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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.

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 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.

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.

With alternative routing, a rejected direct call can try a longer route. The attempt rates now depend on blocking itself, giving an additional feedback absent from fixed routing.

For an explicit example, take three nodes with a unit-resource link between each pair, all of capacity $C=1000$. For every unordered pair, direct calls arrive at rate $v=940$ and have mean holding time one. A call first tries its direct link; if that link is full, it tries the unique two-link route through the third node, and otherwise is lost. Under a symmetric Erlang fixed point approximation, all links have blocking $B$. A given link sees its own direct load $v$. It can also carry overflow from either of the other two pairs; each contributes reduced load $vB(1-B)$, since the direct link must block and the other link of the alternative route must accept. HenceThis is an instance of multiple Erlang fixed points under alternative routing. Put $f(B)=E(1000,940[1+2B(1-B)])-B$. Evaluation using the positive Erlang B formula recurrence $E(0,v)=1$, $E(k,v)=vE(k-1,v)/(k+vE(k-1,v))$ givesThe sign statements can be certified with rational arithmetic at these rational arguments. The intermediate value theorem gives a distinct root in each of $(0,0.01)$, $(0.01,0.10)$ and $(0.10,0.25)$; numerical roots are approximately $0.002748$, $0.071037$ and $0.189785$.

Thus the alternative-routing fixed point is not unique. The mechanism is extra link occupancy from overflow: moderate blocking stimulates two-link calls and can reinforce congestion. This multiplicity belongs to the approximation. The exact finite stochastic model has a unique stationary law on its reachable communicating class; the extra approximate solutions do not create three exact invariant distributions.

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.

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.