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<math> \Rightarrow | <math> \Rightarrow | ||
\begin{cases} | \begin{cases} | ||
| − | + | ||
| − | \mu_1=0 & \mu_2=0 & x_1=7 & x_2=3 & | + | \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 & | + | \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=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 & | + | \mu_1=20 & \mu_2=-8 & x_1=1 & x_2=1 & wrong |
| − | + | ||
\end{cases}</math><br> | \end{cases}</math><br> | ||
In all <math>x^T=[3 -1]</math> is the maximizer of original function.<br> | In all <math>x^T=[3 -1]</math> is the maximizer of original function.<br> | ||
Revision as of 22:38, 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 \\ & 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
$ \begin{cases} \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 \end{cases} $
$ \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.
