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===== <math>\color{blue}\text{Solution 1:}</math> ===== | ===== <math>\color{blue}\text{Solution 1:}</math> ===== | ||
| − | <math>\text{We need to find a direction }d\text{, such that } \exists\alpha_{0}>0,</math> | + | <math>\text{We need to find a direction }d\text{, such that } \exists\alpha_{0}>0,</math> <math>\left( \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right) + \alpha d \text{ for all } \alpha\in\Omega \left[0,\alpha_{0}\right]</math><br> |
| − | + | <math>\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.</math> | |
| − | + | ||
| − | <math>\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</math> | + | |
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<math>\color{blue}\text{Solution 1:}</math> | <math>\color{blue}\text{Solution 1:}</math> | ||
| − | <math>\text{Let} f(x)=-x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2}, g_{1}(x)=-x_{1}, 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> | ||
| + | \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> | ||
| + | <br> | ||
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Revision as of 22:13, 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) <br> \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} $
