Line 38: Line 38:
 
<math>\text{Let } f(x)=-x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2}, \text{ }g_{1}(x)=-x_{1}, \text{ }g_{2}(x)=-x_{2}</math>  
 
<math>\text{Let } f(x)=-x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2}, \text{ }g_{1}(x)=-x_{1}, \text{ }g_{2}(x)=-x_{2}</math>  
  
<math>\text{It's equivalent to minimize } f(x) <br>
+
<math>\text{It's equivalent to minimize } f(x)</math>
\text{                subject to } g_{1}(x)\leq0, g_{2}(x)\leq0,</math><br>  
+
  
<math>\left{ \begin{array}{c} l(x,\mu)=\nabla f(x)+\mu_{1}\nablag_{1}(x)+ \mu_{2}\nablag_{2}(x) \\ d_{2} \end{array}</math><br>  
+
<math>\text{ subject to } g_{1}(x)\leq0, g_{2}(x)\leq0,</math><font color="#ff0000"><span style="font-size: 17px;">'''
 +
'''</span></font>
 +
 
 +
<font color="#ff0000"><span style="font-size: 17px;">'''<math>\left{ \begin{array}{c} l(x,\mu)=\nabla f(x)+\mu_{1}\nablag_{1}(x)+ \mu_{2}\nablag_{2}(x) \\ d_{2} \end{array}</math>
 +
'''</span></font>
 +
 
 +
<font color="#ff0000"><span style="font-size: 17px;"</span></font>
  
 
<br>  
 
<br>  

Revision as of 22:14, 26 June 2012


ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 3, August 2011


 $ \color{blue}\text{1. } \left( \text{20 pts} \right) \text{ Consider the optimization problem, } $

               $ \text{maximize} -x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2} $

               $ \text{subject to } x_{1}\geq0, x_{2}\geq0 $

$ \color{blue}\left( \text{i} \right) \text{ Characterize feasible directions at the point } x^{*}=\left[ \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right] $

$ \color{blue}\text{Solution 1:} $

$ \text{We need to find a direction }d\text{, such that } \exists\alpha_{0}>0, $   $ \left( \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right) + \alpha d \text{ for all } \alpha\in\Omega \left[0,\alpha_{0}\right] $

$ \text{As } x_{1}\geq0, x_{2}\geq0, d= \left( \begin{array}{c} x \\ y \end{array} \right)\text{where } x\in\Re, \text{ and } y\geq0. $


$ \color{blue}\text{Solution 2:} $

$ d\in\Re_{2}, d\neq0 \text{ is a feasible direction at } x^{*} \text{, if } \exists \alpha_{0} \text{ that } \left[ \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right] + \alpha\left[ \begin{array}{c} d_{1} \\ d_{2} \end{array} \right] \in\Omega \text{ for all } 0\leq\alpha\leq\alpha_{0} $ 

$ \because \begin{Bmatrix}x\in\Omega: x_{1}\geq0, x_{2}\geq0\end{Bmatrix} $


$ \therefore d= \left[ \begin{array}{c} d_{1} \\ d_{2} \end{array} \right], d_{1}\in\Re^{2}, d_{2}\neq0 $


$ \color{blue}\left( \text{ii} \right) \text{Write down the second-order necessary condition for } x^{*} \text{. Does the point } x^{*} \text{ satisfy this condition?} $

$ \color{blue}\text{Solution 1:} $

$ \text{Let } f(x)=-x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2}, \text{ }g_{1}(x)=-x_{1}, \text{ }g_{2}(x)=-x_{2} $

$ \text{It's equivalent to minimize } f(x) $

$ \text{ subject to } g_{1}(x)\leq0, g_{2}(x)\leq0, $

$ \left{ \begin{array}{c} l(x,\mu)=\nabla f(x)+\mu_{1}\nablag_{1}(x)+ \mu_{2}\nablag_{2}(x) \\ d_{2} \end{array} $

<span style="font-size: 17px;"</span>



Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin