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The Erlang fixed point approximation treats resource blocking as independent and applies the Erlang loss formula to traffic screened by other resources. Its self-consistency equations have a unique fixed point under fixed routing, but can have several under alternative routing.

For positive integer capacity $C$ and offered load $a>0$, let $m_C(a)$ be the carried load of an Erlang loss resource. Differentiating the upper-truncated Poisson distribution gives $m_C^{\prime }(a)=\mathrm{Var}(N)/a>0$. Differentiating its probability at $C$ gives the displayed derivative of the Erlang loss formula; it is positive because $m_C(a)<C$. Thus blocking increases continuously from zero to one, and the carried load increases from zero to $C$. These properties justify the inverse blocking coordinates in the convex potential for the Erlang fixed point.

For calls arriving at rate ${\nu }_r$ with mean holding time $1/{\mu }_r$, offered traffic is the dimensionless load ${\alpha }_r={\nu }_r/{\mu }_r$, often measured in Erlangs. The Erlang loss formula distinguishes this offered traffic from the mean traffic actually carried.

For fixed routing, let each route $r$ require one unit on every resource in its fixed set, let $C_j\ge 1$ be integer resource capacities, and let ${\alpha }_r$ be the offered load, arrival rate times mean holding time. The Erlang fixed point approximation treats distinct resource availability events as independent. If $B_j$ is the approximate blocking probability at resource $j$, the reduced-load approximation giveswhere the Erlang loss formula isScreening excludes resource $j$ itself: it is the load offered to that resource, not merely the load ultimately admitted there. Approximate route acceptance is ${\prod }_{j\in r}(1-B_j)$. This specifies the approximation, not an assertion of exact independence in the actual loss network.

Here is a constructive variational proof of both existence and uniqueness. For $C\ge 1$ write $e_C(a)=E(a,C)$ and $m_C(a)=a[1-e_C(a)]$, the carried load of an Erlang loss resource. Under its upper-truncated Poisson occupancy law, differentiating the finite sums givesConsequently $e_C$ increases continuously from $0$ to $1$, and $m_C$ increases from $0$ to $C$. Put $y_j=-\log (1-B_j)$ andEach $g_j$ is continuous and strictly increasing, tending to $C_j$. Consider the convex potential for the Erlang fixed pointThe first sum is convex, and each integral is strictly convex because its derivative $g_j$ strictly increases. Hence $F$ is strictly convex. Also each integral grows at least linearly for sufficiently large $y_j$, since $g_j(y_j)\to C_j>0$. Thus $F$ is coercive, its bounded sublevel sets are compact, and it attains a unique minimum $y^{\ast }$.

Its derivative isFor a resource used by some positive-load route, this derivative is negative at $y_j=0$ for every finite choice of the other coordinates, so its minimizing coordinate is positive and satisfies ${\partial }_{j}F=0$. Cancelling $e^{-y_j}$ gives $e_{C_j}^{-1}(B_j)=a_j(B)$, exactly the required fixed-point equation. An unused resource minimizes its integral at $y_j=0$, giving $B_j=0=E(0,C_j)$. Conversely any fixed point has $B_j<1$, since the total offered load is finite; its logarithmic coordinates satisfy these same minimizing conditions. Strict convexity therefore proves one and only one fixed-routing Erlang fixed point. Zero-capacity resources can be removed with their blocked routes, whose acceptance probabilities are already zero.

For a concrete alternative routing counterexample, take three nodes forming a triangle, each link of capacity $1000$, with offered traffic $950$ for each endpoint pair. Try the direct link first; if it is blocked, try the other two links together. In a symmetric reduced-load approximation, a link has its direct load plus two overflow streams, each of load $950B(1-B)$: the preferred link is blocked with probability $B$, and the other link of the alternative path is available with probability $1-B$. The link itself is excluded from screening. Thus a common link blocking probability must satisfyDefine the right side minus $B$ as $G(B)$. The stable Erlang loss formula recursion $b_0=1$, $b_k=a{b}_{k-1}/(k+a{b}_{k-1})$ givesAll four arguments are rational, so the same recursion can certify these signs with rational arithmetic, without trusting rounded values. Continuity and the intermediate value theorem give at least three distinct symmetric fixed points, one in each intervening interval. This is nonuniqueness of the approximation; the exact finite irreducible loss network still has a unique stationary distribution.

A loss network admits a call only when all the resources it requires have available capacity. Calls which fail admission disappear rather than waiting. Under fixed routing, class $r$ has a fixed nonzero integer resource-requirement vector $a_r$, Poisson process rate ${\nu }_r$, and independent exponential distributions with parameter ${\mu }_r$. Let $A$ have columns $a_r$ and let $C$ be the resource-capacity vector. The feasible call counts areWrite ${\alpha }_r={\nu }_r/{\mu }_r$ for the offered load. On $\mathcal{S}$, the Markov transition rates are $q(n,n+e_r)={\nu }_r$ if the new state is feasible, and $q(n,n-e_r)={\mu }_r{n}_r$.

The product-form stationary distribution of a loss network isIndeed, whenever $n+e_r$ is feasible,These pairwise equalities give detailed balance, including the truncated boundary: there is no transition to an infeasible state in either direction. The finite reachable chain is irreducible under positive arrival and holding rates, so this is its unique stationary distribution. It can also be viewed as independent Poisson offered populations conditioned on all capacity constraints. Conditioning usually makes the actual resource occupancies dependent.

An arriving call of class $r$ sees this stationary distribution by Poisson arrivals see time averages. Its acceptance probability iswhere a partition function with a negative capacity is zero. This is the blocking partition ratio for a loss network. Thus its blocking probability is $1-P_r$, admitted arrival rate is ${\nu }_r{P}_r$, and mean number present is ${\alpha }_r{P}_r$, by balancing admitted births with departures. With a single unit-demand resource, the formula reduces to the Erlang loss formulaThe stationary count law is also insensitive to the shape of independent holding-time distributions in the standard fixed-routing model when their means are unchanged. With nonexponential holding times, counts alone need not be Markov; residual holding times are included in the state for this insensitivity assertion. State-dependent admission controls and alternative routing do not automatically preserve the same simple product form.

Let $J$ be a finite collection of resources with positive integer capacities $C_j$, and let $R$ be a finite collection of nonempty routes. In the standard unit-resource loss network, a call on route $r$ needs one unit simultaneously at each resource in $r$; fixed routing means that these requirements do not change when a request is rejected. Suppose route $r$ has an independent Poisson process of arrivals of rate ${\nu }_r$ and independent holding times of mean $1/{\mu }_r$. Its offered traffic is ${\alpha }_r={\nu }_r/{\mu }_r$. The Erlang fixed point approximation approximates resource blocking events as independent, rather than claiming this independence for the exact loss network.

Write $b_j$ for the approximate blocking probability at resource $j$. The reduced-load approximation screens a call by every other resource that it needs, so the offered traffic seen by resource $j$ isFor a single resource with offered traffic $a$ and capacity $C$, the Erlang loss formula isThus the Erlang fixed point approximation is the systemThe resulting route acceptance probability is ${\prod }_{j\in r}(1-b_j)$. In particular, the screening product for $a_j$ excludes $j$ itself; including it would describe a different approximation.

For existence, define $T_j(b)=E(a_j(b),C_j)$ on $[0,1{]}^J$. Each $a_j$ is a continuous polynomial, and the denominator of the Erlang loss formula is at least one. Consequently $T$ is continuous and maps this nonempty compact convex set into itself. The Brouwer fixed-point theorem supplies a fixed point. At any such point $b_j<1$, since finite offered traffic and positive capacity give $E(a,C)<1$.

For uniqueness, we construct a strictly convex function whose minimizer is equivalent to an Erlang fixed point. Put $y_j=-\log (1-b_j)$. First establish the single-resource facts needed for this change of coordinates. For $a>0$, the inverse blocking probability has the expressionThe terms with $k<C$ decrease strictly with $a$, so $E(a,C)$ increases strictly from zero to one. Let $a_C(y)$ be its inverse parametrization defined by $E(a_C(y),C)=1-e^{-y}$, with $a_C(0)=0$. If $K$ has the truncated Poisson distribution with weights proportional to $a^k/k!$, then shifting the sum gives its mean, the carried load of an Erlang loss resource, asDifferentiating the normalized weights showsThe variance is positive because both $K=0$ and $K=1$ have positive probability. Moreover, $m_C(a)\to C$ as $a\to \infty$, since the largest term in the truncated Poisson distribution dominates. Henceis continuous, strictly increasing from zero to $C$.

Now minimize the following convex potential for the Erlang fixed point over the nonnegative orthant:Each exponential term is convex, and each integral is a strictly convex function of its coordinate because its derivative $U_{C_j}$ is strictly increasing. Their sum is therefore strictly convex. It is also a coercive function: once $z$ is large enough, $U_{C_j}(z)\ge C_j/2$, so each integral grows at least linearly in its coordinate, while the exponential terms are nonnegative. Thus $F$ attains exactly one minimum.

The coordinate derivative isAt an interior coordinate of a minimizer it vanishes. At $y_j=0$, it is nonpositive, whereas the one-sided minimum condition requires it to be nonnegative; it therefore vanishes there too. Conversely, a zero gradient is a global minimizer of this convex function. Multiplying ${\partial }_{j}F=0$ by $e^{y_j}$ givesApplying the Erlang loss formula converts this exactly to $b_j=E(a_j(b),C_j)$. The change of coordinates is one-to-one, proving uniqueness under fixed routing. A zero-capacity resource can instead be fixed as blocked with probability one, with all routes using it removed before applying the proof; an unused positive-capacity resource has blocking probability zero.

For alternative routing, consider a symmetric three-node loss network. Each of its three links has capacity $1000$; every node pair offers traffic $950$, tries its direct link first, and, if that link is blocked, tries the two-link path through the third node. Let all approximate link blocking probabilities equal $b$. Each link receives its direct offered traffic and the overflow from the other two node pairs. An overflow is generated with probability $b$ and reaches the link under consideration only when its other required link accepts, with probability $1-b$. The symmetric reduced-load approximation therefore givesFor $g(b)=E(950[1+2b(1-b)],1000)-b$, evaluate the Erlang loss formula by the stable recursion $E(a,0)=1$, $E(a,k)=aE(a,k-1)/(k+aE(a,k-1))$. It givesThese signs can also be certified with exact rational arithmetic, since all four loads are rational. Continuity and the intermediate value theorem give three distinct roots, one in each intervening interval. Thus the alternative-routing approximation need not be unique. This is multiplicity of approximate fixed points, not multiplicity of stationary distributions for the exact finite irreducible Markov chain.

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.

Start from the Erlang loss formula and reverse the order in its denominator:For each fixed $j$, the product tends to $(C/\nu {)}^j$. If $\nu >C$, it is bounded by this same summable geometric series. Extend the summand by zero for $j>NC$ and apply the dominated convergence theorem with counting measure to obtainIf $\nu \le C$, retain the first $K+1$ summands. Their limiting sum is at least $K+1$, so the reciprocal tends to infinity as $K$ can be arbitrarily large. This includes $\nu =C$ without a separate normal approximation. ThereforeHere $C$ and $N$ are positive integers, as appropriate for circuit counts; taking $\lfloor NC\rfloor$ gives the same limit for a positive real capacity scale. This is the proportional scaling limit of the Erlang loss formula.

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.

The alternative routing term feeds additional traffic back into the link. We can prove the existence of three solutions rather than relying on an iteration that happens to find different answers. Choose ${\nu }_N=20N$, $C_N=21N$, and define the continuous residualBy the proportional scaling limit of the Erlang loss formula, at each fixed $B$,At $B=1/100$, the multiplier is $1.0198<21/20$, so $G_{\infty }(1/100)=-1/100$. At $B=1/8$ it is $39/32$, givingAlso $G_N(0)>0$ and $G_N(1)<0$ for every $N$, since a finite-capacity Erlang loss formula gives a blocking probability strictly between zero and one. For all sufficiently large $N$, the signs at $0,1/100,1/8,1$ therefore alternate. Applying the intermediate value theorem in the three disjoint intervening intervals proves at least three distinct fixed points. This supplies a fluid scaling proof of multiple Erlang fixed points.

This multiplicity concerns the Erlang fixed point approximation. It does not imply several equilibrium laws for an exact finite irreducible Markov chain, which has a unique stationary distribution.

For a single resource, assume unit-capacity calls arrive as a Poisson process of rate $a$, holding times have independent exponential distributions with mean $h$, and a call is rejected when all $C$ units are occupied. Let $\nu =ah$. Its occupancy is a birth-death process on $\{0,\dots ,C\}$, with birth rate $a$ below capacity and death rate $k/h$ in state $k$. Detailed balance for a birth-death process givesBy Poisson arrivals see time averages, the Erlang loss formula isThis derivation assumes exponential holding times; the standard insensitivity result extends the same formula to independent holding times of the same mean, but is not needed here.

A loss network with fixed routing has finitely many resources $j$, capacities $C_j$, and call types $r$ with a fixed nonempty set of resources. Type $r$ has Poisson arrivals of rate $a_r$, independent holding times of mean $h_r$, and offered load ${\nu }_r=a_r{h}_r$. It is accepted only if every resource on its route has a free unit, and otherwise is lost. In the displayed equations use the unit-incidence matrix $A_{jr}=1$ for $j\in r$ and zero otherwise. The equations concern this unit-demand model, rather than arbitrary multi-unit requirements.

The Erlang fixed point approximation replaces the correlated resource availabilities by independent ones. A type-$r$ call survives blocking at resources other than $j$ with approximate probability ${\prod }_{i\in r\setminus \{j\}}(1-B_i)$. Treating this filtering as Poisson thinning gives the reduced offered loadApply the single-resource formula to obtain $B_j=E({\rho }_j,C_j)$. Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.

Assume first $C_j\ge 1$. The right side is a continuous map from $[0,1{]}^J$ into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let $m_C(\rho )=\rho [1-E(\rho ,C)]$ be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yieldsThus $E$ increases continuously from zero to one and $m_C$ increases from zero to $C$.

Write $x_j=-\log (1-B_j)\ge 0$, and define ${\rho }_C(x)=E(\cdot ,C{)}^{-1}(1-e^{-x})$ and ${\ell }_C(x)=m_C({\rho }_C(x))$, with ${\ell }_C(0)=0$. Each ${\ell }_C$ is continuous and strictly increasing. A fixed point satisfiesThese are precisely the stationary equations for the convex potential for the Erlang fixed pointThe first term is convex; the second is strictly convex because each ${\ell }_{C_j}$ strictly increases. Hence $F$ is strictly convex. Its gradient cannot vanish at two distinct points: the difference of gradients dotted with the difference of points is strictly positive, with a positive contribution from every changed coordinate in the integral term. This also covers boundary coordinates $x_j=0$, using the right derivative there. Every fixed point has $B_j<1$, since its offered load is finite, so the change of variables is valid. There is exactly one fixed point. If a capacity is zero, its blocking probability is one; delete it and all routes using it before applying the argument to the remaining positive capacities. Unused resources have blocking probability zero.

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.

Consider a triangle of three links, each with capacity $C$. Calls for each pair of its three nodes arrive as independent Poisson processes at rate $\lambda$ and have independent holding times with an exponential distribution of mean one. Try the direct link first; if it is full, try the two-link path through the third node; if either alternative link is full, reject the call. This is alternative routing in a loss network.

In the symmetric Erlang fixed point approximation, let $B$ be each link's blocking probability. A given link receives direct offered traffic $\lambda$. Each of the other two call types contributes overflow traffic $\lambda B$, screened by the availability $1-B$ of the other link on its alternative path. ThusThe extra term expresses a feedback: blocking sends calls onto longer paths, which consume more total capacity and can create still more blocking.

Here is an explicit finite-capacity example, not just a limiting argument. Take $C=1000$ and $\lambda =950$, and let $F(B)=E(950[1+2B(1-B)],1000)-B$. Evaluating the Erlang loss formula givesThe displayed decimals are rounded, but the four signs can be checked exactly using the rational recursion $E(a,0)=1$, $E(a,k)=aE(a,k-1)/(k+aE(a,k-1))$. By the intermediate value theorem, there is a fixed point in each of $(0,0.01)$, $(0.01,0.1)$, and $(0.1,0.4)$. Numerically these three areThus alternative routing can produce multiple Erlang fixed points. This does not imply multiple stationary distributions for the exact finite continuous-time Markov chain, which is irreducible and has a unique stationary distribution; the multiplicity belongs to the approximation. The symmetric alternative-routing model is also discussed in Kelly's review of fixed point models of loss networks.

First consider unit requirements, $A_{jr}\in \{0,1\}$. The Erlang loss formula for a resource with capacity $C$ and offered traffic $a$ isThe Erlang fixed point approximation assumes that resources block independently and that the traffic retained after screening at other resources can be treated as a Poisson process. If $B_j$ is the approximate blocking probability and $q_j=1-B_j$, the reduced-load approximation givesThe link whose load is being calculated is excluded from the screening product. Otherwise one would confuse its offered load with its carried load.

For existence, these equations define a continuous map from $[0,1{]}^J$ into itself. The Brouwer fixed-point theorem gives a fixed point. Assume $C_j\ge 1$; a zero-capacity resource forces rejection of every call needing it and can be removed together with those call types.

For uniqueness, let $K$ have probabilities proportional to $a^k/k!$, $0\le k\le C$, and write $m_C(a)=\mathbb{E}K$. Direct differentiation givesThese identities hold for $a>0$, and $m_C(0)=0$. Also $E(a,C)$ increases from zero to one and $m_C(a)$ increases from zero to $C$.

Put $p_j=-\log q_j$. Define $a_C(p)$ by $E(a_C(p),C)=1-e^{-p}$, and define $h_C(p)=m_C(a_C(p))$. This is a continuous, strictly increasing function on $[0,\infty )$, starting at zero and tending to $C$. Multiplying each Erlang fixed point approximation load equation by $q_j$ transforms it intoThese are exactly the zero-gradient conditions ofEach integral is a strictly convex function; the exponential terms are convex functions. Thus $\Phi$ is a strictly convex function. It is also a coercive function, since $h_{C_j}(p)\to C_j>0$, so it has a unique minimizer. At $p_j=0$, its partial derivative is negative if some positive-traffic route uses $j$, so that coordinate of the minimizer is positive. If no route uses $j$, the unique minimizing coordinate is zero. The minimizer therefore satisfies the equations in every coordinate. Conversely any fixed point has $B_j<1$, finite $p$, and these zero-gradient equations, so must equal that unique minimizer. Hence the Erlang fixed point approximation exists and is unique for fixed routing.

For integer requirements, the common generalized Erlang fixed point approximation usesMultiplication by $q_j$ gives the same equations for $\Phi$, so the existence and uniqueness argument also covers this generalized approximation. It remains an approximation, rather than the exact blocking law for a call requesting several units.

Scale the offered load and capacity together. Reversing the denominator of the Erlang loss formula givesFor each fixed $j$, the summand tends to $(C/\nu {)}^j$. If $\nu >C$, the dominated convergence theorem applies with the summable geometric series bound $(C/\nu {)}^j$. Otherwise each fixed finite partial sum has limiting value at least its number of terms, so the reciprocal diverges. Taking reciprocals proves the formula, including the critical case $\nu =C$.

For rational offered load, every step of the stable recursion for the Erlang loss formula is rational. Comparing the final numerator and denominator with rational thresholds therefore certifies signs without floating-point uncertainty. Opposite certified signs of a continuous fixed-point residual imply a root by the intermediate value theorem; several alternating intervals can prove multiple Erlang fixed points under alternative routing.