Optimali sutartis


Sutartis optimali, jei jos neapsimoka lauzyti.
Tegu sutartinis vektorius $(\bar{y_i}, \bar{x_i})$,
tada "apgavystes" vektorius

\begin{eqnarray}(\bar{\bar{y_i}}, \bar{\bar{x_i}})=\nonumber \\
\arg \max _{(y_...
...i,x_i, \bar{y_j}, \bar{x_j},\ j \neq i),\\
i=1,...,m. \nonumber
\end{eqnarray}


Sutartis $x=(\bar{y_i}, \bar{x_i},\ i=1,...,m)$ optimali, kai

\begin{eqnarray}\min_x f(x)=0,
\end{eqnarray}


kur

\begin{eqnarray}f(x)=\sum_{i=1}^m ((\bar{\bar{y_i}}-\bar{y_i})^2+( \bar{\bar{x_i}}-\bar{x_i})^2).
\end{eqnarray}


$f(x)$ minimizuojama globaliu optimizavinmo metodu.



jonas mockus 2004-03-01