(Created page with "Category:ECE Category:QE Category:problem solving <center> <font size= 4> ECE Ph.D. Qualifying Exam </font size> <font size= 4> Auto...") |
|||
| (10 intermediate revisions by the same user not shown) | |||
| Line 20: | Line 20: | ||
The problem equal to<br> | The problem equal to<br> | ||
Minimize <math>(x_1)^2+(x_2)^2-14x_1-6x_2-7</math><br> | Minimize <math>(x_1)^2+(x_2)^2-14x_1-6x_2-7</math><br> | ||
| − | Subject to <math> | + | Subject to <math>x_1+x_2-2<=0</math> and <math>x_1+2x_2-3<=0</math><br> |
Form the lagrangian function<br> | Form the lagrangian function<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> | <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>\mu_1(x_1+x_2-2)=0</math><br> |
| − | \mu_1(x_1+x_2-2)=0 | + | <math>\mu_2(x_1+2x_2-3)=0</math><br> |
| − | \mu_2(x_1+2x_2-3)=0 | + | <math>\mu_1>=0</math>, <math>\mu_2>=0</math><br> |
| − | \mu_1>=0, \mu_2>=0 | + | |
| − | + | ||
| − | </math><br> | + | |
<math> \Rightarrow | <math> \Rightarrow | ||
\begin{cases} | \begin{cases} | ||
| − | + | \mu_1=0 & \mu_2=0 & x_1=7 & x_2=3 & wrong \\ | |
| − | \mu_1=0 & \mu_2=0 & x_1=7 & x_2=3 & | + | \mu_1=0 & \mu_2=4 & x_1=5 & x_2=-1 & wrong \\ |
| − | \mu_1=0 & \mu_2=4 & x_1=5 & x_2=-1 & | + | |
\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> | ||
| + | ---- | ||
| + | ---- | ||
| + | ===Similar Problem=== | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob5]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_13/AC-3.pdf?dl=1 2013 QE AC3 Prob1]<br> | ||
---- | ---- | ||
[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]] | [[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Latest revision as of 11:48, 25 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.
Similar Problem
2015 QE AC3 Prob5
2013 QE AC3 Prob1
