hw4: done 1 and 2
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@ -104,11 +104,47 @@ $$X \approx \begin{bmatrix}-0.966\\-0.199\\-0.166\end{bmatrix}$$
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To reformulate the problem, we first rewrite the explicit values of $G$, $c$,
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$A$ and $b$:
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$$G = \begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}$$
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$$c = \begin{bmatrix}-8\\-3\\-3\end{bmatrix}$$
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$$G = 2 \cdot \begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}$$
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$$c = \begin{bmatrix}-8\\-3\\-3\end{bmatrix}$$
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$$A = \begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}$$
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$$b = \begin{bmatrix}3\\0\end{bmatrix}$$
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Then, using these variable values and the formulation given on the assignment
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sheet the problem is restated in this new form.
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## Exercise 2.2
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The lagrangian for this problem is the following:
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$$L(x, \lambda) = \frac12\langle x, Gx\rangle + \langle x, c \rangle - \lambda
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(Ax - b) =$$$$= \begin{bmatrix}x_1&x_2&x_3\end{bmatrix}
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\begin{bmatrix}3&0&0\\2&2.5&0\\1&2&2\end{bmatrix}
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\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} +
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\begin{bmatrix}x_1&x_2&x_3\end{bmatrix}
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\begin{bmatrix}-8\\-3\\-3\end{bmatrix} - \lambda
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\left(\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}
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\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} -
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\begin{bmatrix}3\\0\end{bmatrix}\right)$$
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The KKT conditions are the following:
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First we have the condition on the partial derivatives of the Lagrangian w.r.t.
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$X$:
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$$\nabla_x L(x, \lambda) = Gx + c - A^T \lambda = \begin{bmatrix}3 x_1 - 8 +
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\lambda_1\\ 2x_1 + 2.5 x_2 - 3 + \lambda_2\\x_1 + 2x_2 + 2x_3 - 3 + \lambda_1
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+ \lambda_2\end{bmatrix} > 0$$
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Then we have the conditions on the equality constraint:
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$$Ax - b = 0 \Leftrightarrow \begin{bmatrix}x_1 + x_3\\x_2 + x_3\end{bmatrix} =
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\begin{bmatrix}3\\0\end{bmatrix}$$
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Then we have the conditions on the equality constraint:
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$$\lambda^T (Ax - b) = 0 \Leftarrow Ax - b = 0 \text{ which is true if the above
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condition is true.}$$
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Since we have no inequality constraints, we don't need to apply the KKT
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conditions realated to inequality constraints.
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