Optimization with absolute values: Difference between revisions

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==Numerical Example==
==Numerical Example==


<math>\min{|x_1| + 2|x_2| + |x_3|} </math><BR>
<math>\min{|x_1| + 2|x_2| + |x_3|} </math>
<math>\min |x_1| + 2|x_2| + |x_3| </math>
 
<math>  \begin{align}
<math>  \begin{align}
\ s.t. x_1 + x_2 - x_3 \le 10 \\
\ s.t. x_1 + x_2 - x_3 \le 10 \\
Line 34: Line 34:


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

Revision as of 16:48, 20 November 2020

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|} $

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

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 \le x_1 \le U_1 $

$ -U_2 \le x_2 \le U_2 $

$ -U_3 \le x_3 \le U_3 $

The problem has now been reformulated as a linear programming problem that can be solved normally:

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

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

$ -U_1 \le x_1 \le U_1 $

$ -U_2 \le x_2 \le U_2 $

$ -U_3 \le x_3 \le 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 $.