New answers tagged constraint
4
With CPLEX you can use logical constraints.
In OPL for instance you can write
float epsilon=0.01;
dvar float X;
dvar float Y;
subject to
{
(X>=epsilon) => (X>=Y);
}
And if you wonder , logical constraints are available in OPL but also in all APIs.
float epsilon=0.01;
dvar float X; dvar float Y;
subject to { (X>=epsilon) => (X>=Y); }
...
9
Introduce binary variable $Z$ and linear constraints
\begin{align}
X - \epsilon &\le (\bar{X} - \epsilon) Z \tag1 \\
Y - X &\le (\bar{Y} - 0) (1-Z) \tag2 \\
\end{align}
Constraint $(1)$ enforces $X > \epsilon \implies Z = 1$.
Constraint $(2)$ enforces $Z = 1 \implies X \ge Y$.
4
You could try and get rid of variables $Y_{p,t,j,c}$:
remove constraints $(6)-(7)$; constraints $(6)$ simply link $x$ and $Y$ variables, so removing them should not have any impact on the feasibility of the solution (assuming you remove $Y$ variables) ; constraints $(7)$ are not mandatory if my understanding is correct
replace constraints $(5)$ by
$$
\sum_{...
10
Let $x_{p,\ell}$ be the continuous variables in your table. Introduce integer variables $y_{p,\ell}$ and binary variables $z_{p,\ell}$, and impose linear constraints
\begin{align}
-z_{p,\ell} \le x_{p,\ell} - y_{p,\ell} &\le z_{p,\ell} &&\text{for all $p$ and $\ell$} \tag1 \\
\sum_\ell z_{p,\ell} &\le 1 &&\text{for all $p$} \tag2
\...
6
A variable that can assume values of zero or between some lower and upper bound is called a semi-continuous variable. Most high-end solvers have direct support for this type of variable. If not supported, you can model this with an additional binary variable:
$$\begin{align} & \color{darkblue}L\cdot \color{darkred}\delta \le \color{darkred}x \le \color{...
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