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\begin{document}
\lecture{9}{Duality theorem, how to solve dual using solver for primal}{Lapsy
Garg}

\section{Duality}
\emph{Theoram(Duality):} If both primal and dual are feasible and bounded,
then their optimums must be equal.\\
\emph{Proof:}\\
If $x_0$ and $y_o$ are optimum then $c^{T}x_o = y_{o}^{T}b$\\
For any feasible x and y, \\
\begin{center}
$
c^{T}x = y^{T}Ax \leq y^{T}b\\
c^{T}x \leq y^{T}b\\
$
\end{center}

Let $x_o$ be optimum, using the previos discussion
\begin{center}
$
c^{T}x_0 = y^{T}Ax_0 = y^{T}b'\\
$
\end{center}
\\
Extend $y^{T}$ by putting all other y's as 0(zero) to get $c^{T}x_0 = y^{T}b$

\section{Rephrasing}
\begin{center}
$
min\:y^{T}b\\
A^{T}y = c\\
y_i \geq 0\\
$
\end{center}
can be rephrased as\\

\begin{center}
$
max\: -b^{T}y\\
A^{T}y \leq c\\
A^{T}y \geq c\\
y \geq 0\\
$
\end{center}

For reverse part\\
\begin{center}
$
max\:c^{T}x\\
Ax \leq b\\
$
\end{center}
can be rephrased as\\
\begin{center}
$
min\: -x^{T}c\\
Ax+x' = b\\
x'\geq0\\
$
\end{center}
\end{document}