| (4 intermediate revisions by the same user not shown) | |||
| Line 57: | Line 57: | ||
\end{bmatrix} \\ | \end{bmatrix} \\ | ||
\begin{align} | \begin{align} | ||
| − | T(x^*) & = {y: Dh(x^*)y = 0} \\ | + | T(x^*) & = \{y: Dh(x^*)y = 0\} \\ |
| − | & = {y: \begin{bmatrix} | + | & = \{y: \begin{bmatrix} |
1 & 1 & 0 \\ | 1 & 1 & 0 \\ | ||
0 & 1 & 1 | 0 & 1 & 1 | ||
| − | \end{bmatrix} y = 0} \\ | + | \end{bmatrix} y = 0\} \\ |
| − | & = {y: y = \begin{bmatrix} | + | & = \{y: y = \begin{bmatrix} |
1 \\ | 1 \\ | ||
-1 \\ | -1 \\ | ||
1 | 1 | ||
| − | \end{bmatrix} a, a \in \Re } \\ | + | \end{bmatrix} a, a \in \Re \} |
| − | \end{ | + | \end{align} \\ |
| − | </math> | + | \forall y \in T(x^*), y \ne 0: y^T L(x^*,\lambda^*) y = a \begin{bmatrix} 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} |
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 \\ | ||
| + | 0 & 1 & 0 | ||
| + | \end{bmatrix} \begin{bmatrix} | ||
| + | 1 \\ | ||
| + | -1 \\ | ||
| + | 1 | ||
| + | \end{bmatrix} a = \begin{bmatrix} -1 & 2 & -1 \end{bmatrix} \begin{bmatrix} | ||
| + | 1 \\ | ||
| + | -1 \\ | ||
| + | 1 | ||
| + | \end{bmatrix} a^2 = -4 a^2 < 0 \\</math> | ||
| + | |||
| + | <math>\therefore x^* = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}^T</math> is a maximizer | ||
| + | <br> '''Comment: ''' Solution 2 uses the formal procedure of Lagrange Multiplier approach, which is more complicated but would apply to more general cases. Solution 1 is not as general but is simpler for the given problem. They both have the same results. <br> | ||
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | [[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | ||
Latest revision as of 11:22, 25 March 2015
QE2013_AC-3_ECE580-5
Solution 1:
From the constraint, it can be seen that:
$ x_1 = x_3 = -x_2 $
Substitute into the objective function:
$ f(x) = x_2 (x_1 + x_3) = -2 x_2^2 $
Therefore it has a maximizer but no minimizer (f(x) goes to $ -\infty $ as $ |x_2| $ increases)
The maximizer is $ x_1 = x_2 = x_3 = 0 $. There f(x) reaches the maximum value of 0.
Solution 2:
$ f(x) = x_1 x_2 + x_2 x_3 \\ h_1(x) = x_1 + x_2 \\ h_2(x) = x_2 + x_3 \\ l(x,\lambda) = f(x) + \lambda_1 h_1(x) + \lambda_2 h_2(x) = x_1 x_2 + x_2 x_3 + \lambda_1 (x_1 + x_2) + \lambda_2 (x_2 + x_3) \\ \nabla l(x,\lambda) = \begin{bmatrix} x_2 + \lambda_1 \\ x_1 + x_3 + \lambda_1 + \lambda_3 \\ x_2 + \lambda_2 \\ x_1 + x_2 \\ x_2 + x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \\ \Rightarrow x^* = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\ \lambda^* = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ L(x^*,\lambda^*) = F(x^*) + \lambda_1^* H_1(x^*) + \lambda_2^* H_2(x^*) = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \\ \begin{align} T(x^*) & = \{y: Dh(x^*)y = 0\} \\ & = \{y: \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} y = 0\} \\ & = \{y: y = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} a, a \in \Re \} \end{align} \\ \forall y \in T(x^*), y \ne 0: y^T L(x^*,\lambda^*) y = a \begin{bmatrix} 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} a = \begin{bmatrix} -1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} a^2 = -4 a^2 < 0 \\ $
$ \therefore x^* = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}^T $ is a maximizer
Comment: Solution 2 uses the formal procedure of Lagrange Multiplier approach, which is more complicated but would apply to more general cases. Solution 1 is not as general but is simpler for the given problem. They both have the same results.
Back to QE2013 AC-3 ECE580
