Consider the problem −
$min :fleft ( x right )$ such that $x in X$, where X is an open set in $mathbb{R}^n$ and $g_i left ( x right )leq 0, i=1, 2,…,m$
Let $S=left { x in X:g_ileft ( x right )leq 0, forall i right }$
Let $hat{x} in S$ and let $f$ and $g_i,i in I$ are differentiable at $hat{x}$ and $g_i, i in J$ are continuous at $hat{x}$. Furthermore, $bigtriangledown g_ileft ( hat{x} right), i in I$ are linearly independent. If $hat{x}$ solves the above problem locally, then there exists $u_i,i in I$ such that
$bigtriangledown fleft ( xright)+displaystylesumlimits_{iin I} u_i bigtriangledown g_ileft ( hat{x} right)=0$, $::u_i geq 0, i in I$
If $g_i,i in J$ are also differentiable at $hat{x}$. then $hat{x}$, then
$bigtriangledown fleft ( hat{x}right)+displaystylesumlimits_{i= 1}^m u_i bigtriangledown g_ileft ( hat{x} right)=0$
$u_ig_ileft ( hat{x} right)=0, forall i=1,2,…,m$
$u_i geq 0 forall i=1,2,…,m$
Example
Consider the following problem −
$min :fleft ( x_1,x_2 right )=left ( x_1-3right )^2+left ( x_2-2right )^2$
such that $x_{1}^{2}+x_{2}^{2}leq 5$,
$x_1,2x_2 geq 0$ and $hat{x}=left ( 2,1 right )$
Let $g_1left ( x_1, x_2 right)=x_{1}^{2}+x_{2}^{2}-5$,
$g_2left ( x_1, x_2 right)=x_{1}+2x_2-4$
$g_3left ( x_1, x_2 right)=-x_{1}$ and $g_4left ( x_1,x_2 right )=-x_2$
Thus the above constraints can be written as −
$g_1 left ( x_1,x_2 right)leq 0, g_2 left ( x_1,x_2 right) leq 0$
$g_3 left ( x_1,x_2 right)leq 0,$ and $g_4 left ( x_1,x_2 right) leq 0$ Thus, $I=left { 1,2 right }$ therefore, $ u_3=0,:: u_4=0$
$bigtriangledown f left ( hat{x} right)=left ( 2,-2 right), bigtriangledown g_1 left ( hat{x} right)= left ( 4,2 right)$ and
$bigtriangledown g_2left ( hat{x} right ) =left ( 1,2 right )$
Thus putting these values in the first condition of Karush-Kuhn-Tucker conditions, we get −
$u_1=frac{1}{3}$ and $u_2=frac{2}{3}$
Thus Karush-Kuhn-Tucker conditions are satisfied.
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