# Linearize Minimax and Minimin in Objective function?

 1 Dear OR-Exchange Experts, I would like to know if it is possible to linearize the following objective function: $$\text{minimize } Z(\textbf{x}) = m_1 * x_1 + m_2 * x_2 + c_1 * \max((x_1 + x_2) - C_{\epsilon}, 0) - c_2 * \min((x_1 + x_2) + C_{\chi}, 0)$$ subject to $$L_1 \le x_1 \le U_1$$ $$L_2 \le x_2 \le U_2$$ $$L$$ and $$U$$ are the lower and the upper bound. $$x_1$$, $$x_2$$ can adopt any real value between the bounds (either negative or positive values) where the following parameters are constant and positive, i.e. $$m_1, m_2, c_1, c_2, C_{\epsilon}, C_{\chi} \ge 0$$. My first approach is to move the maximum and minimum into a constraint with two new continuous variables of a special ordered set of type 1: $$\{k_1, k_2\}$$ with exactly that order. and add the following two constraints: $$k_1 \ge c_1 * ((x_1 + x_2) - C_{\epsilon})$$ $$k_2 \ge -c_2 * ((x_1 + x_2) + C_{\chi})$$ such that the objective function becomes $$\text{minimize } Z(\textbf{x}) = m_1 * x_1 + m_2 * x_2 + k_1 + k_2$$ I am not sure if it is not too complex. Maybe someone of you know some hints or a better way. I am thankful for any help. Thank you. asked 11 Aug '16, 09:14 Thomas 23●6 accept rate: 0% Paul Rubin ♦♦ 14.6k●5●13

 1 Your method is fine, but you need only constraint $$k_1 \ge 0$$ and $$k_2 \ge 0$$. The SOS1 requirement is correct (given that both $$C_\epsilon$$ and $$C_\chi$$ are nonnegative), but it is unnecessary. answered 11 Aug '16, 18:05 Paul Rubin ♦♦ 14.6k●5●13 accept rate: 19% Thank you very much, Paul. (12 Aug '16, 03:43) Thomas
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