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>
+
<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>
\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.


Back to QE AC question 3, August 2016

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva