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

<span style="font-size: 17px;"</span>$ \text{It is equivalent to minimize } f\left(x\right) \text{, } $ $ \text{ subject to } g_{1}(x)\leq0, g_{2}(x)\leq0 $

$ l\left(x,\mu \right) = \nabla f(x)+\mu_{1}\nabla g_{1}(x)+ \mu_{2}\nabla g_{2}(x)=\left( \begin{array}{c} 2x_{1}-1+x_{2} \\ 1+x_{1} \end{array} \right) + \left( \begin{array}{c} -\mu_{1} \\ 0 \end{array} \right) +\left( \begin{array}{c} 0 \\ -\mu_{2} \end{array} \right) =0 $

$ -\mu_{1}x_{1}-\mu_{2}x_{2}=0 $

$ \because x_{1}=1/2, x_{2}=0 $

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