes with $A_{jr}=1$ contribute. Treating this reduced stream as Poisson leads to $B_j=E(a_j(B),C_j)$. Independence and Poisson thinning here are the approximation, not exact properties of a general loss network.

To construct the convex potential for the Erlang fixed point, write $e_C(a)=E(a,C)$ and $m_C(a)=a[1-e_C(a)]$, the carried load of an Erlang loss resource, equivalently the mean occupancy in the single-link stationary distribution. Differentiating its finite sums gives, for $a>0$,The variance is positive because the finite occupancy law is nondegenerate for $a>0$. Furthermore $e_C$ maps $[0,\infty )$ increasingly onto $[0,1)$, while $m_C$ increases from zero to $C$. SetIn particular $U(0,C)=0$, $U(\cdot ,C)$ is continuous and strictly increasing, and $U(z,C)\to C$ as $z\to \infty$.

Forthe first sum is convex, and each integrated strictly increasing function is strictly convex. Thus $F$ is a strictly convex function on $y\ge 0$. Since $U$ tends to a positive limit on every coordinate, $F$ is a coercive function there; it has a unique minimizer. Its coordinate derivative isThe Erlang fixed point equation is exactly ${\partial }_{j}F=0$. For an unused link, $a_j=0$ and the optimum is $y_j=0$; for a used link with positive offered load, the derivative at zero is negative and the optimum is positive. Hence every fixed point gives the unique minimizer, and the minimizer gives a fixed point. The Erlang fixed point is unique under these fixed-routing assumptions.

A convergent cyclic substitution for the Erlang fixed point updates one coordinate at a time, repeatedly cycling through all links:Always use the most recently available values of the other coordinates. Start, for example, with $B=0$. In $y$ coordinates this is exact coordinate descent for $F$: with other coordinates fixed, the displayed update is the unique coordinate minimizer.

Here is a convergence proof. All iterates stay in the initial compact sublevel set of the coercive function $F$, and their potential values decrease to a limit. Each single-coordinate update is continuous, since its reduced load is continuous and finite and $E(a,C)<1$. Thus a whole-sweep map $T$ is continuous. If a full-sweep subsequence converges to $z$, continuity gives $F(Tz)=F(z)$, because both consecutive potential values converge to the same limit. Strict coordinate convexity means equality can occur only when no coordinate changes during that sweep. Thus $z$ minimizes every coordinate, satisfies the nonnegative-orthant first-order conditions, and is the unique global minimizer of $F$. Every subsequential limit is therefore the same point, proving convergence of the full-sweep iterates and, by continuity of the updates, all intermediate iterates. This argument applies to sequential substitution; it does not assume simultaneous updates converge.

Let $R_s$ be the finite nonempty set of routes for source-sink pair $s$, with fixed nonnegative demand $f_s$. A Wardrop equilibrium is a feasible route-flow vector such that every route carrying positive flow has minimum total delay among the routes for its pair. Ifthis means that, for each $s$, some ${\lambda }_s$ satisfies $c_r(y)\ge {\lambda }_s$ for $r\in R_s$, with equality whenever $x_r>0$.

The link-route incidence matrix has $A_{jr}=1$ when route $r$ uses link $j$ and zero otherwise, or the traversal multiplicity if repeated traversals are permitted. The source-sink route incidence matrix has $H_{sr}=1$ when route $r$ serves pair $s$ and zero otherwise. Thus $Ax=y$ gives link throughputs, and $Hx=f$ gives the fixed demands.

The feasible route-flow set is a product of finite simplexes, hence nonempty and both a compact set and a convex set. The Beckmann potentialis continuous and convex, so it attains a minimum. Its route derivative is ${\partial }_r\Phi =c_r(Ax)$. At a minimum, moving a small amount of flow from any used route to any other route for the same pair cannot decrease the potential. Hence every used route has minimum route delay, exactly the Wardrop equilibrium condition. Conversely, at a Wardrop equilibrium $x$, for any feasible $\tilde{x}$,because the used flow has cost ${\lambda }_s$ and competing routes cost at least ${\lambda }_s$. The first-order inequality for a convex function makes $x$ a global minimum. This proves both existence and the specified optimization characterization.

Since each $D_j$ is strictly increasing, its primitive is a strictly convex function. Therefore the objective is strictly convex as a function of link loads $y$, and the equilibrium throughputs $y$ are unique. The mapping $x\mapsto Ax$ need not be injective on $Hx=f$, so route flows need not be unique.

For a concrete route-flow nonuniqueness at a Wardrop equilibrium example, use one unit-demand pair with two successive stages, each containing two parallel links. There are four routes, indexed by their two link choices. Give every link delay $D(u)=u$. For every $0\le t\le 1/2$,produces load $1/2$ on all four links. Every route has delay one, so each vector is a Wardrop equilibrium with the same unique link loads.