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%\title{Rosso model}
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\noindent
\begin{large}
\textbf{Rosso model} equation is:
\end{large}
\paragraph{}
\noindent
\begin{itemize}
\item
when $ t \leq mean(\lambda)$
 \begin{eqnarray*}
\begin{array}{l}
N_{t}=N_0
\end{array}
\end{eqnarray*}
\item
otherwise
 \begin{eqnarray*}
\begin{array}{l}
 \displaystyle N_{t}= N_{max} / (1+A \times B)
\end{array}
\end{eqnarray*}
where:
 \begin{eqnarray*}
\begin{array}{l}
     \displaystyle A= N_{max}/ N_0 -1 \\[3mm]
     \displaystyle B=exp(-\mu_{max}  \times (t-\lambda)) 
\end{array}
\end{eqnarray*}
\end{itemize}
 $t$ is time. The parameters to estimate are $\mu_{max}$, $\lambda$, $N_0$ and $N_{max}$.
\newline
The noisy output is defined as:
\begin{displaymath}
N_{t}=(1-\%noise) \times N_{t} + \xi (N_{t} \times  \%noise)
\end{displaymath}
where $\xi$ is random numbers from Poisson distribution with mean parameter $N_{t} \times  \%noise$.


\paragraph{}
\begin{flushleft}
\textbf{Example of Rosso curve}\\
Time unit is hour. Maximal time is 500h.
$\mu_{max}=0.05$, $\lambda=40$, $N_{0}=230$  and $N_{max}=4e8$.
\end{flushleft}

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  \centering
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