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<math>\text{ Standard form: optimize} \left(x_{1}-2\right)^{2}+\left(x_{2}-1\right)^{2}</math><br> | <math>\text{ Standard form: optimize} \left(x_{1}-2\right)^{2}+\left(x_{2}-1\right)^{2}</math><br> | ||
− | <math>\text{subject to } g_{1}\left( x \right) x_{1}^{2}-x_{2}\leq0</math> | + | <math>\text{subject to } g_{1}\left( x \right) x_{1}^{2}-x_{2}\leq0</math> |
− | <math>g_{2}\left( x \right) x_{1}+x_{2}-2\leq0</math> | + | <math>g_{2}\left( x \right) x_{1}+x_{2}-2\leq0</math> |
− | <math>g_{3}\left( x \right) -x_{1}\leq0</math> | + | <math>g_{3}\left( x \right) -x_{1}\leq0</math> |
+ | |||
+ | <math>\text{KKT condition: (1) } Dl\left( \mu ,\lambda \right)=Df\left(x \right)+\mu_{1}Dg_{1}\left( x \right)+\mu_{2}Dg_{2}\left( x \right)+\mu_{3}Dg_{3}\left( x \right)</math><br> | ||
+ | |||
+ | <math>=\left [ 2x_{1}-4+2\mu_{1}x_{1}+\mu_{2}-\mu_{3}, 2x_{2}-2-\mu_{1}+\mu_{2} \right ]</math> | ||
+ | |||
+ | <math>\left ( 2 \right ) \mu^{T}g\left ( x \right )=0 \Rightarrow \mu_{1}\left ( x_{2}^2-x_{2} \right )+\mu_{2}\left ( x_{1}+x_{2}-2 \right ) - \mu_{3}x_{1}=0</math> | ||
Revision as of 19:54, 27 June 2012
ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 3, August 2011
$ \color{blue}\text{5. } \left( \text{20 pts} \right) \text{ Consider the optimization problem, } $
$ \text{optimize} \left(x_{1}-2\right)^{2}+\left(x_{2}-1\right)^{2} $
$ \text{subject to } x_{2}- x_{1}^{2}\geq0 $
$ 2-x_{1}-x_{2}\geq0, x_{1}\geq0. $
$ \color{blue} \text{The point } x^{*}=\begin{bmatrix} 0 & 0 \end{bmatrix}^{T} \text{ satisfies the KKT conditions.} $
$ \color{blue}\left( \text{i} \right) \text{Does } x^{*} \text{ satisfy the FONC for minimum or maximum? Where are the KKT multipliers?} $
$ \color{blue}\text{Solution 1:} $
$ \text{ Standard form: optimize} \left(x_{1}-2\right)^{2}+\left(x_{2}-1\right)^{2} $
$ \text{subject to } g_{1}\left( x \right) x_{1}^{2}-x_{2}\leq0 $
$ g_{2}\left( x \right) x_{1}+x_{2}-2\leq0 $
$ g_{3}\left( x \right) -x_{1}\leq0 $
$ \text{KKT condition: (1) } Dl\left( \mu ,\lambda \right)=Df\left(x \right)+\mu_{1}Dg_{1}\left( x \right)+\mu_{2}Dg_{2}\left( x \right)+\mu_{3}Dg_{3}\left( x \right) $
$ =\left [ 2x_{1}-4+2\mu_{1}x_{1}+\mu_{2}-\mu_{3}, 2x_{2}-2-\mu_{1}+\mu_{2} \right ] $
$ \left ( 2 \right ) \mu^{T}g\left ( x \right )=0 \Rightarrow \mu_{1}\left ( x_{2}^2-x_{2} \right )+\mu_{2}\left ( x_{1}+x_{2}-2 \right ) - \mu_{3}x_{1}=0 $
$ \color{blue}\text{Solution 2:} $
Automatic Control (AC)- Question 3, August 2011
Problem 1: https://www.projectrhea.org/rhea/index.php/ECE-QE_AC3-2011_solusion
Problem 2: https://www.projectrhea.org/rhea/index.php/ECE-QE_AC3-2011_solusion-2
Problem 3: https://www.projectrhea.org/rhea/index.php/ECE-QE_AC3-2011_solusion-3
Problem 4: https://www.projectrhea.org/rhea/index.php/ECE-QE_AC3-2011_solusion-4