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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 the standard fixed routing loss network with independent Poisson processes, the stationary occupancy law depends on independent holding-time distributions only through their means. The occupancy process alone need not be a Markov process when those holding times lack the memoryless property.

A loss network admits a call only when all its required resources have enough free capacity. Rejected calls do not queue. Under fixed routing and independent Poisson processes, its exact occupancy law has a product-form stationary distribution of a loss network.

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.

For unit resource requirements, treat the resource-blocking events as approximately independent. Let $B_j$ be the approximate blocking probability at resource $j$. Screening class-$r$ traffic by every resource other than $j$ produces the reduced-load approximationThe modeled resource is excluded from its own screening product. For general integer requirements, the standard generalized approximation is $a_j=(1-B_j{)}^{-1}{\sum }_r{A}_{jr}{\alpha }_r{\prod }_k(1-B_k{)}^{A_{kr}}$; in the unit-requirement case this reduces to the displayed formula. The approximate route acceptance is ${\prod }_j(1-B_j{)}^{A_{jr}}$. The independence used here is an approximation to the correlated exact stationary distribution, not an extra property of that law.

Here is a proof of fixed-routing existence and uniqueness. Suppose first that every retained capacity $C_j$ is a positive integer. Links of zero capacity permanently reject their routes and can be removed. If $K$ has the upper-truncated Poisson distribution with parameter $a$, its mean isThe derivative follows by differentiating the normalized weights: $dP(K=k)/da=P(K=k)(k-m_C(a))/a$. The probability at the upper endpoint similarly givesThus $E(\cdot ,C)$ increases from zero to one, while $m_C$ increases from zero to $C$. This is the monotonicity of the Erlang blocking probability.

Set $y_j=-\log (1-B_j)$. For $y\ge 0$ defineHere the inverse is with respect to offered load, and the value at zero is zero. This function is the carried load evaluated at the inverse blocking coordinate, so it is continuous and strictly increasing, with limit $C_j$ as $y\to \infty$. Consider the convex potential for the Erlang fixed pointEach exponential of a linear function is convex. Each integral is strictly convex because its derivative $u_j$ is strictly increasing. Their sum is strictly convex on the nonnegative orthant. It is also coercive: for large $y_j$, $u_j(y_j)$ is bounded below by a positive constant, so the integral grows without bound; the other terms are nonnegative. Hence $F$ attains a unique minimum.

At a positive coordinate, the first-order condition isAt $y_j=0$, the derivative equals minus the right-hand side and is nonpositive. A boundary minimum can therefore occur there only when that load is zero, in which case the derivative is zero and the same equation holds. Since $1-B_j=e^{-y_j}$, division by this positive acceptance probability makes the first-order condition precisely the reduced-load equations above. Conversely every fixed point is such a minimum. Thus fixed routing has exactly one Erlang fixed point, including zero blocking at unused resources. The proof does not rely on a potentially nonconvergent simultaneous substitution algorithm.

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.

For a standard loss network, let type-$r$ calls arrive as independent external Poisson processes of rates ${\nu }_r$. Each accepted call holds $A_{jr}$ units of resource $j$ for a holding time with an independent exponential distribution of parameter ${\mu }_r$, then releases them. Capacities $C_j$ and requirements are nonnegative integers. Rejected calls leave immediately and do not queue. With fixed routes, the feasible count set isFor a call of each type using at least one finite-capacity resource, this set is finite. The count continuous-time Markov chain has accepted birth rate ${\nu }_r{1}_{\{n+e_r\in \mathcal{F}(C)\}}$ and death rate ${\mu }_r{n}_r$. Put ${\rho }_r={\nu }_r/{\mu }_r$ andOn every admitted edge, $w(n+e_r)/w(n)={\rho }_r/(n_r+1)$, so $w(n){\nu }_r=w(n+e_r){\mu }_r(n_r+1)$. This proves detailed balance and the product-form stationary distribution of a loss network:It is unique when all relevant arrival rates are positive, since departures lead to zero and every feasible state can be built from zero through feasible births.

An external Poisson process arrival sees this stationary prearrival law by Poisson arrivals see time averages. Indeed its constant conditional arrival rate makes the expected count of arrivals in any state set equal to ${\nu }_r$ times the time spent there. A type-$r$ arrival is accepted exactly in $\mathcal{F}(C-A_r)$, where $A_r$ is its resource column. Thus the blocking partition ratio for a loss network isSet $Z$ to zero when reduced capacities have an empty feasible set.

For the triangle, index call types by their endpoint pairs and give physical links integer capacities $C_{12},C_{23},C_{31}$. We assume each call needs one circuit on every traversed link; arrivals of the three types are independent Poisson processes, and holding times have independent exponential distributions with parameters ${\mu }_{12},{\mu }_{23},{\mu }_{31}$ determined by endpoints alone. Calls may be rearranged instantaneously and without cost among their direct and two-link paths, without restarting or changing their holding times. Admit a call if any simultaneous arrangement of all calls is feasible. These assumptions make the three type counts Markovian even though the physical route assignments may change.

The cut feasibility for a rearrangeable triangle gives the exact feasible count set. Define virtual endpoint capacitiesEvery endpoint call must use at least one of the two links incident to that endpoint, so feasibility requiresThese cut inequalities are also sufficient. If each count fits its direct-link capacity, route every call directly. Otherwise at most one count exceeds its own direct capacity: two excesses would violate the cut inequality at their shared endpoint. If, for example, $n_{12}>C_{12}$, route $C_{12}$ such calls directly and the excess $d=n_{12}-C_{12}$ through node three. The other link usages are $n_{31}+d$ and $n_{23}+d$, and the first two cut inequalities bound them by $C_{31}$ and $C_{23}$. The other cases are cyclic permutations. Thus the count process is exactly a virtual fixed routing loss network in which a pair call consumes one unit of each endpoint resource.

WriteThe exact stationary distribution count probabilities are these weights divided by $Z$. The three arriving-call blocking probabilities are thereforeEquivalently, a type-$12$ call is blocked on the union of the events $n_{12}+n_{31}=K_1$ and $n_{12}+n_{23}=K_2$, with the other pairs treated cyclically. The partition ratios count their union without double-counting intersections. These expressions are exact for the stated rearrangeable triangular loss network; a restricted rearrangement policy would require a different admission model.

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.

For a fixed routing loss network, let $C_j$ be finite integer resource capacities and $A_{jr}$ the integer resource requirement of one route-$r$ call. Assume each route uses a resource, independent Poisson processes of calls with rates ${\nu }_r$, independent exponential distributions of holding times with parameters ${\mu }_r$, and immediate rejection when any required resource is unavailable. There is no waiting. The continuous-time Markov chain of call counts has feasible setPut ${\rho }_r={\nu }_r/{\mu }_r$ andThe ratios of adjacent weights satisfy $\pi (n){\nu }_r=\pi (n+e_r){\mu }_r(n_r+1)$ whenever the arrival is feasible. Thus detailed balance proves the product-form stationary distribution of a loss network,An arriving route-$r$ call is admitted precisely when $An\le C-A_{\cdot r}$. By Poisson arrivals see time averages, its exact blocking probability isDefine $Z$ to be zero if a capacity is negative. This is an exact finite sum, not an independence approximation for resource blocking events.

For the triangle, denote the physical link capacities by $C_{12},C_{13},C_{23}$ and the numbers of calls between the pairs by $n_{12},n_{13},n_{23}$. Assume unit capacity per call on each traversed link, unrestricted instantaneous rearrangement of existing calls, no interruption or extra resource cost during rearrangement, and independent exponential distributions of holding times whose parameters depend on the endpoint pair but not on the chosen path. Arrivals for each pair are independent Poisson processes, and admission occurs whenever some routing of all calls is feasible. These assumptions make the count vector a Markov process even though individual paths can change.

The necessary cut constraints areEach counts calls that must cross the two links separating one node from the others. They are also sufficient. If every pair count is at most its direct capacity, route all calls directly. Otherwise at most one pair can exceed its direct capacity: two such excesses would contradict their pairwise cut constraint. Suppose it is $12$, and put $d=n_{12}-C_{12}>0$. Route $C_{12}$ of these calls directly and divert the remaining $d$ through node $3$. Route the other calls directly. The loads on links $13$ and $23$ are respectively $n_{13}+d\le C_{13}$ and $n_{23}+d\le C_{23}$ by the two cut constraints involving $n_{12}$. This is an integer routing of every call. The other cases follow by relabelling.

Consequently the rearrangeable triangular loss network is equivalent, at the level of call counts and blocking, to a fixed routing loss network with three virtual node resources:The columns correspond to $12,13,23$, and a call uses its two endpoint resources. Its normalizer isWith $e_j$ now indexing virtual resources, the three exact answers areThe preceding detailed balance proof applies because departures occur at rates ${\mu }_{ab}{n}_{ab}$ regardless of the current routing. Without unrestricted rearrangement, feasibility can depend on the existing path assignments and this equivalence need not hold.

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.

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.

A loss network models calls that need several resources simultaneously and are rejected, rather than queued, when insufficient capacity remains. Take finite resource and route sets. Let $A_{jr}\ge 0$ be the integer amount of resource $j$ required by a route-$r$ call, $C_j$ its capacity, andUnder fixed routing, each arriving call has a predetermined resource requirement. Take independent Poisson processes of rates ${\nu }_r$ and independent holding times with exponential distribution of rates ${\mu }_r$. A feasible arrival changes $n$ to $n+e_r$ at rate ${\nu }_r$; a departure changes it to $n-e_r$ at rate ${\mu }_r{n}_r$. Put ${\rho }_r={\nu }_r/{\mu }_r$ and assume each route uses a finite positive-capacity resource, so the state space is finite.

The product-form stationary distribution of a loss network isFor any feasible adjacent pair,These detailed balance equations prove stationarity and make the process a reversible Markov chain. Equivalently, Independent random variables with Poisson distributions of means ${\rho }_r$ are conditioned on satisfying the joint capacity constraints. The conditioning makes resource occupancies dependent even though the unconstrained counts are independent.

By Poisson arrivals see time averages, a route-$r$ arrival sees acceptance probability $G(C-A_r)/G(C)$, interpreting the numerator as zero for a negative capacity. Hence its blocking probability is $1-G(C-A_r)/G(C)$ and the expected value of its number in service is ${\rho }_{r}G(C-A_r)/G(C)$. This connects a stationary occupancy law to observable rejection and carried traffic.

The insensitivity of loss networks extends this occupancy formula to independent general holding-time distributions with the same expected values, under the usual fixed resource requirements and admission rule. Counts alone then need not be a Markov chain; residual holding times belong in a Markov description. The invariant occupancy formula survives. This is useful because detailed call-duration distributions can be difficult to estimate, while their means are much easier to measure.

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.