I am an electrical engineer and currently I have to deal with an optimization problem with a very specific requirement:
$\begin{array}{*{20}{c}}
{\mathop {Min}\limits_x }&{f\left( x \right)}\\
{{c_1}:}&{{x_1} + {x_2} \ge 1}\\
{{c_2}:}&{{x_2} + {x_3} \ge 1}\\
{}&{{x_1},{x_2},{x_3} \in \left\{ {0,1} \right\}}
\end{array}$
Particularly, there are some control variables $t_4$ and $t_5$ where ${t_4},{t_5} \in \left\{ {0,1} \right\}$ that literally state the following:
If $t_4=1$ then the constraint $c_1$ is needed to be kept.
If $t_4=0$ then the constraint $c_1$ is not needed to be kept. In other words, $c_1$ becomes redundant and will not participate in the optimization process.
If $t_5=1$ then the constraint $c_2$ is needed to be kept.
If $t_5=0$ then the constraint $c_2$ is not needed to be kept. In other words, $c_2$ becomes redundant and will not participate in the optimization process.
1/ How to linearize these statements and incorporate the corresponding variable $t_4$ and $t_5$ into the optimization problem ?
I am also not sure if we can exploit the fact that all of these variables are binary ${x_1},{x_2},{x_3},{t_4},{t_5} \in \left\{ {0,1} \right\}$ ?
Edited:
$\begin{array}{*{20}{c}}
{\mathop {Min}\limits_x }&{f\left( x \right)}\\
{{c_1}:}&{{x_1} + {x_2} \ge 1}\\
{{c_2}:}&{{x_2} + {x_3} \ge 1}\\
{{c_3}:}&{{x_4} + {x_5} \ge {y_1}}\\
{{c_4}:}&{{x_5} + {x_6} \ge {y_2}}\\
{}&{{x_1},{x_2},{x_3} \in \left\{ {0,1} \right\}}\\
{}&{{y_1},{y_2} \in \left\{ {0,1} \right\}}
\end{array}$
2/ I almost forgot about the case of constraint $c_3$ and $c_4$. Under this case how can we repeat the technique in question 1 for the two binary variable $t_6$, $t_7$ ?
If $t_6=1$ then the constraint $c_3$ is needed to be kept. If $t_6=0$ then the constraint $c_3$ is not needed to be kept. In other words, $c_3$ becomes redundant and will not participate in the optimization process.
Similarly for $t_7$, it is the same for $c_4$.
Thank you for your enthusiasm !
Best Answer
Replace $c_1$ with $c_1':x_1+x_2\geq t_4$. If $t_4=1$, you recover $c_1$; if $t_4=0$, the constraint becomes $x_1+x_2\geq 0$, which is redundant, since $x_1,x_2\in\{0,1\}$. Similarly, replace $c_2$ with $c_2':x_2+x_3\geq t_5$.