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%\title{WeibullT model}
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\noindent
\textbf{WeibullT  model} equation is:
 \begin{eqnarray*}
\begin{array}{l}
log_{10}(N_t)=log_{10}(A*B + C)
\end{array}
\end{eqnarray*}
where:
 \begin{eqnarray*}
\begin{array}{l}
C = 10^{log_{10}(N_{r})}\\
A = 10^{log_{10}(N_0)} - C\\
B= 10^{-(t / \delta)^p}\\
\end{array}
\end{eqnarray*}
$t$ is time and $log_{10}$ is base 10 logarithm. The parameters to estimate are $\delta$, $p$, $log_{10}(N_0)$, $log_{10}(N_{r})$.
\newline
The noisy output is defined as:
\begin{displaymath}
log_{10}(N_t)=\mathcal{N}(log_{10}(N_t), \%noise)
\end{displaymath}
i.e random number from the normal distribution with mean parameter $log_{10}(N_t)$ and standard deviation parameter $\%noise$.


\paragraph{}
\begin{flushleft}
\textbf{Example of WeibullT curve}\\
Time unit is mn.  Maximal time is 50mn.
$\delta=10.03$, $p=1.8$, $log_{10}(N_0)=9.63$, $log_{10}(N_{r})=-4.08$
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