For a single resource, assume unit-capacity calls arrive as a Poisson process of rate , holding times have independent exponential distributions with mean , and a call is rejected when all units are occupied. Let . Its occupancy is a birth-death process on , with birth rate below capacity and death rate in state . Detailed balance for a birth-death process gives
By Poisson arrivals see time averages, the Erlang loss formula is
This 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 , capacities , and call types with a fixed nonempty set of resources. Type has Poisson arrivals of rate , independent holding times of mean , and offered load . 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 for 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- call survives blocking at resources other than with approximate probability . Treating this filtering as Poisson thinning gives the reduced offered load
Apply the single-resource formula to obtain . Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.
Assume first . The right side is a continuous map from into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yields
Thus increases continuously from zero to one and increases from zero to .
Write , and define and , with . Each is continuous and strictly increasing. A fixed point satisfies
These are precisely the stationary equations for the convex potential for the Erlang fixed point
The first term is convex; the second is strictly convex because each strictly increases. Hence 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 , using the right derivative there. Every fixed point has , 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.

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