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Claudio Maggioni 2021-07-12 22:25:52 +02:00
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@ -121,12 +121,6 @@ central path C converges to anything then $\tau$ approaches zero, therefore
leading the path to the constrained minimizer while keeping $x$ and $s$ leading the path to the constrained minimizer while keeping $x$ and $s$
positive but at the same time minimizing their pairwise products to zero. positive but at the same time minimizing their pairwise products to zero.
In practice, most central path implementation use Newton steps toward points on
$\mathcal{C}$ for which $\tau > 0$, rather than pure Newton steps for $F$. This
is usually possible since those newton steps are biased to stay in the
hyperspace region where $x > 0$ and $s > 0$ even without enforcing these
constraints.
# Exercise 2 # Exercise 2
## Exercise 2.1 ## Exercise 2.1
@ -213,27 +207,8 @@ constrained minimizer.
# Exercise 3 # Exercise 3
## Exercise 3.1 ## Exercise 3.1
<!--
I consider the given problem, which is exactly the same as one of the problems
of the previous assignment (Homework 4):
$$\min_{x} f(x) = 3x^2_1 + 2x_1x_2 + x_1x_3 + Yes, the problem can be solved with Uzawa's method since the problem can be
2.5x^2_2 + 2x_2x_3 + 2x^2_3 - 8x_1 - 3x_2 - 3x_3
$$$$\text{ subject to } x_1 + x_3 = 3 \;\;\; x_2 + x_3 = 0$$
defining $x$ as $(x_1,\,x_2,\,x_3)^T$, that can be written in the form of a
quadratic minimization problem:
$$\min_{x} f(x) = \dfrac{1}{2} \langle x,\, Gx\rangle + \langle x,\, c\rangle \\
\text{ subject to } Ax = b$$
Where $G\in \mathbb{R}^{n\times n}$ is a symmetric positive definite matrix,
$x$, $c \in \mathbb{R}^n$. The equality constraints are defined in terms of the
matrix $A\in \mathbb{R}^{m\times n}$, with $m \leq n$ and vector $b \in
\mathbb{R}^m$. Here, matrix $A$ has full rank.
-->
Yes, the problem can be solved with _Uzzawa_'s method since the problem can be
reformulated as a saddle point system. The KKT conditions of the problem can be reformulated as a saddle point system. The KKT conditions of the problem can be
reformulated as a matrix-vector to vector equality in the following way: reformulated as a matrix-vector to vector equality in the following way:
@ -256,7 +231,7 @@ g\\
h h
\end{bmatrix}$$ \end{bmatrix}$$
This is the system the _Uzzawa_ method will solve. Therefore, we need to check This is the system the Uzawa method will solve. Therefore, we need to check
if the matrix: if the matrix:
$$K = \begin{bmatrix}G & A^T \\ A& 0\end{bmatrix} = \begin{bmatrix} $$K = \begin{bmatrix}G & A^T \\ A& 0\end{bmatrix} = \begin{bmatrix}
@ -279,7 +254,7 @@ $$\begin{bmatrix}
8.7663\end{bmatrix}$$ 8.7663\end{bmatrix}$$
Therefore, the system is indeed a saddle point system and it can be solved with Therefore, the system is indeed a saddle point system and it can be solved with
_Uzzawa_'s method. Uzawa's method.
## Exercise 3.2 ## Exercise 3.2