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aggiunta parte sezione 3.2.4
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@ -19,7 +19,7 @@
\title{Gran Compendio OLI} \title{Gran Compendio OLI}
\author{\href{https://github.com/meschio94/Gran-Compendio-OLI}{Meschio}} \author{\href{https://github.com/meschio94/Gran-Compendio-OLI}{Meschio}}
\date{2021\\V1.0b} \date{2021\\V1.0c}
\begin{document} \begin{document}
\maketitle \maketitle
@ -461,7 +461,33 @@ max: &8u_1 - 2u_2 \\
\raisebox{-.5\height}{\includegraphics[width=5cm]{immagini/ProblemaDuale1.jpg}} \raisebox{-.5\height}{\includegraphics[width=5cm]{immagini/ProblemaDuale1.jpg}}
\end{center} \end{center}
Osservando l'ultima immagine per chiarezza col risultato finale che si riconduce al sistema iniziale. Osservando l'ultima immagine per chiarezza col risultato finale che si riconduce al sistema iniziale.\\
Se volessi risolvere il sistema invece, partendo dal sistema già orientato nel verso giusto e il suo problema duale:\\
\begin{align*}
min: &2x_1 + 4x_2 - 3x_3 & & & max: &5u_1 - 10u_2 \\
&x_1 - 3x_2 - x_3 \geqq -5 & & & &u_1 - u_2 \leqq 2\\
&-2x_1 - 2x_2 + x_3 \geqq -10 & & & &-3u_1 - 2u_2 \leqq 4\\
&x_1,x_2,x_3 \geqq 0 & & & &-u_1 + u_2 \leqq -3\\
& & & & &u_1,u_2 \geqq 0\\
\end{align*}
\begin{align*}
\left\{ \begin{array}{rcl} (x_1 -3x_2 - x_3 + 5)u_1 = 0 \\ (-2x_1 -2x_2 - x_3 + 10)u_2 = 0 \\ (u_1 -2u_2 - 2 )x_1 = 0 \\ (-3u_1 -2u_2 - 4 )x_2 = 0 \\ (-u_1 +u_2 + 3 )x_3 = 0 \end{array}\right. & &\Rightarrow & &\left\{ \begin{array}{rcl} (0)u_1 = 0 \\ (0)u_2 = 0 \\ u_1 -2u_2 = 2 \\ (-3u_1 -2u_2 - 2 )0 = 0 \\ (-u_1 +u_2) = -3 \end{array}\right.\\
\end{align*}
$u = (4,1) z_D= -30$
\begin{itemize}
\item Moltiplico ogni vincolo del primale per la sua variabile corrispondente del
duale e lo pongo uguale a zero, prima porto a sinistra il termine noto.\\
\item Moltiplico ogni vincolo del duale per la sua variabili corrispondente del
primale e lo pongo uguale a zero, prima porto a sinistra il termine noto.
\item Metto tutto a sistema e semplifico, ad esempio sappiamo i valori delle
variabili della funzione obiettivo del duale, inoltre i vincoli che ci portano alla
soluzione ottima saranno uguale a zero e i restanti diversi da \item Ottengo un sistema che posso risolvere, e dovrebbero essere presenti solo
variabili del problema primale, risolvo il sistema e ottengo i valori delle
variabili della funzione obiettivo del problema primale
\end{itemize}
\subsection{Rappresentazione Grafica} \subsection{Rappresentazione Grafica}
Anche tu sei a digiuno di grafici ed hai un incontrollato appetito di disegnarne altri ?\\ Anche tu sei a digiuno di grafici ed hai un incontrollato appetito di disegnarne altri ?\\

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\contentsline {subsection}{\numberline {6.1}PLC degenerate}{57}{subsection.6.1}% \contentsline {subsection}{\numberline {6.1}PLC degenerate}{58}{subsection.6.1}%
\contentsline {subsection}{\numberline {6.2}PLC sensitivty}{58}{subsection.6.2}% \contentsline {subsection}{\numberline {6.2}PLC sensitivty}{59}{subsection.6.2}%
\contentsline {subsection}{\numberline {6.3}B\&B}{58}{subsection.6.3}% \contentsline {subsection}{\numberline {6.3}B\&B}{59}{subsection.6.3}%
\contentsline {subsection}{\numberline {6.4}PLC minimization}{58}{subsection.6.4}% \contentsline {subsection}{\numberline {6.4}PLC minimization}{59}{subsection.6.4}%
\contentsline {subsection}{\numberline {6.5}cutting Plane}{58}{subsection.6.5}% \contentsline {subsection}{\numberline {6.5}cutting Plane}{59}{subsection.6.5}%
\contentsline {subsection}{\numberline {6.6}PLC dual}{59}{subsection.6.6}% \contentsline {subsection}{\numberline {6.6}PLC dual}{60}{subsection.6.6}%
\contentsline {subsection}{\numberline {6.7}Dijkstra}{59}{subsection.6.7}% \contentsline {subsection}{\numberline {6.7}Dijkstra}{60}{subsection.6.7}%
\contentsline {subsection}{\numberline {6.8}B\&B knapsack}{59}{subsection.6.8}% \contentsline {subsection}{\numberline {6.8}B\&B knapsack}{60}{subsection.6.8}%
\contentsline {subsection}{\numberline {6.9}Branch \& cut}{59}{subsection.6.9}% \contentsline {subsection}{\numberline {6.9}Branch \& cut}{60}{subsection.6.9}%
\contentsline {subsection}{\numberline {6.10}PLC}{60}{subsection.6.10}% \contentsline {subsection}{\numberline {6.10}PLC}{61}{subsection.6.10}%