hw4: done 1 and 2

Claudio_Maggioni_4/Claudio_Maggioni_4.md

@@ -104,11 +104,47 @@ $$X \approx \begin{bmatrix}-0.966\\-0.199\\-0.166\end{bmatrix}$$
 To reformulate the problem, we first rewrite the explicit values of $G$, $c$,
 $A$ and $b$:
 
-$$G = \begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}$$
+$$G = 2 \cdot \begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}$$
 $$c = \begin{bmatrix}-8\\-3\\-3\end{bmatrix}$$
 $$A = \begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}$$
 $$b = \begin{bmatrix}3\\0\end{bmatrix}$$
 
+Then, using these variable values and the formulation given on the assignment
+sheet the problem is restated in this new form.
+
+## Exercise 2.2
+
+The lagrangian for this problem is the following:
+
+$$L(x, \lambda) = \frac12\langle x, Gx\rangle + \langle x, c \rangle - \lambda
+(Ax - b) =$$$$= \begin{bmatrix}x_1&x_2&x_3\end{bmatrix}
+  \begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}
+    \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} +
+      \begin{bmatrix}x_1&x_2&x_3\end{bmatrix}
+        \begin{bmatrix}-8\\-3\\-3\end{bmatrix} - \lambda
+          \left(\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}
+            \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} -
+              \begin{bmatrix}3\\0\end{bmatrix}\right)$$
 
 
+The KKT conditions are the following:
 
+First we have the condition on the partial derivatives of the Lagrangian w.r.t.
+$X$:
+
+$$\nabla_x L(x, \lambda) = Gx + c - A^T \lambda = \begin{bmatrix}3 x_1 - 8 +
+  \lambda_1\\ 2x_1 + 2.5 x_2 - 3 + \lambda_2\\x_1 + 2x_2 + 2x_3 - 3 + \lambda_1
++ \lambda_2\end{bmatrix} > 0$$
+
+Then we have the conditions on the equality constraint:
+
+$$Ax - b = 0 \Leftrightarrow \begin{bmatrix}x_1 + x_3\\x_2 + x_3\end{bmatrix} =
+  \begin{bmatrix}3\\0\end{bmatrix}$$
+
+Then we have the conditions on the equality constraint:
+
+$$\lambda^T (Ax - b) = 0 \Leftarrow Ax - b = 0 \text{ which is true if the above
+condition is true.}$$
+
+Since we have no inequality constraints, we don't need to apply the KKT
+conditions realated to inequality constraints.