The constraint looks something like this

$$b_1 x_1 + b_2 x_2 = 100$$

where $b_1, b_2 \in \{0,1\}$ and $x_1, x_2 \in \mathbb R$ have lower and upper bounds. Also, $b_1 + b_2 = 1$.


closed as unclear what you're asking by José Carlos Santos, Saad, Rhys, Namaste, Hurkyl Mar 28 '18 at 15:42

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  • $\begingroup$ Is that the only problematic constraint? $\endgroup$ – Rodrigo de Azevedo Mar 28 '18 at 11:00
  • $\begingroup$ No..there are other constraints but those are linear $\endgroup$ – Anoop Kumar Mangaraj Mar 28 '18 at 11:21
  • $\begingroup$ Are the $b_i$´s parameters or variables? $\endgroup$ – callculus Mar 28 '18 at 11:23
  • $\begingroup$ This can be linearized as: $$\begin{align} &(1-\delta)100 + \delta L_1 \le x_1 \le (1-\delta) 100 + \delta U_1\\ &\delta100 + (1-\delta) L_2 \le x_2 \le \delta 100 + (1-\delta) U_2\\& \delta \in \{0,1\}\\&x_i \in [L_i,U_i] \end{align}$$ $\endgroup$ – Erwin Kalvelagen Mar 28 '18 at 16:00

If I understand your question correctly, the problematic constraint is, essentially, the disjunction

$$x_1 = 100 \lor x_2 = 100$$

Let the other equality constraints be of the form $\rm A x = b$. Hence,

$$\begin{array}{rl} \mathrm A \mathrm x = \mathrm b \land \left( x_1 = 100 \lor x_2 = 100 \right) &\equiv \left( \mathrm A \mathrm x = \mathrm b \land x_1 = 100 \right) \lor \left( \mathrm A \mathrm x = \mathrm b \land x_2 = 100 \right)\\ &\equiv \left( \begin{bmatrix} \mathrm A\\ \mathrm e_1^\top\end{bmatrix} \mathrm x = \begin{bmatrix} \mathrm b\\ 100\end{bmatrix} \right) \lor \left( \begin{bmatrix} \mathrm A\\ \mathrm e_2^\top\end{bmatrix} \mathrm x = \begin{bmatrix} \mathrm b\\ 100\end{bmatrix} \right)\end{array}$$

which defines the union of two hyperplanes. I assume there are also non-negativity constraints.

You can optimize a linear objective over each (convex) feasible region and then take the minimum or maximum of each optimal value. In other words, solve two linear programs.


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