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## [Help-glpk] Piecewise Linear Function

 From: Simone Atzeni Subject: [Help-glpk] Piecewise Linear Function Date: Fri, 27 Feb 2009 15:41:31 +0300

Hi,

I have a piecewise linear function like this:

$h_i({\bf x},{\bf u})= \begin{cases} A_1({\bf x}) + {\bf u} (B_1({\bf x}) - A_1({\bf x})), \text{se 0 \le {\bf u} < 1} \\ B_1({\bf x}) + ({\bf u}-1) (B_2({\bf x}) - B_1({\bf x})), \text{se 1 \le {\bf u} < 2} \\ B_2({\bf x}) + ({\bf u}-2) (B_3({\bf x}) - B_2({\bf x})), \text{se {\bf u} \ge 2} \\ \end{cases}$

where:

\begin{itemize}

\item $A_1({\bf x}) = 3x$

\item $B_1({\bf x}) = 4x$

\item $B_2({\bf x}) = 5x$

\item $B_3({\bf x}) = 6x$

\end{itemize}

This function represents the constraints in a MILP.

To solve this MILP I have to convex my function, but I don't know like do it.

Somebody can help me?

Thanks

Simone


 Hi,I have a piecewise linear function like this:$h_i({\bf x},{\bf u})= \begin{cases} A_1({\bf x}) + {\bf u} (B_1({\bf x}) - A_1({\bf x})), & \text{se 0 \le {\bf u} < 1} \\ B_1({\bf x}) + ({\bf u}-1) (B_2({\bf x}) - B_1({\bf x})), & \text{se 1 \le {\bf u} < 2} \\ B_2({\bf x}) + ({\bf u}-2) (B_3({\bf x}) - B_2({\bf x})), & \text{se {\bf u} \ge 2} \\ \end{cases}$ where:\begin{itemize} \item $A_1({\bf x}) = 3x$\item $B_1({\bf x}) = 4x$\item $B_2({\bf x}) = 5x$\item $B_3({\bf x}) = 6x$\end{itemize} This function represents the constraints in a MILP.To solve this MILP I have to convex my function, but I don't know like do it.Somebody can help me?ThanksSimone