Line 20: Line 20:
 
We form the lagrangian:<br>
 
We form the lagrangian:<br>
 
<math>l(x,\lambda)=x_1x_2+\lambda_1(x_1+x_2+x_3-1)+\lambda_2(x_1+x_2-x_3)</math><br>
 
<math>l(x,\lambda)=x_1x_2+\lambda_1(x_1+x_2+x_3-1)+\lambda_2(x_1+x_2-x_3)</math><br>
<math>\begin{cases}\nabla_xL=\begin{bmatrix} x_2+\lambda_1+\lambda_2 \\ x_1+\lambda_1+\lambda_2 \\ \lambda_1+\lambda_2\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\
+
<math>\begin{cases}
 +
\nabla_xl=\begin{bmatrix} x_2+\lambda_1+\lambda_2 \\ x_1+\lambda_1+\lambda_2 \\ \lambda_1+\lambda_2\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\
 
x_1+x_2+x_3-1=0 \\
 
x_1+x_2+x_3-1=0 \\
 
x_1+x_2-x_3=0
 
x_1+x_2-x_3=0

Revision as of 22:24, 18 February 2019


ECE Ph.D. Qualifying Exam

Automatic Control (AC)

Question 3: Optimization

August 2016 Problem 4


Solution

We form the lagrangian:
$ l(x,\lambda)=x_1x_2+\lambda_1(x_1+x_2+x_3-1)+\lambda_2(x_1+x_2-x_3) $
$ \begin{cases} \nabla_xl=\begin{bmatrix} x_2+\lambda_1+\lambda_2 \\ x_1+\lambda_1+\lambda_2 \\ \lambda_1+\lambda_2\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ x_1+x_2+x_3-1=0 \\ x_1+x_2-x_3=0 \end{cases} $
No valid solution for lagrangian condition
Such that the problem can not be optimized


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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