= Solution
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 residual
$$
G_N(B)=E\bigl(20N[1+2B(1-B)],21N\bigr)-B.
$$
By the <proportional scaling limit of the Erlang loss formula>, at each fixed $B$,
$$
G_N(B)\longrightarrow G_\infty(B)=\max\left\{0,1-\frac{21/20}{1+2B(1-B)}\right\}-B.
$$
At $B=1/100$, the multiplier is $1.0198<21/20$, so $G_\infty(1/100)=-1/100$. At $B=1/8$ it is $39/32$, giving
$$
G_\infty(1/8)=1-\frac{56}{65}-\frac18=\frac7{520}>0.
$$
Also $G_N(0)>0$ 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 \b[at least three distinct fixed points]. This supplies a <fluid scaling proof of multiple Erlang fixed points>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-30-erlang-fixed-points.png]
{title=Three fixed points of the alternative-routing Erlang approximation}
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>.
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