= Solution
For a renewal gap $T$ of mean $m$, the equilibrium first-delay density is $\mathbb P(T>t)/m$. Here $m=3/\lambda$ and the <Erlang distribution> survivor is $e^{-\lambda t}(1+\lambda t+\lambda^2t^2/2)$. The <equilibrium delay of an Erlang renewal process> is therefore
$$
\boxed{f_0(t)=\frac\lambda3e^{-\lambda t}\left(1+\lambda t+\frac{\lambda^2t^2}{2}\right).}
$$
This is an equal mixture of shapes one, two and three. For each equilibrium stream use this first delay, followed by independent shape-three gaps. A joint stationary version of the cyclic routing is obtained by choosing the initial position in the three-label cycle uniformly at random, independently of the Poisson arrivals. Its marginal streams have the stated equilibrium law, but remain dependent on one another.
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