First, we can remark that (Sep. 2) is feasible by simply taking the point $(0, \mathbf{1})$ and bounded because the objective function is non-negative for $z \succeq0$. \\
Let $(\omega, z)$ a feasible point. Suppose that there exists $i$ such that $z_i > \mathcal{L}(\omega, x_i, y_i)$. Then we let $z'$ such that $z'_j = z_j$ for $j \neq i$ and $z'_i =\mathcal{L}(\omega, x_i, y_i)$. We have in particular : \vspace{-2mm}
And the other constraints are also verified because $z$ is feasible. Because the factor of $z_i$ is strictly positive in the objective function and because $z'_i < z_i$, the value associated to the point $(\omega, z')$ is strictly inferior to the value of $(\omega, z)$. \\
Furthermore the other constraints are also verified because $z$ is feasible. Because the factor of $z_i$ is strictly positive in the objective function and because $z'_i < z_i$, the value associated to the point $(\omega, z')$ is strictly inferior to the value of $(\omega, z)$. \\
Then for $(\omega^*, z^*)$ optimal we have $z_i^*\leqslant\mathcal{L}(\omega^*, x_i, y_i)$ for every $i$. Furthermore, because of constraints, this inequality is in fact an equality. So : \vspace{-2mm}
$$ z_i^*=\mathcal{L}(\omega^*, x_i, y_i)$$
In this case the optimal value is :
...
...
@@ -222,7 +222,7 @@
\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) $\inf_{a \in\mathcal{P}}- a^\trans x$:
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) $\inf_{a \in\mathcal{P}}- a^\trans x$:
$$ g(a, z)=\inf_a -a^\trans x + z^\trans C^\trans a - z^\trans d =\syst{ll}{
-z^\trans d &\text{If }\, Cz = x \\
-\infty&\text{Otherwise}
...
...
@@ -250,7 +250,7 @@
We just proved the equivalence \eqref{equiv} that allows us to conclude :
\begin{center}
\fbox{$\displaystyle\begin{array}{llr}
\max_x & c^\trans x \\
\min_x & c^\trans x \\
\text{s.t.}&\sup_{a \in\mathcal{P}} a^\trans x \leqslant b
This gradient depends only on $x_i$ then the minimum of $L$ w.r.t $x$ is obtained by optimizing on each $x_i$ independently. We can remark that if $\nu_i > 0$, then $L(\alpha e_i, \lambda, \nu)$ tends to $-\infty$ when $\alpha$ tends to $+\infty$. So, for $L$ to be bounded, we need $\nu\preceq0$. In this case $L$ is convex in $x$ and we just need to find the vanishing point of the gradient to obtain the minimum of $L$:
$$\nabla_{x_i} L(x, \lambda, \nu)=0\quad\Leftrightarrow\quad\left(\nu_i < 0\,\text{ and }\, x_i =\dfrac{c_i + a_i^\trans\lambda+\nu_i}{2\nu_i}\right)\;\text{or}\;\left(\nu_i =0\,\text{ and }\, c_i + a_i^\trans\lambda=0\right)$$
Because $g$ depends on $\nu_i$ only in the term $g_i(\lambda, \nu_i)$ we can simplify the expression of the Lagrange dual by maximizing each $g_i$ with respect to $\nu_i$ and removing each $\nu_i$ from the dual variables. \\
We compute the Lagrange dual of the LP relaxation:
$$ h(x, \alpha, \beta, \gamma)=\inf_x \; c^\trans x +\alpha^\trans Ax -\alpha^\trans b -\beta^\trans x +\gamma^\trans x -\gamma^\trans\mathbf{1}=\syst{ll}{
-\alpha^\trans b -\gamma^\trans\mathbf{1}&\text{If }\; c + A^\trans\alpha+\gamma=\beta\\
-\infty&\text{Otherwise}
}$$
This gives the follwing dual problem:
$$\begin{array}{llr}
\max_{\alpha, \beta, \gamma}&-\alpha^\trans b -\gamma^\trans\mathbf{1}\\
\text{s.t.}& c + A^\trans\alpha+\gamma=\beta\\
&\alpha, \beta, \gamma\succeq0
\end{array}$$
We can inject the non-negativity of $\beta$ in the equality constraint and remove this variable:
$$\begin{array}{llr}
\max_{\alpha, \gamma}&-\alpha^\trans b -\gamma^\trans\mathbf{1}\\
\text{s.t.}& c + A^\trans\alpha+\gamma\succeq0\\
&\alpha, \gamma\succeq0
\end{array}$$
We also remark that the objective function is strictly decreasing with respect to $\gamma$. Then the optimal is obtain for $\gamma$ as small as possible when $\alpha$ is fixed. This leads us to: