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%\title{LnBaranyi-Roberts model}
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\noindent
\begin{large}
\textbf{LnBaranyi-Roberts model} equation is:
\end{large}
\paragraph{}
\noindent
 \begin{eqnarray*}
\begin{array}{l}
     \displaystyle \ln(N_t)= \ln(N_0) + (\mu_{max}  \times  A) - \ln(B)
\end{array}
\end{eqnarray*}
\noindent
where
 \begin{eqnarray*}
\begin{array}{l}
     \displaystyle A= t +  \frac{ln(C)}{\mu_{max}} \\[3mm]
     \displaystyle B= 1 + \frac{exp(\mu_{max}  \times A) -1 }{exp(ln(N_{max}) - ln( N_0))} \\[3mm]
\end{array}
\end{eqnarray*}
with
 \begin{eqnarray*}
\begin{array}{l}
     \displaystyle C=exp(- \mu_{max}  \times t) + exp(- \mu_{max}  \times \lambda) - exp(- \mu_{max}  \times t - \mu_{max}  \times \lambda)  \\
\end{array}
\end{eqnarray*}
\noindent
 $t$ is time and $ln$ is natural logarithm. The parameters  to estimate are $\mu_{max}$, $\lambda$, $ln(N_0)$ and $ln(N_{max})$.
\newline
The noisy output is defined as:
\begin{displaymath}
\ln(N_t)=\mathcal{N}(\ln(N_t), \%noise)
\end{displaymath}
i.e random number from the normal distribution with mean parameter $\ln(N_t)$ and standard deviation parameter $\%noise$.

\paragraph{}
\begin{flushleft}
\textbf{Example of LnBaranyi-Roberts curve}\\
Time unit is hour. Maximal time is 500h.
$\mu_{max}=0.05$, $\lambda=40$, $ln(N_{0})=5.4$  and $ln(N_{max})=20$.
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