diff --git a/Claudio_Maggioni_5/Claudio_Maggioni_5.md b/Claudio_Maggioni_5/Claudio_Maggioni_5.md
index 1ecd25c..e3a7f41 100644
--- a/Claudio_Maggioni_5/Claudio_Maggioni_5.md
+++ b/Claudio_Maggioni_5/Claudio_Maggioni_5.md
@@ -23,24 +23,127 @@ header-includes:
 The resulting MATLAB plot of each constraint and of the feasible region is shown
 below:
 
-![Plot of feasible region and constraints](./ex2-1.png)
+![Plot of feasible region and constraints\label{fig:a}](./ex2-1.png)
+
+## Exercise 2.2
+
+<!--The Simplex method is used to solve linear programs which are defined as
+follows:
+
+$$\min c^Tx, \text{ subject to } Ax = b, x > 0$$
+
+And when we have inequalities constraints such as:
+
+$$\min c^Tx,\text{ subject to }Ax \leq b$$
+
+We can introduce slack variable to convert the inequalities into equalities:
+
+$$\min c^Tx,\text{ subject to }Ax + z = 0, z>0$$
+
+Each iterate generated by the simplex method is a basic feasible point which
+is a-->
+
+According to Nocedal, a vector $x$ is a basic feasible point if it is in the
+feasible region and if there exists a subset $\beta$ of the
+index set $1, 2, \ldots, n$ such that:
+
+- $\beta$ contains exactly $m$ indices, where $m$ is the number of rows of $A$;
+- For any $i \notin \beta$, $x_i = 0$, meaning the bound $x_i \geq 0$ can be
+  inactive only if $i \in \beta$;
+- The $m$ x $m$ matrix $B$ defined by $B = [A_i]_{i \in \beta}$ (where $A_i$ is
+  the i-th column of A) is non-singular, i.e. all columns corresponding to the
+  indices in $\beta$ are linearly independent from each other.
+
+The geometric interpretation of basic feasible points is that all of them
+are verticies of the polytope that bounds the feasible region. We will use this
+proven property to manually solve the constrained minimization problem presented
+in this section by aiding us with the graphical plot of the feasible region in
+figure \ref{fig:a}.
+
+<!--
+And as already been
+said, the basic feasible point is the basic feasible solution for the problem.
+Regarding the section 13.3, which outline how the steps are done, we know that
+most steps consist of a move from one vertex to an adjacent one for which the
+basis $\beta$ differs exactly one component.The step is an edge along which the
+objective function is reduced.
+
+Then, one major issue at every simplex iteration
+is to decide which index must be removed from the basis. As the book specify,
+unless the step is a direction of unboundedness, a single index must be removed
+by replacing it with another from outside $\beta$.
+
+From a geometrical point of view,
+a move, is along an edge of the feasible polytope that decreases $c^Tx$ and we
+continue moving along that edge until we find a new vertex. One we find a vertex
+we defined the new constraint $x_p > 0$ which is one of the component $x_p,p \in \beta$
+decreased to zero. Afterwards we can remove the index $p$ from the basis $\beta$ and
+replace it with $q$. A more detailed visualisation can be see in the following
+Figure 1 taken from the book.-->
 
 ## Exercise 2.3
 
+Since the geometrical interpretation of the definition of basic feasible point
+states that these point are non-other than the vertices of the feasible region,
+we first look at the plot above and to these points (i.e. the verticies of the
+bright green non-trasparent region). Then, we look which constraint boundaries cross these
+edges, and we formulate an algebraic expression to find these points. In
+clockwise order, we have:
+
+- The lower-left point at the origin, given by the boundaries of the constraints
+ $x_1 \geq 0$ and $x_2 \geq 0$:
+ $$x^*_1 = \begin{bmatrix}0\\0\end{bmatrix}$$
+- The top-left point, at the intersection of constraint boundaries $x_1 \geq 0$ and
+  $-3x_1 + 2x_2 \leq 3$:
+  $$x_1 = 0 \;\;\; 2x_2 = 3 \Leftrightarrow x_2 = \frac32 \;\;\; x^*_2 =
+  \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix}$$
+- The top-center-left point at the intersection of constraint boundaries $-3x_1 +
+  2x_2 \leq 3$ and $2x_1 + 3x_2 \leq 6$:
+  $$-3x_1 + 2x_2 + 3x_1 + \frac92 x_2 = \frac{13}{2} x_2 = 3 + 9 = 12
+  \Leftrightarrow x_2 = 12 \cdot \frac2{13} = \frac{24}{13}$$$$
+  -3x_1 + 2 \cdot \frac{24}{13} = 3 \Leftrightarrow x_1 = \frac{39 - 48}{13}
+  \cdot \frac{1}{-3} = \frac{3}{13} \;\;\; x^*_3 = \frac{1}{13} \cdot
+  \begin{bmatrix}3\\24\end{bmatrix}$$
+- The top-center-right point at the intersection of constraint boundaries $2x_1
+  + 3x_2 \leq 6$ and $2x_1 + x_2 \leq 4$:
+  $$2x_1 + 3x_2 - 2x_1 - x_2 = 2x_2 = 6 - 4 = 2 \Leftrightarrow x_2 = 1 \;\;\;
+  2x_1 + 1 = 4 \Leftrightarrow x_1 = \frac32 \;\;\; x^*_4 = \frac12 \cdot
+  \begin{bmatrix}3\\2\end{bmatrix}$$
+- The right point at the intersection of $2x_1 + x_2 \leq 4$ and $x_2 \geq
+  0$:
+  $$x_2 = 0 \;\;\; 2x_1 + 0 = 4 \Leftrightarrow x_1 = 2 \;\;\; x^*_5 =
+  \begin{bmatrix}2\\0\end{bmatrix}$$
+
+Therefore, $x^*_1$ to $x^*_5$ are all of the basic feasible points for this
+constrained minimization problem.
+
 We then compute the objective function value for each basic feasible point
 found, The smallest objective value will correspond with the constrained
 minimizer problem solution.
 
 $$
-x_1 = \begin{bmatrix}0\\0\end{bmatrix} \;\;\; f(x_1) = 4 \cdot 0 + 3 \cdot 0 =
+x^*_1 = \begin{bmatrix}0\\0\end{bmatrix} \;\;\; f(x^*_1) = 4 \cdot 0 + 3 \cdot 0 =
 0$$$$
-x_2 = \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix} \;\;\;
-f(x_2) = 4 \cdot 0 + 3 \cdot \frac32 = \frac92$$$$
-x_3 = \frac{1}{13} \cdot \begin{bmatrix}3\\24\end{bmatrix} \;\;\; f(x_3) = 4
+x^*_2 = \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix} \;\;\;
+f(x^*_2) = 4 \cdot 0 + 3 \cdot \frac{3}{2} = \frac92$$$$
+x^*_3 = \frac{1}{13} \cdot \begin{bmatrix}3\\24\end{bmatrix} \;\;\; f(x^*_3) = 4
 \cdot \frac{3}{13} + 3 \cdot \frac{24}{13} = \frac{84}{13}$$$$
-x_4 = \frac12 \cdot \begin{bmatrix}3\\2\end{bmatrix} \;\;\; f(x_4) = 4 \cdot
-frac32 + 3 \cdot 1 = 9$$$$
-x_5 = \begin{bmatrix}2\\0\end{bmatrix} \;\;\; 4 \cdot 2 + 1 \cdot 0 = 8$$
+x^*_4 = \frac12 \cdot \begin{bmatrix}3\\2\end{bmatrix} \;\;\; f(x^*_4) = 4 \cdot
+\frac32 + 3 \cdot 1 = 9$$$$
+x^*_5 = \begin{bmatrix}2\\0\end{bmatrix} \;\;\; f(x^*_5) = 4 \cdot 2 + 1 \cdot 0 = 8$$
+
+Therefore, $x^* = x^*_1 = \begin{bmatrix}0 & 0\end{bmatrix}^T$ is the global
+constrained minimizer.
+
+# Exercise 3
+
+## Exercise 3.1
+
+## Exercise 3.2
+
+The MATLAB code used to find the solution can be found under section 3.2 of the
+`main.m` script. The solution is:
+
+$$x=\begin{bmatrix}0.7692\\-2.2308\\2.2308\end{bmatrix} \;\;\; \lambda=
+\begin{bmatrix}-10.3846\\2.1538\end{bmatrix}$$
 
-Therefore, $x^* = x_1$ is the global constrained minimizer with $\lambda^* = \lambda_1 = NaN$ as
-the slack variable value.
diff --git a/Claudio_Maggioni_5/Claudio_Maggioni_5.pdf b/Claudio_Maggioni_5/Claudio_Maggioni_5.pdf
index 62a821a..23abb2b 100644
Binary files a/Claudio_Maggioni_5/Claudio_Maggioni_5.pdf and b/Claudio_Maggioni_5/Claudio_Maggioni_5.pdf differ
diff --git a/Claudio_Maggioni_5/main.asv b/Claudio_Maggioni_5/main.asv
deleted file mode 100644
index 55f0518..0000000
--- a/Claudio_Maggioni_5/main.asv
+++ /dev/null
@@ -1,79 +0,0 @@
-clc
-clear
-close all
-
-%% Exercise 2.1
-
-syms x1 x2
-c1 = 2 * x1 + 3 * x2 - 6;
-c2 = -3 * x1 + 2 * x2 - 3;
-c3 = 2 * x2 - 5;
-c4 = 2 * x1 + x2 - 4;
-
-c1 = solve(c1 == 0, x2, 'Real', true);
-c2 = solve(c2 == 0, x2, 'Real', true);
-c3 = solve(c3 == 0, x2, 'Real', true);
-c4 = solve(c4 == 0, x2, 'Real', true);
-
-i1 = solve(c1 == c2, x1, 'Real', true);
-i2 = solve(c1 == c4, x1, 'Real', true);
-i3 = solve(c4 == 0, x1, 'Real', true);
-
-px = double([0 0 i1 i2 i3]);
-pysym = [0 subs(c2, x1, 0) subs(c1, x1, i1) subs(c4, x1, i2) subs(c4, x1, i3)];
-py = double(pysym); 
-
-xl = -0.05;
-xh = 2.05;
-
-orange = [232/255 128/255 18/255];
-grey = [0.5 0.5 0.5];
-colors = [orange; 0 0 1; 1 0 0; 0 1 0];
-dirs = [30 30 30 30];
-
-axis([-0.05 2.05 -0.15 5.15])
-i = 1;
-hold on
-for c = [c1 c2 c3 c4]
-    pl = patch([xl, xl, xh, xh], ...
-      double([-0.15, subs(c, x1, xl), subs(c, x1, xh), -0.15]), ...
-      colors(i, :));
-    pl.EdgeColor = colors(i, :);
-    pl.FaceAlpha = .2;
-    pl.EdgeAlpha = .2;
-    i = i + 1;
-end
-
-pl = patch([0, 0, xh, xh], ...
-              double([0, 5.15, 5.15, 0]), ...
-              grey);
-pl.EdgeColor = grey;
-pl.FaceAlpha = .2;
-pl.EdgeAlpha = .2;
-
-pl = patch(px, py, 'green');
-pl.EdgeColor = 'green';
-
-legend('2x1 + 3x2 <= 6', '-3x1 + 2x2 <= 3', '2x2 <= 5', ...
-    '2x1 + x2 <= 4', 'x1 > 0 and x2 > 0', 'feasible region');
-hold off
-
-%% gsppn
-
-for i=1:5
-    obj = 4 * px(i) + 3 * py(i);
-    fprintf("x1=%g x2=%g y=%g\n", px(i), py(i), obj);
-end
-    
-
-%% Exercise 3.2
-
-G = [6 2 1; 2 5 2; 1 2 4];
-c = [-8; -3; -3];
-A = [1 0 1; 0 1 1];
-b = [3; 0];
-
-[x, lambda] = uzawa(G, c, A, b, [0;0;0], [0;0], 1e-8, 100);
-display(x);
-display(lambda);
-