,\quad \mathbb{C}_{3}=\left( 
\begin{array}{rrr}
0 & 2 & -2 \\ 
0 & -1 & 1 \\ 
0 & -1 & 1\end{array}\right) \ .
\label{A17}$$ For an $n$-step process we have $
\mathbb{M}^{n}=\mathbb{C}_{1}+\left( \chi _{2}\right) ^{n}\mathbb{C}_{2}+n\left( \chi _{2}\right) ^{n}\mathbb{C}_{3}.$ Setting $\chi_{2} = \exp \left( -1/T_2 \right) $ we get $$\mathbb{M}^{n}=\mathbb{C}_{1}+\exp \left({-\frac{n}{T_{2}}}\right)\mathbb{C}_{2}
+ n\exp\left({-\frac{n}{T_{2}}}\right)  \mathbb{C}_{3}
\label{A18}$$ with the characteristic decay time $T_{2}=-\left( \ln \chi _{2}\right)
^{-1}=\left( \ln 10\right) ^{-1}\approx  0.43$ for the exponential decay modes. Due to the factor $n$ that multiplies the exponential in the third term, it is going to decay at a lower pace than the second one, thus dominating the evolution of $\mathbb{M}^{n}$ toward the stationary matrix $\mathbb{C}_{1}$ as $n\rightarrow \infty$.

Entropy
-------

The complexity of the flow of vehicles in a network, as the one represented by the digraph (\[fig01\]), has a measure that can be evaluated by calculating the entropy at each moment $n$. Since $\sum_{j=1}^{N}\left( \mathbb{M}^{n}\right) _{ij}=1$ ($N$ stands for the dimension of the square matrix), for the SM we define the partial entropy associated to each row $i$ as $$S_{i}\left( n\right) =-\sum_{j=1}^{N}\left( \mathbb{M}^{n}\right) _{i,j}\ln
\left( \mathbb{M}^{n}\right) _{i,j}\qquad \mathrm{for\quad }i=1,...,N;
\label{Ent1}$$ noting that if no vehicle migrates from one site to the other, then $\left( 
\mathbb{M}^{n}\right) 