Commit df9af168 authored by nanored's avatar nanored

ex 4 dm 2

parent cffb4be8
......@@ -8,6 +8,7 @@
\usepackage{kpfonts}
\usepackage{tikz}
\usepackage{bbm}
\usepackage{hyperref}
\title{
\noindent\rule{\linewidth}{0.4pt}
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\newcommand{\trans}{\mathsf{T}}
\newcommand{\syst}[2]{\left\{ \begin{array}{#1} #2 \end{array} \right.}
\newcommand{\diag}{\text{\textbf{diag}}~}
\newcommand{\pmat}[1]{\begin{pmatrix} #1 \end{pmatrix}}
\newcommand{\R}{\mathbb{R}}
\begin{document}
......@@ -153,7 +157,10 @@
Where $h$ is a strictly concave function. We compute the gradient of $h$ with respect to $y$ :
$$ \nabla_y h(\nu, y) = \nu - 2 y $$
Then $h$ is maximal for $y = \frac{1}{2} \nu$ and we obtain :
$$ \| \nu \|_2^{2 \, *} = \frac{1}{4} \| \nu \|_2^2 $$
\begin{equation}
\label{conj2}
\| \nu \|_2^{2 \, *} = \frac{1}{4} \| \nu \|_2^2
\end{equation}
Then our dual can be written :
\begin{center}
\fbox{$ \displaystyle \text{(Dual RLS)} \quad \begin{array}{llr}
......@@ -174,9 +181,9 @@
In this case the optimal value is :
$$ p_2^* = \dfrac{1}{n \tau} \sum_{i = 1}^n \mathcal{L}(\omega^*, x_i, y_i) + \dfrac{1}{2} \| \omega^* \|_2^2 = \dfrac{1}{\tau} p_1(\omega^*) $$
Where $p_1(\omega)$ is the value of the point $\omega$ in the problem (Sep. 1). \\
Finally we can remark that for every $\omega$, the point $\left( \omega, \left( \mathcal{L}(\omega, x_i, y_i) \right)_{i = 1}^n \right)$ is feasible for the problem (Sep. 2). We showed before that the optimal value can be expressed in this form then :
Finally we can remark that for every $\omega$, the point $\left( \omega, \left( \mathcal{L}(\omega, x_i, y_i) \right)_{i = 1}^n \right)$ is feasible for the problem (Sep. 2). We showed before that the optimal value can be expressed in this form then:
$$ p_2^* = \inf_\omega \dfrac{1}{\tau} p_1(\omega) = \dfrac{1}{\tau} p_1^* = \dfrac{1}{\tau} p_1(\omega^*) $$
We can now conclude :
We can now conclude:
\begin{center}
\fbox{\parbox{0.93\linewidth}{\centering
The problem (Sep. 2) as a solution $(\omega^*, z^*)$ of value $p_2^*$ such that $\omega^*$ is optimal for (Sep. 1), $z^*$ is the vector of losses and $p_1^* = \tau p_2^*$.
......@@ -184,6 +191,73 @@
\end{center}
\paragraph{2.}
We first compute the dual function :
$$ \begin{array}{lll}
g(\omega, z, \lambda, \pi) & = & \inf_{\omega, z} \dfrac{1}{n \tau} \mathbf{1}^\trans z + \dfrac{1}{2} \| \omega \|_2^2 + \mathbf{1}^\trans \lambda - \lambda^\trans \diag(y) X \omega - \lambda^\trans z - \pi^\trans z \\
& = & \mathbf{1}^\trans \lambda - \dfrac{1}{2} \left\| 2 X^\trans \diag(y) \lambda \right\|_2^{2 \, *} + \inf_z \left( \dfrac{1}{n \tau} \mathbf{1} - \lambda - \pi \right)^\trans z
\end{array} $$
Where $X \in \R^{n \times d}$ is the matrix with $x_i^\trans$ as $i$-th row and $ \| . \|_2^{2 \, *}$ is the conjugate of the 2-norm as seen in the previous exercise (eq \eqref{conj2}):
$$ X = \pmat{x_1^\trans \\ \vdots \\ x_n^\trans } \qquad \| v \|_2^{2 \, *} = \dfrac{1}{4} \| v \|_2^2 $$
We then obtain:
$$ g(\omega, z, \lambda, \pi) = \syst{ll}{
\mathbf{1}^\trans \lambda - \dfrac{1}{2} \left\| X^\trans \diag(y) \lambda \right\|_2^2 & \text{If } \; \lambda = \frac{1}{n \tau} \mathbf{1} - \pi \\[2mm]
- \infty & \text{Otherwise}
} $$
This gives the following dual problem:
$$ \begin{array}{llr}
\max_{\lambda, \pi} & \mathbf{1}^\trans \lambda - \dfrac{1}{2} \left\| X^\trans \diag(y) \lambda \right\|_2^2 \\[2mm]
\text{s.t.} & \lambda = \frac{1}{n \tau} \mathbf{1} - \pi \\[1mm]
& \lambda \succeq 0, \; \pi \succeq 0
\end{array} $$
We can inject the inequality on $\pi$ in the first equality and remove this variable from the dual:
\begin{center}
\fbox{$ \text{(Dual Sep. 2)} \quad \begin{array}{llr}
\max_{\lambda} & \mathbf{1}^\trans \lambda - \dfrac{1}{2} \left\| X^\trans \diag(y) \lambda \right\|_2^2 \\[2mm]
\text{s.t.} & 0 \preceq \lambda \preceq \frac{1}{n \tau} \mathbf{1}
\end{array} $}
\end{center}
\section*{Exercise 4 (Robust linear programming)}
\paragraph{}
We denote by $p^*(x)$ the value:
$$ p^*(x) = \sup_{a \in \mathcal{P}} a^\trans x = - \inf_{a \in \mathcal{P}} - a^\trans x $$
Then $p(x)$ is the opposite of the optimal value of a linear program. We know that strong duality holds in this case and we compute the dual function associated to (P) $\inf_{a \in \mathcal{P}} - a^\trans x$:
$$ g(a, z) = \inf_a -a^\trans x + z^\trans C^\trans a - z^\trans d = \syst{ll}{
-z^\trans d & \text{If } \, Cz = x \\
- \infty & \text{Otherwise}
} $$
This gives the following dual of (P):
$$ (D) \quad \begin{array}{llr}
\max_z & - d^\trans z \\
\text{s.t.} & Cz = x \\
& z \succeq 0
\end{array} $$
The optimal value of this dual is $d^*(x) = -p^*(x)$ by strong duality.
\paragraph{}
Because the objective function of the two LP we want to show the equivalence are the same, it is sufficient to show that :
\begin{equation}
\label{equiv}
p^*(x) \leqslant b \; \Leftrightarrow \; \exists z \succeq 0, \, C z = x \text{ and } d^\trans z \leqslant b
\end{equation}
\textit{Remark that I wrote $Cz =x$ instead of $C^\trans z = x$ as suggested in the exercise statement, which is certainly wrong} \\
Now we suppose that $x$ satisfies $p^*(x) \leqslant b$. We know that there exists $z^* \succeq 0$ such that:
$$ C z^* = x \qquad \text{And} \qquad - d^\trans z^* = d^*(x) = - p^*(x) \geqslant -b \; \Leftrightarrow \; d^\trans z^* \leqslant b $$
Inversely, now we suppose that $p^*(x) > b$. Let $z \succeq 0$ such that $Cz = x$. Thus, $z$ is in the polytope of (D). This gives :
$$ - d^\trans z \leqslant d^*(x) = -p^*(x) < -b \; \Leftrightarrow \; d^\trans z > b $$
Then there is no $z \succeq 0$ such that $C z = x \text{ and } d^\trans z \leqslant b$. \\
We just proved the equivalence \eqref{equiv} that allows us to conclude :
\begin{center}
\fbox{$ \displaystyle \begin{array}{llr}
\max_x & c^\trans x \\
\text{s.t.} & \sup_{a \in \mathcal{P}} a^\trans x \leqslant b
\end{array} \qquad \Leftrightarrow \qquad \begin{array}{llr}
\min_{x, z} & c^\trans x \\
\text{s.t.} & d^\trans z \leqslant b \\
& Cz = x \\
& z \succeq 0
\end{array} $}
\end{center}
\end{document}
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