Guys, I am trying to find ways of linearizing the operator Special Ordered Set Type 1 (SOS1). In order to understand what is my goal, I will, firstly, describe the problem I am facing. Let's consider the below mathematical program.
$\min_{x \in \mathbb{R}^n}$ $f(x)$
s.t. $h(x) = 0$
$l \leqslant x \leqslant u$
Where $f(x)$, and $h(x)$ are smooth and $\mathcal{C}^1$ vector-based functions. The lagrangean function of the above program, would be:
$\mathcal{L}(x, \pi, \lambda, \beta) = f(x) + \pi g(x) + \lambda (l - x) + \beta (x - u)$.
Resulting in the below multi-objective program:
$\max_{\pi, \lambda \geqslant 0, \beta \geqslant 0}$ $\min_{x \mathbb{R}^n} \mathcal{L}(x, \pi, \lambda, \beta)$
Which the optimal solution would satisfy the following optimality conditions:
$\nabla_{x} \mathcal{L} = \nabla_x f + \pi \nabla_x h - \lambda + \beta = 0$
$\nabla_{\pi} \mathcal{L} = h(x) = 0$
$\nabla_{\lambda} \mathcal{L} = l - x \leqslant 0$
$\nabla_{\beta} \mathcal{L} = x - u \leqslant 0$
$x^T \nabla_{x} \mathcal{L} = 0$
$\pi^T \nabla_{\pi} \mathcal{L} = 0$
$\lambda^T \nabla_{\lambda} \mathcal{L} = 0$
$\beta^T \nabla_{\beta} \mathcal{L} = 0$
$\lambda, \beta \geqslant 0$
If we desire to linearize the complementarity (or orthogonality) constraints of $\lambda$ and $\beta$, we would have to impose some constraints conotating that:
$\lambda = 0 \vee l - x = 0$
$\beta = 0 \vee x - u = 0$
For this porpuse, I know that there are SOS1 approaches based techniques, cf. here, and here, but AFAIK, SOS1 is based on branching rules, cf. here, and here. My question is, is there any way of linearizing SOS1, or simply orthogonality constraints for bounds, without relying on imposing branching rules?
Thanks for the attention, case there is any error or missing point in my statement, please, let me know.
Given that $\lambda \ge 0,$ $$\lambda = 0 \vee l - x = 0$$ can be linearized as $$\lambda \le M z$$ $$L (1-z) \le l - x \le U (1-z)$$ where $z$ is a new binary variable and $M, L, U$ are suitably large positive constants.