For me $a,c\in[0,1]$ and $\epsilon>0$ is small (say $0.01$).
Is it possible to set this condition
$$c\leq\max\bigg(\frac{(a-\epsilon)}{(1-\epsilon)}, 0\bigg)$$
in real linear program?
This does not work $$\frac{(a-\epsilon)}{(1-\epsilon)}\leq d$$ $$0\leq d$$ $$c\leq d.$$
Essentially I want $d=0$ if $\frac{(a-\epsilon)}{(1-\epsilon)}<0$ else set $d=\frac{(a-\epsilon)}{(1-\epsilon)}$ and set $c\leq d$ which is equivalent to $c\leq d$ where $d=0$ if $a<\epsilon$ else $d=\frac{(a-\epsilon)}{(1-\epsilon)}$. How do I do that?
The nonlinear constraint $$x\leq\max\bigg(\frac{y-\epsilon}{1-\epsilon}, 0\bigg)$$ where $x,y$ are variables and $\epsilon$ is a parameter, cannot be represented with purely linear constraints since it is not convex. You need an extra binary variable $z$ that indicates which of the two terms in the max is active, and a prespecified upper bound $M$ on the absolute value of the first term of the max expression. Then you can state this problem as: $$\begin{align}x &\leq \frac{y-\epsilon}{1-\epsilon} + zM \\ x &\leq 0 + (1-z)M \\ z &\in \{0,1\}. \end{align}$$