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Inserita descrizione Knapsack pag 53

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@ -37,6 +37,7 @@ Non è un sostituto all'esercizio e lo studio, l'esame richiede una buona dose d
\tableofcontents \tableofcontents
\newpage
\section{Modelli} \section{Modelli}
Mi dispiace, qua non troverete guide per fare i modelli, sfortunatamente questi van fatti alla nausea cercando di capire di volta in volta e costruendosi il metodo, riporto solamente delle fattispecie di problemi con soluzioni di tipologie degne di nota. Mi dispiace, qua non troverete guide per fare i modelli, sfortunatamente questi van fatti alla nausea cercando di capire di volta in volta e costruendosi il metodo, riporto solamente delle fattispecie di problemi con soluzioni di tipologie degne di nota.
@ -315,7 +316,7 @@ min: &8x_1 + 3x_2 + x_3 \\
&x_1,x_2,x_3^+,x_3^- \geqq 0 &x_1,x_2,x_3^+,x_3^- \geqq 0
\end{align*} \end{align*}
Le variabili free son state sdoppiate, $x_3 \Rightarrow x_3^+,x_3^-$ e sostituite nel sistema, la $x_3^+$ ha mantenuto sempre il suo segno, mentre la $x_3^-$ aveva l'opposto della sua controparte positiva.\\ Le variabili free son state sdoppiate, $x_3 \Rightarrow x_3^+,x_3^-$ e sostituite nel sistema, la $x_3^+$ ha mantenuto sempre il suo segno, mentre la $x_3^-$ aveva l'opposto della sua controparte positiva.\\
Nel secondo vincolo avevamo un uguaglianza, abbiamo quindi dovuto porre il segno a $\geqq$ e aggiungere una nuova riga, con gli stessi termini moltiplicati per $-1$ e segno $\leqq$. Nel primo vincolo avevamo un'uguaglianza, abbiamo quindi dovuto porre il segno a $\geqq$ e aggiungere una nuova riga, con gli stessi termini moltiplicati per $-1$ e segno $\leqq$.
\newpage \newpage
\section{Matrici : LP, ILP} \section{Matrici : LP, ILP}
@ -330,7 +331,7 @@ Abbiamo 3 strumenti, che non sono equivalenti (quasi), vediamo la mappa concettu
\end{center} \end{center}
Questo sarebbe il flusso logico per approcciare il problema, in realtà il 2 fasi può sempre sostituire il simplesso duale, non è vero il contrario, in ogni caso il 2 fasi essendo più lungo è sconveniente usarlo sempre. Questo sarebbe il flusso logico per approcciare il problema, in realtà il 2 fasi può sempre sostituire il simplesso duale, non è vero il contrario, in ogni caso il 2 fasi essendo più lungo è sconveniente usarlo sempre.
NB. La condizione di scelta del primo ramo "no" (costi ridotti negativi), non è discriminante della radice (termini noti positivi), ovvero, il simplesso standard, \underline{può} essere usato se i suoi costi ridotti sono negativi, è condizione necessaria che abbia i termini noti positivi. NB: la condizione di scelta del primo ramo "no" (costi ridotti negativi), non è discriminante della radice (termini noti positivi), ovvero, il simplesso standard, \underline{può} essere usato se i suoi costi ridotti sono negativi, è condizione necessaria che abbia i termini noti positivi.
\subsection{LP} \subsection{LP}
Linear Programming, in italiano Programmazione Lineare PL Linear Programming, in italiano Programmazione Lineare PL
\subsubsection{Simplesso} \subsubsection{Simplesso}
@ -1090,16 +1091,34 @@ levels: justify your choice.
} }
\\ \\
Soluzione:\\ Soluzione:\\
Per prima cosa dobbiamo ordinare gli oggetti per $ \frac{valore}{peso} $ in ordine crescente, facendo la frazione $ \frac{pj}{wj} $, una volta ordinati, in questo esercizio è già stato fatto dal prof, calcoliamo l'Upper bound del nodo p1 con Dantzig's bound\\ Per prima cosa dobbiamo ordinare gli oggetti per $ \frac{valore}{peso} $ in ordine decrescente, facendo la frazione $ \frac{pj}{wj} $, una volta ordinati, in questo esercizio è già stato fatto dal prof, calcoliamo l'Upper bound del nodo p1 con Dantzig's bound\\
\begin{center} \begin{center}
UB=$ \displaystyle \sum_{j=1}^{s-1} p_j + p_jx_s (\displaystyle \sum_{j=1}^{s-1} p_j + \lfloor p_j x_s \rfloor$ if integer $p_j) $ UB=$ \displaystyle \sum_{j=1}^{s-1} p_j + p_jx_s (\displaystyle \sum_{j=1}^{s-1} p_j + \lfloor p_j x_s \rfloor$ if integer $p_j) $
\end{center} \end{center}
~\\
Dalla figura seguente vediamo che arrivati a P4, ci fermiamo che non possiamo più inserire altri oggetti, e abbiamo la nostra prima soluzione P4 = \{1,1,1,0,0\} con valore 33, risalendo a P3 ci spostiamo a dx non prendendo l'ultimo oggetto (il terzo per la precisione, valore 8 peso 10), come ogni volta che non prendiamo un oggetto, ricalcoliamo UB, in questo caso abbiamo sempre valore 33, e non ha senso esplorare ulteriormente questo ramo.\\
\begin{center} \begin{center}
\raisebox{-.5\height}{\includegraphics[width=7cm]{immagini/grafi/3_2_1_0.jpg}} \raisebox{-.5\height}{\includegraphics[width=7cm]{immagini/grafi/3_2_1_0.jpg}}
\end{center} \end{center}
~\\
Dalla figura vediamo che arrivati a P4, ci fermiamo che non possiamo più inserire altri oggetti, e abbiamo la nostra prima soluzione P4 = \{1,1,1,0,0\} con valore 33, risalendo a P3 ci spostiamo a dx non prendendo l'ultimo oggetto (il terzo per la precisione, valore 8 peso 10), come ogni volta che non prendiamo un oggetto, ricalcoliamo UB, in questo caso abbiamo sempre valore 33, e non ha senso esplorare ulteriormente questo ramo ~\\
La sequenza delle operazioni è la seguente:
\begin{itemize}
\item Ordino gli oggetti in ordine decrescente di rapporto profitto/peso.
\item Scrivo il primo nodo e calcolo $U_0$ ed ipotizzo $C_0$ come massimo valore ammissibile, cioè C
\item Inizio l'esplorazione dell'albero:
\begin{itemize}
\item se vado a sinistra eredito $U_x$ (Upper Bound) e ricalcolo $\overline{c}_x$
\item se vado a destra eredito $\overline{c}_x$ (Capacità residua) e ricalcolo $U_x$
\end{itemize}
\item Per il calcolo di $U_x$ sommo i $p_j$ di tutti gli oggetti per i quali la somma dei $w_j$ risulta minore di C (Capacità del Knapsack). Se ho della capacità residua $\overline{c}$ allora la moltiplico per il valore $p_j/w_j$ e prendo il lower bound $\lfloor \alpha \rfloor$.
\item Se arrivo in fondo ad un ramo allora ho una soluzione intera, cioè senza parti di oggetti, e quindi calcolo Z (come valore) e X (come elenco di oggetti).
\item Ad ogni nodo, se $U_x$ è minore o uguale alla migliore Z trovata fino a quel momento, allora non proseguo nella esplorazione di quel ramo perchè non troverò un risultato migliore.
\item Ad ogni nodo, se andando a sinistra la $w_j$ dell'oggetto è inferiore alla capacità residua $\overline{c}$ allora mi fermo e non inserisco l'oggetto.
\item Se arrivo in fondo ad un ramo oppure mi sono fermato per poca capacità residua $\overline{c}$, allora risalgo l'albero fino a che non ho la possibilità di andare a destra.
\item Ho concluso il problema quando tutti i percorsi sono stati esplorati, ed ho trovato la $Z^*$ e la $X^*$.
\end{itemize}
\newpage \newpage
\section{GPLK} \section{GPLK}

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\contentsline {section}{\numberline {1}Modelli}{3}{section.1}% \contentsline {section}{\numberline {1}Modelli}{4}{section.1}%
\contentsline {subsection}{\numberline {1.1}Cenni di Base}{3}{subsection.1.1}% \contentsline {subsection}{\numberline {1.1}Cenni di Base}{4}{subsection.1.1}%
\contentsline {subsection}{\numberline {1.2}Esercizi}{5}{subsection.1.2}% \contentsline {subsection}{\numberline {1.2}Esercizi}{6}{subsection.1.2}%
\contentsline {subsubsection}{\numberline {1.2.1}Problema con Delta}{5}{subsubsection.1.2.1}% \contentsline {subsubsection}{\numberline {1.2.1}Problema con Delta}{6}{subsubsection.1.2.1}%
\contentsline {subsubsection}{\numberline {1.2.2}Problema con massima distanza}{6}{subsubsection.1.2.2}% \contentsline {subsubsection}{\numberline {1.2.2}Problema con massima distanza}{7}{subsubsection.1.2.2}%
\contentsline {subsubsection}{\numberline {1.2.3}Problema Variabile Triplo indice}{7}{subsubsection.1.2.3}% \contentsline {subsubsection}{\numberline {1.2.3}Problema Variabile Triplo indice}{8}{subsubsection.1.2.3}%
\contentsline {subsubsection}{\numberline {1.2.4}Problema di Trasporto}{9}{subsubsection.1.2.4}% \contentsline {subsubsection}{\numberline {1.2.4}Problema di Trasporto}{10}{subsubsection.1.2.4}%
\contentsline {subsubsection}{\numberline {1.2.5}Problema di Stoccaggio}{11}{subsubsection.1.2.5}% \contentsline {subsubsection}{\numberline {1.2.5}Problema di Stoccaggio}{12}{subsubsection.1.2.5}%
\contentsline {subsubsection}{\numberline {1.2.6}Problema di Stoccaggio 2}{12}{subsubsection.1.2.6}% \contentsline {subsubsection}{\numberline {1.2.6}Problema di Stoccaggio 2}{13}{subsubsection.1.2.6}%
\contentsline {section}{\numberline {2}Forme: Standard, Canonica, Generale}{13}{section.2}% \contentsline {section}{\numberline {2}Forme: Standard, Canonica, Generale}{14}{section.2}%
\contentsline {subsection}{\numberline {2.1}Standard $\Rightarrow $ Canonica}{13}{subsection.2.1}% \contentsline {subsection}{\numberline {2.1}Standard $\Rightarrow $ Canonica}{14}{subsection.2.1}%
\contentsline {subsection}{\numberline {2.2}Generale $\Rightarrow $ Standard}{13}{subsection.2.2}% \contentsline {subsection}{\numberline {2.2}Generale $\Rightarrow $ Standard}{14}{subsection.2.2}%
\contentsline {subsection}{\numberline {2.3}Esercizio Trasformazione Gran Fritto Misto}{14}{subsection.2.3}% \contentsline {subsection}{\numberline {2.3}Esercizio Trasformazione Gran Fritto Misto}{15}{subsection.2.3}%
\contentsline {section}{\numberline {3}Matrici : LP, ILP}{15}{section.3}% \contentsline {section}{\numberline {3}Matrici : LP, ILP}{16}{section.3}%
\contentsline {subsection}{\numberline {3.1}Fondamenti Concettuali}{15}{subsection.3.1}% \contentsline {subsection}{\numberline {3.1}Fondamenti Concettuali}{16}{subsection.3.1}%
\contentsline {subsubsection}{\numberline {3.1.1}Quale Simplesso?}{15}{subsubsection.3.1.1}% \contentsline {subsubsection}{\numberline {3.1.1}Quale Simplesso?}{16}{subsubsection.3.1.1}%
\contentsline {subsection}{\numberline {3.2}LP}{15}{subsection.3.2}% \contentsline {subsection}{\numberline {3.2}LP}{16}{subsection.3.2}%
\contentsline {subsubsection}{\numberline {3.2.1}Simplesso}{15}{subsubsection.3.2.1}% \contentsline {subsubsection}{\numberline {3.2.1}Simplesso}{16}{subsubsection.3.2.1}%
\contentsline {subsubsection}{\numberline {3.2.2}Simplesso Duale}{16}{subsubsection.3.2.2}% \contentsline {subsubsection}{\numberline {3.2.2}Simplesso Duale}{17}{subsubsection.3.2.2}%
\contentsline {subsubsection}{\numberline {3.2.3}2Fasi}{17}{subsubsection.3.2.3}% \contentsline {subsubsection}{\numberline {3.2.3}2Fasi}{18}{subsubsection.3.2.3}%
\contentsline {subsubsection}{\numberline {3.2.4}Duale del problema}{19}{subsubsection.3.2.4}% \contentsline {subsubsection}{\numberline {3.2.4}Duale del problema}{20}{subsubsection.3.2.4}%
\contentsline {subsection}{\numberline {3.3}Rappresentazione Grafica}{22}{subsection.3.3}% \contentsline {subsection}{\numberline {3.3}Rappresentazione Grafica}{23}{subsection.3.3}%
\contentsline {subsection}{\numberline {3.4}ILP}{24}{subsection.3.4}% \contentsline {subsection}{\numberline {3.4}ILP}{25}{subsection.3.4}%
\contentsline {subsubsection}{\numberline {3.4.1}Tagli di Gomory}{24}{subsubsection.3.4.1}% \contentsline {subsubsection}{\numberline {3.4.1}Tagli di Gomory}{25}{subsubsection.3.4.1}%
\contentsline {subsection}{\numberline {3.5}Esercizi Particolari}{26}{subsection.3.5}% \contentsline {subsection}{\numberline {3.5}Esercizi Particolari}{27}{subsection.3.5}%
\contentsline {subsubsection}{\numberline {3.5.1}Simplesso con variabili free}{26}{subsubsection.3.5.1}% \contentsline {subsubsection}{\numberline {3.5.1}Simplesso con variabili free}{27}{subsubsection.3.5.1}%
\contentsline {subsubsection}{\numberline {3.5.2}Problema PLC con simplesso e gomory in salsa teriyaki}{28}{subsubsection.3.5.2}% \contentsline {subsubsection}{\numberline {3.5.2}Problema PLC con simplesso e gomory in salsa teriyaki}{29}{subsubsection.3.5.2}%
\contentsline {subsubsection}{\numberline {3.5.3}Senstivity Analysis}{30}{subsubsection.3.5.3}% \contentsline {subsubsection}{\numberline {3.5.3}Senstivity Analysis}{31}{subsubsection.3.5.3}%
\contentsline {section}{\numberline {4}Grafi}{32}{section.4}% \contentsline {section}{\numberline {4}Grafi}{33}{section.4}%
\contentsline {subsection}{\numberline {4.1}GT}{32}{subsection.4.1}% \contentsline {subsection}{\numberline {4.1}GT}{33}{subsection.4.1}%
\contentsline {subsubsection}{\numberline {4.1.1}Dijkstra}{32}{subsubsection.4.1.1}% \contentsline {subsubsection}{\numberline {4.1.1}Dijkstra}{33}{subsubsection.4.1.1}%
\contentsline {subsubsection}{\numberline {4.1.2}Shortest Path Tree}{34}{subsubsection.4.1.2}% \contentsline {subsubsection}{\numberline {4.1.2}Shortest Path Tree}{35}{subsubsection.4.1.2}%
\contentsline {subsection}{\numberline {4.2}SST}{35}{subsection.4.2}% \contentsline {subsection}{\numberline {4.2}SST}{36}{subsection.4.2}%
\contentsline {subsubsection}{\numberline {4.2.1}Cenni di Base}{35}{subsubsection.4.2.1}% \contentsline {subsubsection}{\numberline {4.2.1}Cenni di Base}{36}{subsubsection.4.2.1}%
\contentsline {subsubsection}{\numberline {4.2.2}SST Prim's}{36}{subsubsection.4.2.2}% \contentsline {subsubsection}{\numberline {4.2.2}SST Prim's}{37}{subsubsection.4.2.2}%
\contentsline {subsubsection}{\numberline {4.2.3}SST Da Matrice trovare la soluzione ottimale}{37}{subsubsection.4.2.3}% \contentsline {subsubsection}{\numberline {4.2.3}SST Da Matrice trovare la soluzione ottimale}{38}{subsubsection.4.2.3}%
\contentsline {subsection}{\numberline {4.3}Max Flow}{38}{subsection.4.3}% \contentsline {subsection}{\numberline {4.3}Max Flow}{39}{subsection.4.3}%
\contentsline {subsubsection}{\numberline {4.3.1}Cenni di Base}{38}{subsubsection.4.3.1}% \contentsline {subsubsection}{\numberline {4.3.1}Cenni di Base}{39}{subsubsection.4.3.1}%
\contentsline {subsubsection}{\numberline {4.3.2}Flow Network}{39}{subsubsection.4.3.2}% \contentsline {subsubsection}{\numberline {4.3.2}Flow Network}{40}{subsubsection.4.3.2}%
\contentsline {subsection}{\numberline {4.4}DP}{41}{subsection.4.4}% \contentsline {subsection}{\numberline {4.4}DP}{42}{subsection.4.4}%
\contentsline {subsubsection}{\numberline {4.4.1}DP Knapsack 0-1 Dynamic Programming}{41}{subsubsection.4.4.1}% \contentsline {subsubsection}{\numberline {4.4.1}DP Knapsack 0-1 Dynamic Programming}{42}{subsubsection.4.4.1}%
\contentsline {subsubsection}{\numberline {4.4.2}DP: SSP Bellman's-Ford}{43}{subsubsection.4.4.2}% \contentsline {subsubsection}{\numberline {4.4.2}DP: SSP Bellman's-Ford}{44}{subsubsection.4.4.2}%
\contentsline {subsection}{\numberline {4.5}ILP}{45}{subsection.4.5}% \contentsline {subsection}{\numberline {4.5}ILP}{46}{subsection.4.5}%
\contentsline {subsubsection}{\numberline {4.5.1}ILP standard B\&B}{45}{subsubsection.4.5.1}% \contentsline {subsubsection}{\numberline {4.5.1}ILP standard B\&B}{46}{subsubsection.4.5.1}%
\contentsline {subsubsection}{\numberline {4.5.2}ILP esercizio standard B\&B}{49}{subsubsection.4.5.2}% \contentsline {subsubsection}{\numberline {4.5.2}ILP esercizio standard B\&B}{50}{subsubsection.4.5.2}%
\contentsline {subsubsection}{\numberline {4.5.3}ILP esercizio B\&B 0-1}{51}{subsubsection.4.5.3}% \contentsline {subsubsection}{\numberline {4.5.3}ILP esercizio B\&B 0-1}{52}{subsubsection.4.5.3}%
\contentsline {section}{\numberline {5}GPLK}{53}{section.5}% \contentsline {section}{\numberline {5}GPLK}{54}{section.5}%
\contentsline {subsection}{\numberline {5.1}Dal modello al codice}{53}{subsection.5.1}% \contentsline {subsection}{\numberline {5.1}Dal modello al codice}{54}{subsection.5.1}%
\contentsline {subsubsection}{\numberline {5.1.1}Modello Easy}{53}{subsubsection.5.1.1}% \contentsline {subsubsection}{\numberline {5.1.1}Modello Easy}{54}{subsubsection.5.1.1}%
\contentsline {subsubsection}{\numberline {5.1.2}Caso Particolare 1 Graffa}{54}{subsubsection.5.1.2}% \contentsline {subsubsection}{\numberline {5.1.2}Caso Particolare 1 Graffa}{55}{subsubsection.5.1.2}%
\contentsline {subsubsection}{\numberline {5.1.3}Caso Particolare 2 Sommatoria doppio insieme}{56}{subsubsection.5.1.3}% \contentsline {subsubsection}{\numberline {5.1.3}Caso Particolare 2 Sommatoria doppio insieme}{57}{subsubsection.5.1.3}%
\contentsline {subsection}{\numberline {5.2}Dal codice al modello}{57}{subsection.5.2}% \contentsline {subsection}{\numberline {5.2}Dal codice al modello}{58}{subsection.5.2}%
\contentsline {subsubsection}{\numberline {5.2.1}Normale}{57}{subsubsection.5.2.1}% \contentsline {subsubsection}{\numberline {5.2.1}Normale}{58}{subsubsection.5.2.1}%
\contentsline {section}{\numberline {6}Domande varie di teoria}{59}{section.6}% \contentsline {section}{\numberline {6}Domande varie di teoria}{60}{section.6}%
\contentsline {subsection}{\numberline {6.1}PLC degenerate}{59}{subsection.6.1}% \contentsline {subsection}{\numberline {6.1}PLC degenerate}{60}{subsection.6.1}%
\contentsline {subsection}{\numberline {6.2}PLC sensitivty}{60}{subsection.6.2}% \contentsline {subsection}{\numberline {6.2}PLC sensitivty}{61}{subsection.6.2}%
\contentsline {subsection}{\numberline {6.3}B\&B}{60}{subsection.6.3}% \contentsline {subsection}{\numberline {6.3}B\&B}{61}{subsection.6.3}%
\contentsline {subsection}{\numberline {6.4}PLC minimization}{61}{subsection.6.4}% \contentsline {subsection}{\numberline {6.4}PLC minimization}{62}{subsection.6.4}%
\contentsline {subsection}{\numberline {6.5}cutting Plane}{61}{subsection.6.5}% \contentsline {subsection}{\numberline {6.5}cutting Plane}{62}{subsection.6.5}%
\contentsline {subsection}{\numberline {6.6}PLC dual}{62}{subsection.6.6}% \contentsline {subsection}{\numberline {6.6}PLC dual}{63}{subsection.6.6}%
\contentsline {subsection}{\numberline {6.7}Dijkstra}{62}{subsection.6.7}% \contentsline {subsection}{\numberline {6.7}Dijkstra}{63}{subsection.6.7}%
\contentsline {subsection}{\numberline {6.8}B\&B knapsack}{63}{subsection.6.8}% \contentsline {subsection}{\numberline {6.8}B\&B knapsack}{64}{subsection.6.8}%
\contentsline {subsection}{\numberline {6.9}Branch \& cut}{63}{subsection.6.9}% \contentsline {subsection}{\numberline {6.9}Branch \& cut}{64}{subsection.6.9}%
\contentsline {subsection}{\numberline {6.10}PLC}{63}{subsection.6.10}% \contentsline {subsection}{\numberline {6.10}PLC}{64}{subsection.6.10}%
\contentsline {subsection}{\numberline {6.11}Soluzione Base}{64}{subsection.6.11}% \contentsline {subsection}{\numberline {6.11}Soluzione Base}{65}{subsection.6.11}%
\contentsline {subsection}{\numberline {6.12}Ford-Fulkerson}{64}{subsection.6.12}% \contentsline {subsection}{\numberline {6.12}Ford-Fulkerson}{65}{subsection.6.12}%
\contentsline {subsection}{\numberline {6.13}PLI minimization \& relaxation}{65}{subsection.6.13}% \contentsline {subsection}{\numberline {6.13}PLI minimization \& relaxation}{66}{subsection.6.13}%
\contentsline {subsection}{\numberline {6.14}B\&B knapsack 0-1}{65}{subsection.6.14}% \contentsline {subsection}{\numberline {6.14}B\&B knapsack 0-1}{66}{subsection.6.14}%
\contentsline {subsection}{\numberline {6.15}NP problem}{66}{subsection.6.15}% \contentsline {subsection}{\numberline {6.15}NP problem}{67}{subsection.6.15}%