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%\title{Double-Weibull model}
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\noindent
\begin{large}
\textbf{Double-Weibull model} equation is:
\end{large}
\noindent
 \begin{eqnarray*}
\begin{array}{l}
 \displaystyle \log_{10}(N_t) = \log_{10}(N0) - \log_{10}(1+10^\alpha) +  \log_{10}(10^A+10^B)
\end{array}
\end{eqnarray*}
\noindent
where
 \begin{eqnarray*}
\begin{array}{l}
\displaystyle A= -(t/\delta1)^p + \alpha \\[3mm]
\displaystyle B= -(t/\delta2)^p
\end{array}
\end{eqnarray*}

 $t$ is time, $log_{10}$ is base 10 logarithm. The parameters  to estimate are $\log_{10}(N0)$, $\alpha$, $p$, $\delta1$
and  $\delta2$.
\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 Double-Weibull curve}\\
Time unit is hour. Maximal time is 24h.
$\log_{10}(N0)=5.45$,  $\alpha=3.15$, $p=3.67$, $\delta1=5.45$
and  $\delta2=17.21$
\end{flushleft}

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