From 099283c4d398c9ad382559882c72165e8e309078 Mon Sep 17 00:00:00 2001
From: NotXia <35894453+NotXia@users.noreply.github.com>
Date: Sat, 4 May 2024 19:05:51 +0200
Subject: [PATCH] Add CDMO2 linear programming and simplex

---
 .../module2/cdmo2.tex                         |   3 +
 .../module2/img/lp_brewery_problem.png        | Bin 0 -> 16961 bytes
 .../module2/img/lp_unbounded.png              | Bin 0 -> 19482 bytes
 .../module2/sections/_lp.tex                  | 480 ++++++++++++++++++
 4 files changed, 483 insertions(+)
 create mode 100644 src/year1/combinatorial-decision-making-and-optimization/module2/img/lp_brewery_problem.png
 create mode 100644 src/year1/combinatorial-decision-making-and-optimization/module2/img/lp_unbounded.png
 create mode 100644 src/year1/combinatorial-decision-making-and-optimization/module2/sections/_lp.tex

diff --git a/src/year1/combinatorial-decision-making-and-optimization/module2/cdmo2.tex b/src/year1/combinatorial-decision-making-and-optimization/module2/cdmo2.tex
index d6d1190..6d58097 100644
--- a/src/year1/combinatorial-decision-making-and-optimization/module2/cdmo2.tex
+++ b/src/year1/combinatorial-decision-making-and-optimization/module2/cdmo2.tex
@@ -5,11 +5,14 @@
 \def\lastupdate{{PLACEHOLDER-LAST-UPDATE}}
 
 \def\calT{\mathcal{T}}
+\def\calB{\mathcal{B}}
+\def\calN{\mathcal{N}}
 
 
 \begin{document}
     
     \makenotesfront
     \input{./sections/_smt.tex}
+    \input{./sections/_lp.tex}
 
 \end{document}
\ No newline at end of file
diff --git a/src/year1/combinatorial-decision-making-and-optimization/module2/img/lp_brewery_problem.png b/src/year1/combinatorial-decision-making-and-optimization/module2/img/lp_brewery_problem.png
new file mode 100644
index 0000000000000000000000000000000000000000..bf8640dcf0c59e98ea2b4d7ee39a08f30fa900cc
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+\chapter{Linear programming}
+
+
+\begin{description}
+    \item[Linear programming (LP)] \marginnote{Linear programming (LP)}
+        Optimization problem defined through a system of linear constraints (equalities and inequalities) 
+        and an objective function.
+
+        In the Cartesian plane, equalities represent hyperplanes and inequalities are half-spaces.
+
+        \begin{remark}
+            LP is useful for optimal allocation with limited number of resources.
+        \end{remark}
+
+    
+    \item[Feasible solution] \marginnote{Feasible solution}
+        Assignment satisfying all the constraints.
+
+        In the Cartesian plane, it is represented by any point within 
+        the intersection of all half-spaces defined by the inequalities.
+
+    \item[Feasible region] \marginnote{Feasible region}
+        Set of all feasible solutions.
+        It is a convex polyhedron that can be empty, bounded or unbounded.
+
+        \begin{remark}
+            The optimal solution is always at one of the intersection points of the constraints within the feasible region (i.e. vertexes of the polyhedron).
+
+            The number of vertexes is finite but might grow exponentially.
+        \end{remark}
+
+    
+    \item[Canonical form] \marginnote{Canonical form}
+        A linear programming problem is in canonical form if it is defined as:
+        \[ 
+            \begin{split}
+                \max \sum_{j=1}^{n} c_j x_j \text{ subject to } &\sum_{j=1}^{n} a_{i,j} x_j \leq b_i \text{ for } 1 \leq i \leq m \,\,\land \\
+                    & x_j \geq 0 \text{ for } 1 \leq j \leq n
+            \end{split}
+        \]
+        where:
+        \begin{itemize}
+            \item $m$ is the number of constraints.
+            \item $n$ is the number of non-negative variables.
+            \item $a_{i,j}, b_i, c_j \in \mathbb{R}$ are known parameters (given by the definition of the problem).
+            \item $\sum_{j=1}^{n} c_j x_j$ is the objective function to maximize.
+            \item $\sum_{j=1}^{n} a_{i,j} x_j \leq b_i$ are $m$ linear inequalities.
+        \end{itemize}
+
+        In matrix form:
+        \[ \max \{ \vec{cx} \} \text{ subject to } \matr{A}\vec{x} \leq \vec{b} \land \vec{x} \geq \nullvec \]
+        where:
+        \begin{itemize}
+            \item $\vec{c} = \begin{bmatrix} c_1 & \hdots & c_n \end{bmatrix}$.
+            \item $\vec{x} = \begin{bmatrix} x_1 & \hdots & x_n \end{bmatrix}^T$.
+            \item $\vec{b} = \begin{bmatrix} b_1 & \hdots & b_m \end{bmatrix}^T$.
+            \item $\matr{A} = \begin{bmatrix} a_{1,1} & \hdots & a_{1,n} \\ \vdots & \ddots & \vdots \\ a_{m,1} & \hdots & a_{m,n} \end{bmatrix}$
+        \end{itemize}
+
+
+    \item[Standard form] \marginnote{Standard form}
+        A linear programming problem is in standard form if it only has equality constraints (excluded those on single variables):
+        \[ 
+            \begin{split}
+                \max \sum_{j=1}^{n} c_j x_j \text{ subject to } &\sum_{j=1}^{n} a_{i,j} x_j = b_i \text{ for } 1 \leq i \leq m \,\,\land \\
+                    & x_j \geq 0 \text{ for } 1 \leq j \leq n
+            \end{split}
+        \]
+        In matrix form: $\max \{ \vec{cx} \} \text{ subject to } \matr{A}\vec{x} = \vec{b} \land \vec{x} \geq \nullvec$.
+
+        \begin{remark}[Canonical to standard form]
+            Any LP problem with $m$ constraints in canonical form has an equivalent standard form with $m$ slack variables $y_1, \dots, y_m \geq 0$
+            such that:
+            \[ 
+                \forall i \in \{ 1, \dots, m \}:
+                \left( \sum_{j=1}^{n} a_{i,j} x_j \leq b_i \right) \Rightarrow 
+                \left( \sum_{j=1}^{n} a_{i,j} x_j + y_i= b_i \land y_i \geq 0 \right)
+            \]
+        \end{remark}
+\end{description}
+
+\begin{example}[Brewery problem]
+    The definition of the problem in canonical form is the following:
+    \begin{center}
+        \begin{tabular}{lccccc}
+            $\max$ & $13A$ & $+$ & $23B$ \\
+            subj. to & $5A$ & $+$ & $15B$ & $\leq$ & 480 \\
+                & $4A$ & $+$ & $4B$ & $\leq$ & 160 \\
+                & $35A$ & $+$ & $20B$ & $\leq$ & 1190 \\
+                & $A$ & , & $B$ & $\geq$ & 0 \\
+        \end{tabular}
+    \end{center}
+
+    \begin{figure}[H]
+        \centering
+        \includegraphics[width=0.3\linewidth]{./img/lp_brewery_problem.png}
+        \caption{Feasible regions and vertexes of the polyhedron}
+    \end{figure}
+
+    In standard form, it becomes:
+    \begin{center}
+        \begin{tabular}{lccccccccccc}
+            $\max$ & $13A$ & $+$ & $23B$ \\
+            subj. to & $5A$ & $+$ & $15B$ & $+$ & $S_C$ & & & & & $=$ & 480 \\
+                & $4A$ & $+$ & $4B$ & & & $+$ & $S_H$ & & & $=$ & 160 \\
+                & $35A$ & $+$ & $20B$ & & & & & $+$ & $S_M$ & $=$ & 1190 \\
+                & $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & $\geq$ & 0 \\
+        \end{tabular}
+    \end{center}
+\end{example}
+
+
+
+\section{Simplex algorithm}
+
+Algorithm that starts from an extreme point of the polyhedron and iteratively moves to a neighboring vertex as long as the objective function does not decrease.
+
+
+\subsection{Basis}
+
+\begin{description}
+    \item[Basis] \marginnote{Basis}
+        Given an LP problem $\mathcal{P}$ in standard form with $m$ constraints and $n$ variables (note: in standard form, it holds that $m \leq n$) and
+        its constraint matrix $\matr{A} \in \mathbb{R}^{m \times n}$,
+        a (ordered) basis $\calB = \{ x_{i_1}, \dots, x_{i_m} \}$ is a subset of $m$ of the $n$ variables such that 
+        the columns $\matr{A}_{i_1}, \dots, \matr{A}_{i_m}$ of $\matr{A}$ form a $m \times m$ invertible matrix $\matr{A}_\calB$.
+
+        Variables in $\calB$ are basic variables while 
+        $\calN = \{ x_1, \dots, x_n \} \smallsetminus \calB$ are non-basic variables.
+
+        $\mathcal{P}$ can be rewritten by separating basic and non-basic variables:
+        \[ 
+            \max \{ \vec{c}_\calB \vec{x}_\calB  + \vec{c}_\calN\vec{x}_\calN \} \text{ subject to } 
+                \matr{A}_\calB \vec{x}_\calB + \matr{A}_\calN \vec{x}_\calN = \vec{b} \,\land\, \vec{x}_\calB, \vec{x}_\calN \geq \nullvec 
+        \]
+
+    \item[Basic solution] \marginnote{Basic solution}
+        By constraining $\vec{x}_\calN = \nullvec$, an LP problem becomes:
+        \[ 
+            \max \{ \vec{c}_\calB \vec{x}_\calB \} \text{ subject to } 
+                \matr{A}_\calB \vec{x}_\calB = \vec{b} \,\land\, \vec{x}_\calB \geq \nullvec 
+        \]
+        As $\matr{A}_\calB$ is invertible by definition, it holds that:
+        \[ 
+            \vec{x}_\calB = \matr{A}_\calB^{-1} \vec{b} \hspace{2em}\text{ and }\hspace{2em} 
+            \max \{ \vec{c}_\calB \vec{x}_\calB \} = \max \{ \vec{c}_\calB \matr{A}_\calB^{-1} \vec{b} \} 
+        \]
+        $\vec{x}_\calB$ is a basic solution for $\calB$.
+
+    \item[Basic feasible solution (BFS)] \marginnote{Basic feasible solution (BFS)}
+        Given a basic solution $\vec{x}_\calB$ for $\calB$,
+        it is feasible iff:
+        \[ \forall_{i=1}^m \vec{x}_{\calB_i} \geq 0 \]
+
+        \begin{description}
+            \item[Non-degenerate BFS] \marginnote{Non-degenerate BFS}
+                A basic feasible solution is non-degenerate iff $\forall_{i=1}^m \vec{x}_{\calB_i} > 0$.
+                \begin{remark}
+                    A non-degenerate BFS is represented by a unique basis.
+                \end{remark}
+        \end{description}
+
+        \begin{remark}
+            The simplex algorithm iteratively moves through basic feasible solutions $\tilde{\vec{x}}_1, \tilde{\vec{x}}_2, \dots$
+            such that $\vec{c}\tilde{\vec{x}}_k \geq \vec{c}\tilde{\vec{x}}_{k-1}$.
+        \end{remark}
+\end{description}
+
+
+\subsection{Tableau}
+\marginnote{Tableau}
+
+Tabular representation to describe the steps of the simplex algorithm.
+A variable $Z$ is introduced to represent the value of the objective function (which can be seen as a conversion of the objective function into a constraint).
+
+The tableau of an LP problem in standard form is divided into three sections:
+\begin{enumerate}
+    \item Objective function, where the coefficients of the variables are called reduced costs\marginnote{Reduced costs}.
+    \item Equality constraints.
+    \item Variable constraints.
+\end{enumerate}
+
+\begin{example}[Brewery problem]
+    In standard form, the brewery problem is defined as:
+    \begin{center}
+        \begin{tabular}{lccccccccccc}
+            $\max$ & $13A$ & $+$ & $23B$ \\
+            subj. to & $5A$ & $+$ & $15B$ & $+$ & $S_C$ & & & & & $=$ & 480 \\
+                & $4A$ & $+$ & $4B$ & & & $+$ & $S_H$ & & & $=$ & 160 \\
+                & $35A$ & $+$ & $20B$ & & & & & $+$ & $S_M$ & $=$ & 1190 \\
+                & $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & $\geq$ & 0 \\
+        \end{tabular}
+    \end{center}
+
+    As a tableau, assuming an initial basis $\calB = \{ S_C, S_H, S_M \}$, the problem is represented as:
+    \begin{center}
+        \begin{tabular}{cccccccccccccc}
+            \toprule
+            $13A$ & $+$ & $23B$ & & & & & & & $-$ & $Z$ & $=$ & 0 \\
+            \midrule
+            $5A$ & $+$ & $15B$ & $+$ & $S_C$ & & & & & & & $=$ & 480 \\
+            $4A$ & $+$ & $4B$ & & & $+$ & $S_H$ & & & & & $=$ & 160 \\
+            $35A$ & $+$ & $20B$ & & & & & $+$ & $S_M$ & & & $=$ & 1190 \\
+            \midrule
+            $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & & & $\geq$ & 0 \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+    The reduced costs are $\{ 13, 23, 0, 0, 0 \}$.
+\end{example}
+
+
+\subsection{Pivoting}
+
+Given a basis $\calB$, it is possible to insert a new variable $x^\text{in} \notin \calB$ into it and remove an old one $x^\text{out} \in \calB$ 
+to increase (or leave unchanged) the objective function:
+
+\begin{descriptionlist}
+    \item[Entering variable] \marginnote{Entering variable}
+        $x^\text{in}$ should be the variable in $\calN$ with the highest improvement on the objective function.
+
+    \item[Leaving variable] \marginnote{Leaving variable}
+        $x^\text{out}$ should be chosen to ensure that the new basis $\calB' = \calB \cup \{ x^\text{in} \} \smallsetminus \{ x^\text{out} \}$
+        is still a feasible basis.
+
+        \begin{description}
+            \item[Minimum ratio rule] \marginnote{Minimum ratio rule}
+                For each constraint $i$ (i.e. $i$-th row of the system $\matr{A}\vec{x} = \vec{b}$), it is possible to compute the ratio:
+                \[ \frac{\vec{b}_i}{\vec{\alpha}_i^\text{in}} \]
+                where:
+                \begin{itemize}
+                    \item $\vec{\alpha}_i^\text{in}$ is the coefficient associated to the entering variable $x^\text{in}$ in the $i$-th constraint.
+                    \item $\vec{b}_i$ is the known term of the $i$-th constraint.
+                \end{itemize}
+
+                The index $i$ of the leaving variable $x^\text{out} = \calB_i$ is determined as: 
+                \[ \arg\min_i \frac{\vec{b}_i}{\vec{\alpha}_i^\text{in}} \]
+        \end{description}
+\end{descriptionlist}
+
+Once $x^\text{in}$ and $x^\text{out} = \calB_i$ has been determined, 
+$x^\text{in}$ is isolated in the equation of the $i$-th constraint and it is substituted in all the others.
+
+\begin{example}[Brewery problem]
+    The initial tableau of the brewery problem with $\calB = \{ S_C, S_H, S_M \}$ and $\calN = \{ A, B \}$ is:
+    \begin{center}
+        \begin{tabular}{cccccccccccccc}
+            \toprule
+            $13A$ & $+$ & $23B$ & & & & & & & $-$ & $Z$ & $=$ & 0 \\
+            \midrule
+            $5A$ & $+$ & $15B$ & $+$ & $S_C$ & & & & & & & $=$ & 480 \\
+            $4A$ & $+$ & $4B$ & & & $+$ & $S_H$ & & & & & $=$ & 160 \\
+            $35A$ & $+$ & $20B$ & & & & & $+$ & $S_M$ & & & $=$ & 1190 \\
+            \midrule
+            $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & & & $\geq$ & 0 \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \begin{enumerate}
+        \item 
+            It can be easily seen that $x^\text{in} = B$ should be the entering variable.
+        
+            For the leaving variable, the ratios are:
+            \[ \arg\min \left\{ \frac{480}{15}, \frac{160}{4}, \frac{1190}{20} \right\} = \arg\min \left\{ 32, 40, 59.5 \right\} = 1 \]
+            Therefore, the leaving variable is $x^\text{out} = \calB_1 = S_C$.
+        
+            We now isolate $B$ from the first constraint:
+            \[ 
+                \begin{split}
+                    5A + 15B + S_C = 480 &\iff \frac{1}{3}A + B + \frac{1}{15}S_C = 32 \\
+                        &\iff B = 32 - \frac{1}{3}A - \frac{1}{15}S_C \\
+                \end{split}
+            \]
+
+            The tableau with $\calB' = \{ B, S_H, S_M \}$ and $\calN' = \{ A, S_C \}$ is updated as:
+            \begin{center}
+                \begin{tabular}{cccccccccccccc}
+                    \toprule
+                    $\frac{16}{3}A$ & & & $-$ & $\frac{23}{15}S_C$ & & & & & $-$ & $Z$ & $=$ & $-736$ \\
+                    \midrule
+                    $\frac{1}{3}A$ & $+$ & $B$ & $+$ & $\frac{1}{15}S_C$ & & & & & & & $=$ & $32$ \\
+                    $\frac{8}{3}A$ & & & $-$ & $\frac{4}{15}S_C$ & $+$ & $S_H$ & & & & & $=$ & $32$ \\
+                    $\frac{85}{3}A$ & & & $-$ & $\frac{4}{3}S_C$ & & & $+$ & $S_M$ & & & $=$ & $550$ \\
+                    \midrule
+                    $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & & & $\geq$ & 0 \\
+                    \bottomrule
+                \end{tabular}
+            \end{center}
+
+        \item 
+            Now, $x^\text{in} = A$ is the variable that increases the objective function the most.
+
+            For the leaving variable, the ratios are:
+            \[ \arg\min \left\{ \frac{32}{1/3}, \frac{32}{8/3}, \frac{550}{85/3} \right\} = 2 \]
+            Therefore, the leaving variable is $x^\text{out} = \calB'_2 = S_H$.
+
+            $A$ isolated in the second constraint brings to:
+            \[ 
+                \begin{split}
+                    A = \frac{3}{8} (32 + \frac{4}{15}S_C - S_H) \,\iff\, A = 12 + \frac{1}{10}S_C - \frac{3}{8}S_H
+                \end{split}    
+            \]
+
+            The tableau with $\calB'' = \{ A, B, S_M \}$ and $\calN'' = \{ S_C, S_H \}$ is updated as:
+            \begin{center}
+                \begin{tabular}{cccccccccccccc}
+                    \toprule
+                     & & & $-$ & $S_C$ & $-$ & $2S_H$ & & & $-$ & $Z$ & $=$ & $-800$ \\
+                    \midrule
+                    & & $B$ & $+$ & $\frac{1}{10}S_C$ & $+$ & $\frac{1}{8}S_H$ & & & & & $=$ & $28$ \\
+                    $A$ & & & $-$ & $\frac{1}{10}S_C$ & $+$ & $\frac{3}{8}S_H$ & & & & & $=$ & $12$ \\
+                    & & & $-$ & $\frac{25}{6}S_C$ & $-$ & $\frac{85}{8}S_H$ & $+$ & $S_M$ & & & $=$ & $210$ \\
+                    \midrule
+                    $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & & & $\geq$ & 0 \\
+                    \bottomrule
+                \end{tabular}
+            \end{center}
+
+            We cannot continue anymore as the reduced costs $\{ 0, 0, -1, -2, 0 \}$ are $\leq 0$ (i.e. cannot improve the objective function anymore).
+    \end{enumerate}
+\end{example}
+
+
+\subsection{Optimality}
+\marginnote{Optimality}
+
+When any substitution worsens the objective function, the current assignment is optimal.
+In the tableau, this happens when all the reduced costs are $\leq 0$.
+
+\begin{remark}
+    For any optimal solution, there is at least a basis such that the reduced costs are $\leq 0$.
+    Therefore, this is a sufficient condition.
+\end{remark}
+
+\begin{remark}
+    The fact that reduced costs are $\leq 0$ is not a necessary condition.
+\end{remark}
+
+\begin{example}[Brewery problem]
+    The tableau at the last iteration, with $\calB'' = \{ A, B, S_M \}$ and $\calN'' = \{ S_C, S_H \}$, is the following:
+    \begin{center}
+        \begin{tabular}{cccccccccccccc}
+            \toprule
+             & & & $-$ & $S_C$ & $-$ & $2S_H$ & & & $-$ & $Z$ & $=$ & $-800$ \\
+            \midrule
+            & & $B$ & $+$ & $\frac{1}{10}S_C$ & $+$ & $\frac{1}{8}S_H$ & & & & & $=$ & $28$ \\
+            $A$ & & & $-$ & $\frac{1}{10}S_C$ & $+$ & $\frac{3}{8}S_H$ & & & & & $=$ & $12$ \\
+            & & & $-$ & $\frac{25}{6}S_C$ & $-$ & $\frac{85}{8}S_H$ & $+$ & $S_M$ & & & $=$ & $210$ \\
+            \midrule
+            $A$ & , & $B$ & , & $S_C$ & , & $S_H$ & , & $S_M$ & & & $\geq$ & 0 \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    All reduced costs are $\leq 0$ and an optimal solution has been reached:
+    \begin{itemize}
+        \item $S_C = 0$ and $S_H = 0$ (variables in $\calN''$).
+        \item $A = 12$, $B = 28$ and $S_M = 210$ (variables in $\calB''$. Obtained by isolating them in the constraints).
+        \item $- S_C - 2S_H - Z = -800 \iff Z = 800 - S_C - 2S_H \iff Z = 800$ (objective function).
+    \end{itemize}
+\end{example}
+
+\begin{remark}
+    The points in the feasible region of a problem $\mathcal{P}$ are:
+    \[ \mathcal{F}_\mathcal{P} = \{ \vec{x} \in \mathbb{R}^n \mid \matr{A}\vec{x} \leq \vec{b} \land \vec{x} \geq \nullvec \} \]
+\end{remark}
+
+\begin{description}
+    \item[Optimal region] \marginnote{Optimal region}
+        For an LP problem $\mathcal{P}$, its set of solutions is defined as:
+        \[ \mathcal{O}_\mathcal{P} = \{ x^* \in \mathcal{F}_\mathcal{P} \mid \forall \vec{x} \in \mathcal{F}_\mathcal{P}: \vec{c}\vec{x}^* \geq \vec{c}\vec{x} \} \]
+
+        Trivially, it holds that $\mathcal{O}_\mathcal{P} \subseteq \mathcal{F}_\mathcal{P}$ and 
+        $\mathcal{F}_\mathcal{P} = \varnothing \Rightarrow \mathcal{O}_\mathcal{P} = \varnothing$.
+
+        \begin{theorem}
+            If $\mathcal{O}_\mathcal{P}$ is finite, then $\vert \mathcal{O}_\mathcal{P} \vert = 1$ 
+            (therefore, if $\vert \mathcal{O}_\mathcal{P} \vert > 1$, then $\mathcal{O}_\mathcal{P}$ is infinite).
+        \end{theorem}
+\end{description}
+
+\begin{description}
+    \item[Unbounded problem] \marginnote{Unbounded problem} 
+        An LP problem $\mathcal{P}$ is unbounded if it does not have an optimal solution (i.e. $\mathcal{F}_\mathcal{P}$ is an unbounded polyhedron).
+
+        With the tableau formulation, if a column has reduced cost $> 0$ and all the constraint coefficients are $\leq 0$, then the problem is unbounded.
+
+        \begin{example}
+            Given the following tableau:
+            \begin{center}
+                \begin{tabular}{cccccccccccc}
+                    \toprule
+                    & $x$ & $-$ & $y$ & & & & & $-$ & $Z$ & $=$ & $-1$ \\
+                    \midrule
+                    $-$ & $x$ & $-$ & $y$ & $+$ & $S_1$ & & & & & $=$ & $0$ \\
+                    $-$ & $2x$ & $-$ & $y$ & & & $+$ & $S_2$ & & & $=$ & $1$ \\
+                    \midrule
+                    & $x$ & , & $y$ & , & $S_1$ & , & $S_2$ & & & $\geq$ & 0 \\
+                    \bottomrule
+                \end{tabular}
+            \end{center}
+            The unboundedness of the problem can be detected from the first column.
+
+            \begin{figure}[H]
+                \centering
+                \includegraphics[width=0.3\linewidth]{./img/lp_unbounded.png}
+            \end{figure}
+    \end{example}
+\end{description}
+
+
+\subsection{Algorithm}
+
+Given an LP problem $\mathcal{P}$ in standard form, the steps of the simplex algorithm are:
+\begin{enumerate}
+    \item Set $k=0$, find a feasible basis $\calB_k$ and reformulate $\mathcal{P}$ according to it.
+    \item If the basis feasible solution is optimal, return.
+    \item If $\mathcal{P}$ is unbounded, return.
+    \item Select an entering variable $x^\text{in}$.
+    \item Select a leaving variable $x^\text{out}$.
+    \item Let $\calB_{k+1} = \calB_k \cup \{ x^\text{in} \} \smallsetminus \{ x^\text{out} \}$ and reformulate $\mathcal{P}$ according to the new basis.
+    \item Set $k = k + 1$ and go to Point 2.
+\end{enumerate}
+
+\begin{description}
+    \item[Properties] \phantom{}
+        \begin{itemize}
+            \item If all basis feasible solutions are non-degenerate, the simplex algorithm always terminates (as the solution is always strictly improving).
+            \item In the general case, the algorithm might stall by ending up in a loop.
+            \item The worst-case time complexity is $O(2^n)$. The average case is polynomial.
+        \end{itemize}
+\end{description}
+
+\begin{remark}
+    If the problem has lots of vertexes, the interior point method (polynomial complexity) or approximation algorithms should be preferred.
+\end{remark}
+
+
+\subsection{Two-phase method}
+\marginnote{Two-phase method}
+
+Method that solves an LP problem by first finding an initial feasible basis $\calB_0$ and determining if the LP problem is unsatisfiable.
+
+Given an LP problem $\mathcal{P}$ ($\max\{ \vec{cx} \} \text{ subject to } \matr{A}\vec{x} = \vec{b}$ with $m$ constraints and $n$ variables), 
+the two-phase method works as follows:
+\begin{descriptionlist}
+    \item[Phase 1] 
+        Define a new artificial problem $\mathcal{P}'$ from $\mathcal{P}$ with new variables $s_1, \dots, s_m$ as follows:
+        \[
+            \begin{split}
+                \max\left\{ -\sum_{i=1}^{m} s_i \right\} \text{ subject to } 
+                    &\sum_{j=1}^{n} a_{i,j} x_j + s_i = b_i \text{ for } i \in \{ k \in \{ 1, \dots, m \} \mid b_k \geq 0 \} \,\land \\
+                    &\sum_{j=1}^{n} a_{i,j} x_j - s_i = b_i \text{ for } i \in \{ k \in \{ 1, \dots, m \} \mid b_k < 0 \} \,\land \\
+                    & s_i, x_j \geq 0
+            \end{split}
+        \]
+
+        \begin{remark}
+            It holds that $-\sum_{i=1}^{m} s_i \leq 0$ and 
+            $\calB' = \{ s_1, \dots, s_m \}$ is always a feasible basis corresponding to the basis feasible solution $x_j = 0$, $s_i = \vert b_i \vert$
+        \end{remark}
+
+        The problem $\mathcal{P}'$ with basis $\calB'$ can be solved through the simplex method.
+
+        \begin{theorem}
+            Let $\mathcal{F}_\mathcal{P}$ be the feasible region of $\mathcal{P}$. It holds that:
+            \[ \mathcal{F}_\mathcal{P} \neq \varnothing \,\iff\, \sum_{i=1}^{m} s_i = 0 \]
+
+            In other words:
+            \begin{itemize}
+                \item If the optimal solution of $\mathcal{P}'$ is  $< 0$, then $\mathcal{P}$ is unsatisfiable.
+                \item Otherwise, the basis $\calB_{\mathcal{P}'}$ corresponding to the optimal solution of $\mathcal{P}'$ 
+                    can be used as the initial basis of $\mathcal{P}$, after removing the variables $s_i$.
+            \end{itemize}
+        \end{theorem}
+
+    \item[Phase 2] 
+        Solve $\mathcal{P}$ through the simplex algorithm using as initial basis $\calB_{\mathcal{P}'}$.
+\end{descriptionlist}
\ No newline at end of file