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%\title{BiPhasicS model}
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\noindent
\begin{large}
\textbf{BiPhasicS model} equation is:
\end{large}
\paragraph{}
\noindent
 \begin{eqnarray*}
\begin{array}{l}
     \displaystyle \log_{10}(N_t) = log_{10}(N_0) + log_{10}((A + B) \times C)
\end{array}
\end{eqnarray*}
\noindent
where
 \begin{eqnarray*}
\begin{array}{l}
      \displaystyle A = f \times D \\[3mm]
      \displaystyle B = (1-f)  \times exp( -k_{max2} \times t) \\[3mm]
\displaystyle C = E / (1+(E-1) \times D) 
\end{array}
\end{eqnarray*}
with
 \begin{eqnarray*}
\begin{array}{l}
\displaystyle D=  exp( -k_{max1} \times t) \\[3mm]
\displaystyle E =  exp( -k_{max1} \times S1) \\[3mm]
\end{array}
\end{eqnarray*}
\noindent
 $t$ is time, $log_{10}$ is base 10 logarithm. The parameters  to estimate are $f$, $k_{max1}$, $k_{max2}$,  $S1$ and $log_{10}(N_0)$.
\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 BiPhasicS curve}\\
Time unit is mn. Maximal time is 60mn.
$f=0.77$, $k_{max1}=0.71$, $k_{max2}=0.71$,  $S1=12$ and $log_{10}(N_0)=9.47$
\end{flushleft}

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