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Claudio Maggioni 2021-07-12 22:25:52 +02:00
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2 changed files with 3 additions and 28 deletions

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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 $s$
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 2.1
@ -213,27 +207,8 @@ constrained minimizer.
# Exercise 3
## Exercise 3.1
<!--
I consider the given problem, which is exactly the same as one of the problems
of the previous assignment (Homework 4):
$$\min_{x} f(x) = 3x^2_1 + 2x_1x_2 + x_1x_3 +
2.5x^2_2 + 2x_2x_3 + 2x^2_3 - 8x_1 - 3x_2 - 3x_3
$$$$\text{ subject to } x_1 + x_3 = 3 \;\;\; x_2 + x_3 = 0$$
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
Yes, the problem can be solved with Uzawa's method since 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:
@ -256,7 +231,7 @@ g\\
h
\end{bmatrix}$$
This is the system the _Uzzawa_ method will solve. Therefore, we need to check
This is the system the Uzawa method will solve. Therefore, we need to check
if the matrix:
$$K = \begin{bmatrix}G & A^T \\ A& 0\end{bmatrix} = \begin{bmatrix}
@ -279,7 +254,7 @@ $$\begin{bmatrix}
8.7663\end{bmatrix}$$
Therefore, the system is indeed a saddle point system and it can be solved with
_Uzzawa_'s method.
Uzawa's method.
## Exercise 3.2