📜  Karush-Kuhn-Tucker最优性必要条件

📅  最后修改于: 2020-11-25 04:55:41             🧑  作者: Mango


考虑问题-

$ min \:f \ left(x \ right)$使得$ x \ in X $,其中X是$ \ mathbb {R} ^ n $和$ g_i \ left(x \ right)\ leq中的开放集0,i = 1,2,…,m $

设$ S = \ left \ {x \ in X:g_i \ left(x \ right)\ leq 0,\ forall i \ right \} $

令S中的$ \ hat {x} \和I $中的$ f $和$ g_i,i \ i在$ \ hat {x} $和$ g_i,i \ in J $中是可微的,在$ \ hat处是连续的{x} $。此外,$ \ bigtriangledown g_i \ left(\ hat {x} \ right),i \ in I $是线性独立的。如果$ \ hat {x} $在本地解决了上述问题,则在I $中存在$ u_i,i \

$ \ bigtriangledown f \ left(x \ right)+ \ displaystyle \ sum \ limits_ {i \ in I} u_i \ bigtriangledown g_i \ left(\ hat {x} \ right)= 0 $,$ \:\:u_i \ geq 0,我\ in I $

如果$ g_i,i \ in J $也可以在$ \ hat {x} $处微分。然后$ \ hat {x} $,然后

$ \ bigtriangledown f \ left(\ hat {x} \ right)+ \ displaystyle \ sum \ limits_ {i = 1} ^ m u_i \ bigtriangledown g_i \ left(\ hat {x} \ right)= 0 $

$ u_ig_i \ left(\ hat {x} \ right)= 0,\ forall i = 1,2,…,m $

$ u_i \ geq 0 \ forall i = 1,2,…,m $

考虑以下问题-

$ min \:f \ left(x_1,x_2 \ right)= \ left(x_1-3 \ right)^ 2 + \ left(x_2-2 \ right)^ 2 $

这样$ x_ {1} ^ {2} + x_ {2} ^ {2} \ leq 5 $,

$ x_1,2x_2 \ geq 0 $和$ \ hat {x} = \ left(2,1 \ right)$

令$ g_1 \ left(x_1,x_2 \ right)= x_ {1} ^ {2} + x_ {2} ^ {2} -5 $,

$ g_2 \ left(x_1,x_2 \ right)= x_ {1} + 2x_2-4 $

$ g_3 \ left(x_1,x_2 \ right)=-x_ {1} $和$ g_4 \ left(x_1,x_2 \ right)=-x_2 $

因此上述约束可以写成-

$ 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,$和$ g_4 \ left(x_1,x_2 \ right)\ leq 0 $因此,$ I = \ left \ {1,2 \ right \} $ ,$ 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 )$和

$ \ bigtriangledown g_2 \ left(\ hat {x} \ right)= \ left(1,2 \ right)$

因此,将这些值置于Karush-Kuhn-Tucker条件的第一个条件中,我们得到-

$ u_1 = \ frac {1} {3} $和$ u_2 = \ frac {2} {3} $

因此满足了Karush-Kuhn-Tucker条件。