hw4: done 1,2.1

Claudio_Maggioni_4/Claudio_Maggioni_4.md

@@ -88,12 +88,26 @@ X = X_4 \approx \begin{bmatrix}-0.966\\-0.199\\-0.166\end{bmatrix}$$
 $$f(X_1) = -1.1304 \;\; f(X_2) = -1.59219 \;\; f(X_3) = 4.64728 \;\; f(X_4) =
 -5.36549$$
 
-We therefore choose $(\lambda_4, X_4)$ since $f(X_4)$ is the smallest objective
+## Exercise 1.3
+
+To find the optimal solution, we choose $(\lambda_4, X_4)$ since $f(X_4)$ is the
+smallest objective
 value out of all the feasible points. Therefore, the solution to the
 minimization problem is:
 
 $$X \approx \begin{bmatrix}-0.966\\-0.199\\-0.166\end{bmatrix}$$
 
+# Exercise 2
+
+## Exercise 2.1
+
+To reformulate the problem, we first rewrite the explicit values of $G$, $c$,
+$A$ and $b$:
+
+$$G = \begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}$$
+  $$c = \begin{bmatrix}-8\\-3\\-3\end{bmatrix}$$
+$$A = \begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}$$
+$$b = \begin{bmatrix}3\\0\end{bmatrix}$$