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===Solution=== | ===Solution=== | ||
| − | Let <math>t_1=x_1-2</math>, <math>t_2=x_2+1</math><br> | + | Let <math>t_1=x_1-2 </math>, <math>t_2=x_2+1</math><br> |
so that <math>g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0</math> would have some convex property<br> | so that <math>g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0</math> would have some convex property<br> | ||
with <math>f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1</math><br> | with <math>f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1</math><br> | ||
| − | <math>D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix}=\dfrac{1}{27}\begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix}</math> | + | <math>D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix}=\dfrac{1}{27}\begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix}</math><br> |
| − | ---- | + | It is easy to see that <math>\begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix}</math> is n.d .<br> |
| + | Such that function at <math>[2 -1]^T</math> is strictly locally concave.<br> | ||
| + | |||
| + | ---- | ||
| + | ===Similar Problem=== | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob1]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob3]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2014/AC-3.pdf?dl=1 2014 QE AC3 Prob2]<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 10:42, 25 February 2019
Automatic Control (AC)
Question 3: Optimization
August 2016 Problem 3
Solution
Let $ t_1=x_1-2 $, $ t_2=x_2+1 $
so that $ g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0 $ would have some convex property
with $ f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1 $
$ D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix}=\dfrac{1}{27}\begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix} $
It is easy to see that $ \begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix} $ is n.d .
Such that function at $ [2 -1]^T $ is strictly locally concave.
Similar Problem
2015 QE AC3 Prob1
2015 QE AC3 Prob3
2014 QE AC3 Prob2
