How to take a dual

by math_explorer, Oct 28, 2016, 5:44 AM

The dual of the linear program
\begin{align*}
v^* &= \min \vec{\textbf{c}_1}^T\vec{\textbf{x}_1}+\vec{\textbf{c}_2}^T\vec{\textbf{x}_2}+\vec{\textbf{c}_3}^T\vec{\textbf{x}_3}\\
A_{11}\vec{\textbf{x}_1}+A_{12}\vec{\textbf{x}_2}+A_{13}\vec{\textbf{x}_3} &= \vec{\textbf{b}_1}\\
A_{21}\vec{\textbf{x}_1}+A_{22}\vec{\textbf{x}_2}+A_{23}\vec{\textbf{x}_3} &\geq \vec{\textbf{b}_2}\\
A_{31}\vec{\textbf{x}_1}+A_{32}\vec{\textbf{x}_2}+A_{33}\vec{\textbf{x}_3} &\leq \vec{\textbf{b}_3}\\
\vec{\textbf{x}_1} &\ge 0\\
\vec{\textbf{x}_2} &\le 0\\
\vec{\textbf{x}_3} &\;\text{(no restriction)}
\end{align*}is (modulo human error)
\begin{align*}
w^* &= \max \vec{\textbf{y}_1}\vec{\textbf{b}_1}+\vec{\textbf{y}_2}\vec{\textbf{b}_2}+\vec{\textbf{y}_3}\vec{\textbf{b}_3} \\
A_{11}^T\vec{\textbf{y}_1}+A_{21}^T\vec{\textbf{y}_2}+A_{31}^T\vec{\textbf{y}_3} &\leq \vec{\textbf{c}_1}\\
A_{12}^T\vec{\textbf{y}_1}+A_{22}^T\vec{\textbf{y}_2}+A_{32}^T\vec{\textbf{y}_3} &\ge \vec{\textbf{c}_2}\\
A_{13}^T\vec{\textbf{y}_1}+A_{23}^T\vec{\textbf{y}_2}+A_{33}^T\vec{\textbf{y}_3} &= \vec{\textbf{c}_3}\\
\vec{\textbf{y}_1} &\;\text{(no restriction)}\\
\vec{\textbf{y}_2} &\geq 0\\
\vec{\textbf{y}_3} &\leq 0
\end{align*}
I don't have anything else to say here

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