Optimization with absolute values: Difference between revisions
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Steward: Fengqi You | Steward: Fengqi You | ||
==Numerical Example== | |||
<math>\ Min |x_1| + 2|x_2| + |x_3| </math><br> | |||
<math>\ s.t. x_1 + x_2 - x_3 \le 10</math> | |||
<math> x_1 - 3x_2 + 2x_3= 12</math> | |||
We replace the absolute value quantities with a single variable: | |||
<math>|x_1| = U_1 </math> | |||
<math>|x_2| = U_2</math> | |||
<math>|x_3| = U_3</math> | |||
We must introduce additional constraints to ensure we do not lose any information by doing this substitution: | |||
<math> -U_1 \le x_1 \le U_1 </math> | |||
<math> -U_2 \le x_2 \le U_2 </math> | |||
<math> -U_3 \le x_3 \le U_3 </math> | |||
The problem has now been reformulated as a linear programming problem that can be solved normally: | |||
<math>\ Min U_1 + 2U_2 + U_3 </math><br> | |||
<math>\ s.t. x_1 + x_2 - x_3 \le 10</math> | |||
<math> x_1 - 3x_2 + 2x_3 = 12</math> | |||
<math> -U_1 \le x_1 \le U_1 </math> | |||
<math> -U_2 \le x_2 \le U_2 </math> | |||
<math> -U_3 \le x_3 \le U_3 </math> | |||
The optimum value for the objective function is <math>6</math>, which occurs when <math>x_1 = 0 </math> and <math>x_2 = 0 </math> and <math>x_3 = 6 </math>. | |||
Revision as of 16:29, 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| $
$ \ s.t. x_1 + x_2 - x_3 \le 10 $
$ x_1 - 3x_2 + 2x_3= 12 $
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 $
$ \ s.t. x_1 + x_2 - x_3 \le 10 $
$ x_1 - 3x_2 + 2x_3 = 12 $
$ -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 $.