Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 30 / 3

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.