| Line 25: | Line 25: | ||
The KKT condition takes the form<br> | The KKT condition takes the form<br> | ||
<math> | <math> | ||
| − | |||
\nabla_xl(x,\mu)= | \nabla_xl(x,\mu)= | ||
{begin{bmatrix} | {begin{bmatrix} | ||
| Line 33: | Line 32: | ||
\begin{bmatrix} | \begin{bmatrix} | ||
0 \\ 0 | 0 \\ 0 | ||
| − | \end{bmatrix}} | + | \end{bmatrix}}</math><br> |
| − | \mu_1(x_1+x_2-2)=0 | + | <math>\mu_1(x_1+x_2-2)=0</math><br> |
| − | \mu_2(x_1+2x_2-3)=0 | + | <math>\mu_2(x_1+2x_2-3)=0</math><br> |
| − | \mu_1>=0, \mu_2>=0 | + | <math>\mu_1>=0</math>, <math>\mu_2>=0</math><br> |
| − | + | ||
| − | </math><br> | + | |
<math> \Rightarrow | <math> \Rightarrow | ||
\begin{cases} | \begin{cases} | ||
Revision as of 22:46, 18 February 2019
Automatic Control (AC)
Question 3: Optimization
August 2016 Problem 5
Solution
The problem equal to
Minimize $ (x_1)^2+(x_2)^2-14x_1-6x_2-7 $
Subject to $ x_1+x_2-2<=0 $ and $ x_1+2x_2-3<=0 $
Form the lagrangian function
$ l(x,\mu)=(x_1)^2+(x_2)^2-14x_1-6x_2-7+\mu_1(x_1+x_2-2)+\mu_2(x_1+2x_2-3) $
The KKT condition takes the form
$ \nabla_xl(x,\mu)= {begin{bmatrix} 2x_1-14+\mu_1+\mu_2 \\ 2x_2-6+\mu_1+2\mu_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}} $
$ \mu_1(x_1+x_2-2)=0 $
$ \mu_2(x_1+2x_2-3)=0 $
$ \mu_1>=0 $, $ \mu_2>=0 $
$ \Rightarrow \begin{cases} \mu_1=0 & \mu_2=0 & x_1=7 & x_2=3 & wrong \\ \mu_1=0 & \mu_2=4 & x_1=5 & x_2=-1 & wrong \\ \mu_1=8 & \mu_2=4 & x_1=3 & x_2=-1 & f(x)=-33 \\ \mu_1=20 & \mu_2=-8 & x_1=1 & x_2=1 & wrong \end{cases} $
In all $ x^T=[3 -1] $ is the maximizer of original function.
