Line 24: Line 24:
 
<math>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)</math><br>
 
<math>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)</math><br>
 
The KKT condition takes the form<br>
 
The KKT condition takes the form<br>
<math>\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}</math><br>
+
<math>\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}</math><br>
 
<math>\mu_1(x_1+x_2-2)=0</math><br>
 
<math>\mu_1(x_1+x_2-2)=0</math><br>
 
<math>\mu_2(x_1+2x_2-3)=0</math><br>
 
<math>\mu_2(x_1+2x_2-3)=0</math><br>

Revision as of 21:47, 18 February 2019


ECE Ph.D. Qualifying Exam

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.


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