# Difference between revisions of "Optimization with absolute values"

Authors: Matthew Chan (mdc297), Yilian Yin (), Brian Amado (ba392), Peter (pmw99), Dewei Xiao (dx58) - SYSEN 5800 Fall 2020

Steward: Fengqi You

## Numerical Example

$\min {|x_{1}|+2|x_{2}|+|x_{3}|}$ $\min |x_{1}|+2|x_{2}|+|x_{3}|$ {\begin{aligned}\ s.t.x_{1}+x_{2}-x_{3}\leq 10\\x_{1}-3x_{2}+2x_{3}=12\end{aligned}} We replace the absolute value quantities with a single variable:

$|x_{1}|=U_{1}$ $|x_{2}|=U_{2}$ $|x_{3}|=U_{3}$ We must introduce additional constraints to ensure we do not lose any information by doing this substitution:

$-U_{1}\leq x_{1}\leq U_{1}$ $-U_{2}\leq x_{2}\leq U_{2}$ $-U_{3}\leq x_{3}\leq U_{3}$ The problem has now been reformulated as a linear programming problem that can be solved normally:

$\displaystyle \min{ U_1 + 2U_2 + U_3}$

\displaystyle \begin{align} \s.t. x_1 + x_2 - x_3 \le 10 \\ x_1 - 3x_2 + 2x_3= 12 \end{align}

$-U_{1}\leq x_{1}\leq U_{1}$ $-U_{2}\leq x_{2}\leq U_{2}$ $-U_{3}\leq x_{3}\leq U_{3}$ The optimum value for the objective function is $6$ , which occurs when $x_{1}=0$ and $x_{2}=0$ and $x_{3}=6$ .