in value at risk problem , I encountered this scenario based problem :
$\min \{c^Tx+min\{t: pr(\sum_s b^s y^s <t )\ge 1-\alpha\}$
subject to
$Ax \le d $
$B^s x+D^s y^s \le h^s $
$x,y^s \ge0$
where " pr "is probability. s Is scenario.
in value at risk we usually have $ min\{t: pr(\sum_s b^s y^s <t )\ge 1-\alpha$
I can't linearized this problem.
How can I solve? How to be linearized?
my try is:
suppose $min\{t: pr(\sum_s b^s y^s <t )\ge 1-\alpha\} = T$
then we have
$\min \{c^Tx+T\}$
subject to
$ T \ge pr(\sum_s b^s y^s <t )\ge 1-\alpha$
$Ax \le d $
$B^s x+D^s y^s \le h^s $
$x,y^s \ge0$
then we muust linearized it, so
$\min \{c^Tx+T$
subject to
$ pr(\sum_s b^s y^s <t )\ge 1-\alpha$
$ pr(\sum_s b^s y^s <t )\le T$
$Ax \le d $
$B^s x+D^s y^s \le h^s $
$x,y^s \ge0$
then according to chance constrain linearization
$\min \{c^Tx+T\}$
subject to
$\sum_s b^s y^s <t +M (1- \delta_1 ^s)$
$\sum_s p^s \delta_1 ^s \ge 1- \alpha $
$ \sum_s b^s y^s <t +M (1- \delta_2 ^s)$
$\sum_s p^s \ \delta_2 ^s\le T $
$Ax \le d $
$B^s x+D^s y^s \le h^s $
$x,y^s \ge0 $ $ \delta_1 ^s, \delta_2 ^s \in \{0,1\}$
I'm going to answer here based on my best interpretation of what you're asking.
The problem you wish to solve is
\begin{equation} \begin{array}{rl} \min\ & c^\text{T}x + \text{VaR}_{1-\alpha}(b(\xi)^\text{T}y) \\ \text{s.t.}\ & B(\xi)x+D(\xi)y\leqslant h(\xi) \\ & x,y\geqslant 0 \end{array} \end{equation}
where $\xi$ is a random variable with finite support (i.e. has a finite number $N$ of realizations, which we index $s=1,\dots,N$). An equivalent formulation on this problem is
\begin{equation} \begin{array}{rl} \min\ & c^\text{T}x + t \\ \text{s.t.}\ & t\geqslant\text{VaR}_{1-\alpha}(b(\xi)^\text{T}y) \\ & B(\xi)x+D(\xi)y\leqslant h(\xi) \\ & x,y\geqslant 0 \end{array} \end{equation}
Now, the definition of VaR is
$$ \text{VaR}_{1-\alpha}(b(\xi)^\text{T}y)=F_Y^{-1}(\alpha), $$
where $F_Y$ is the CDF of the random variable $b(\xi)^\text{T}y$. Since CDFs are non-decreasing by definition, the following are equivalent:
$$ t\geqslant\text{VaR}_{1-\alpha}(b(\xi)^\text{T}y)\iff F_Y(t)\geqslant\alpha\iff\text{Pr}[b(\xi)^\text{T}y\leqslant{t}]\geqslant\alpha. $$
Basically, we have reduced our problem to the following program:
\begin{equation} \begin{array}{rl} \min\ & c^\text{T}x + t \\ \text{s.t.}\ & \text{Pr}[b(\xi)^\text{T}y\leqslant{t}]\geqslant\alpha \\ & B(\xi)x+D(\xi)y\leqslant h(\xi) \\ & x,y\geqslant 0 \end{array} \end{equation}
The problem is that this formulation doesn't make much sense. If we want to solve a problem with recourse then the constraint
$$ \text{Pr}[b(\xi)^\text{T}y\leqslant{t}]\geqslant\alpha $$
doesn't make much sense, since we are choosing $y$ after the random variable $\xi$ is realized. On the other hand, if we are doing a true chance constraint, then we must choose $y$ before $\xi$ is realized, in which case, it's not clear what the constraint
$$ B(\xi)x+D(\xi)y\leqslant h(\xi) $$
means. Do we want to enforce this for all realizations of $\xi$? Just some? Do we want to enforce it as a chance constraint?
(It's sort of uncommon for scenario-based recourse models and chance constraints to be used in the same model).