On the convexity of the Engset formula

September 28, 2026

Introduction and result

The Engset formula gives a way of estimating how likely it is that a pool of servers will all be busy when traffic from one of finitely many sources arrives. The finitude of the number of sources of traffic is exactly what makes the Engset formula special; in the limit of the number of traffic sources, there are cheaper approximations.

Questions like this one are the subject of the field of queueing theory. For an introduction, I recommend reading the Wikipedia pages on queueing theory and the Engset formula and following the references therein.

This short note focuses on a specific form of the Engset formula which is typically used when only the following quantities are known:

• the (integer) number of servers $m \geq 1$;
• the (integer) number of sources of traffic $N > m$;
• the offered traffic per-source $\alpha > 0$.

In this case, the Engset formula is an implicit equation of the form $P = f(P)$. The $P$ that solves this equation is the so-called "blocking probability" (i.e., the probability that all resources are busy). The function $f$ is given by $$ f(P) = \frac{ \binom{N - 1}{m} }{ \sum_{j = 0}^m \binom{N - 1}{j} \left( 1/\alpha + P - 1 \right)^{m - j} }. $$

The form above appears in a note by Engset¹ and is reproduced in Driksna et al.² along with numerical tables. It also appears in various other works³⁴⁵.

It turns out that the Engset formula is convex! Specifically...

Theorem. $f$ is convex on the nonnegative half-line $[0, \infty)$.

Why care about convexity?

Over a decade ago, Tommy Carpenter and I collaborated on studying the Engset formula above. Our project was born of an observation by Tommy: applying a fixed point iteration to $f$ often resulted in divergence.

The resulting paper⁶ characterized the instability of the fixed point iteration and suggested Newton's method as an alternative to compute the blocking probability. Specifically, the paper established that $f$ is convex on the nonnegative half-line whenever $\alpha \leq 1$. Convexity was in turn used to guarantee the convergence of Newton's method independent of the initial guess.

Even though we didn't have a proof at the time of writing, we conjectured⁶ based on numerical evidence that convexity (and hence convergence) should be independent of the choice of $\alpha > 0$.

Proof

Throughout, we use $(y_k)$ as shorthand for a sequence of numbers $y_0, y_1, \ldots$

The following specializes a more general result by Keilson⁷. It is used to prove the subsequent lemma.

Proposition. Let $K$ be a nonnegative integer random variable with probability mass function $k \mapsto p_k$. If the sequence $(p_k)$ is log-concave, then $2 (\mathbb{E} K)^2 \geq \mathbb{E} \left[ K \left(K - 1 \right) \right]$.

Lemma. Let $m$ be a positive integer, $(a_k)$ be a positive and log-concave sequence, and $$ Q(x) = \sum_{k = 0}^m a_k x^k. $$ Then, $1 / Q$ is convex on the nonnegative half-line.

Proof. Fix a positive number $x$ and let $K_x$ be the nonnegative integer random variable with PMF $$ \mathbb{P}(K_x = k) = \frac{a_k x^k}{Q(x)}. $$ By direct computation, $$ \mathbb{E} K_x = \frac{x Q^\prime(x)}{Q(x)} \text{ and } \mathbb{E} \left[ K_x \left( K_x - 1 \right) \right] = \frac{x^2 Q^{\prime \prime}(x)}{Q(x)}. $$ Since $(a_k)$ is log-concave, so too is the PMF of $K_x$. Applying Keilson's result at each positive $x$, $$ 2 (Q^\prime)^2 - Q Q^{\prime \prime} \geq 0 \text{ on } (0, \infty). $$ Therefore, $$ \left( \frac{1}{Q} \right)^{\prime \prime} = \frac{2 \left(Q^\prime\right)^2 - Q Q^{\prime \prime}}{Q^3} \geq 0 \text{ on } (0, \infty). \blacksquare $$

We are now ready to prove the main result.

Proof (Engset formula convexity). If $m = N - 1$, then $f(P) = (1/\alpha + P)^{-m}$ and the desired result is trivial. Therefore, proceed assuming $m < N - 1$. Let $$ Q(x) =\sum_{j=0}^{m}\binom{N-1}{j}(x-1)^{m-j}. $$ It follows that, by power series manipulations, $$ Q(x) = [z^m] \frac{(1 + z)^{N - 1}}{1 - \left(x - 1\right) z} = [z^m] \sum_{k \geq 0} x^k z^k \left(1 + z\right)^{N - k - 2} = \sum_{k \geq 0} \binom{N - k - 2}{m - k} x^k. $$ Note that $Q$ satisfies the requirements of the previous lemma. The desired result then follows by simple algebra. In particular, the Engset formula can be written as the reciprocal of $Q$ evaluated at $x = 1 / \alpha + P$. $\blacksquare$

References