Incorrect solution for symmetric problems in fmincon

1 次查看(过去 30 天)
If I maximize XX(1)^2+XX(2)^2 subject to x1 + x2 <=1 and use starting value X0=[0.5,0.5] I get as solution X=[0.5,0.5], although the two optima are X=[1,0] and X=[0,1].
Any clue how to prevent this from happening? (Other than using an asymmetric starting value). I already tried changing algorithm to sqp but that doesn't help.
See code:
function [XX,VAL] = test_con_opt()
clc;
close all;
dbstop if error;
sum_x = 1;
AA = [1,1];
bb = sum_x; %Inequality constraint: x1 + x2 <= sum_x
lb = [0,0];
pwr = 2;
%X0 = [0.25,0.75];
%X0 = [0.75,0.025];
X0 = [0.5,0.5];
[XX,mVAL] = fmincon(@(XX)obj_fun(pwr,XX(1),XX(2)),X0,AA,bb,[],[],lb);
VAL = - mVAL;
end
function [mVAL] = obj_fun(pwr,x1,x2)
mVAL = - (x1^pwr + x2^pwr);
end

采纳的回答

Alan Weiss
Alan Weiss 2019-4-22
fmincon is a gradient-based algorithm. When your initial point is [0.5,0.5], the gradient is zero, and fmincon stops, since it is at a stationary point.
In general, you can take random initial points, which are unlikely to be exact stationary points (assuming that stationary points are isolated).
Alan Weiss
MATLAB mathematical toolbox documentation
  3 个评论
Matt J
Matt J 2021-12-18
编辑:Matt J 2021-12-18
It seems to me that it could conclude that from a comparison of the gradient of objective & linear constraint.
On the contrary, the gradient of the objective at your initial point is grad=[-1,-1]. Therefore in both feasible directions u= [1,-1] and u=[-1,1], the directional derivative is zero. So, the algorithm cannot see a feasible direction of increase or decrease.
Sargondjani
Sargondjani 2021-12-18
Thanks Matt. I understand now it's really a second order thing.

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Solver Outputs and Iterative Display 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by