error in minimize function

2015 6 16
0 个回答
更新时间：2015 6 17
1 次查看（30 天）

0 个投票

Hi I want to calculate minimum of this function but when I using nonlinear constrain, i have this error
x=fmincon(fx,[1,1],[],[],[],[],[1 0.99],[1 1.1],@confun)
No feasible solution found.
fmincon stopped because the size of the current step is less than the default value of the step size tolerance but constraints are not satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
x =
1.000000000000000 1.099999986802130

5 个评论
显示 3更早的评论 隐藏 3更早的评论

Walter Roberson
Walter Roberson 2015-6-17
We are going to need to see your confun. fx might help as well.
You realize that you set the lower bound and upper bound to the first element to both be 1, so the first element will always be exactly 1?
Mohammad
Mohammad 2015-6-17
编辑：Mohammad 2015-6-17
yes I want the first element be 1. my function is :
f=@(G,r)abs((G*r)/4 + ((G^2*r^2)/16 + (G*r)/4 - 1/4)^(1/2) + 1/2) + abs((2004*G + 200*G*r + 1001*G^2*r + 200)/(6006*G + 600) - ((1003*G + 200*G*r + G^2*r)/(6006*G + 600) - (2004*G + 200*G*r + 1001*G^2*r + 200)^2/(9*(2002*G + 200)^2))/(((G/(4004*G + 400) + (2004*G + 200*G*r + 1001*G^2*r + 200)^3/(27*(2002*G + 200)^3) - ((1003*G + 200*G*r + G^2*r)*(2004*G + 200*G*r + 1001*G^2*r + 200))/(6*(2002*G + 200)^2))^2 + ((1003*G + 200*G*r + G^2*r)/(6006*G + 600) - (2004*G + 200*G*r + 1001*G^2*r + 200)^2/(9*(2002*G + 200)^2))^3)^(1/2) + G/(4004*G + 400) + (2004*G + 200*G*r + 1001*G^2*r + 200)^3/(27*(2002*G + 200)^3) - ((1003*G + 200*G*r + G^2*r)*(2004*G + 200*G*r + 1001*G^2*r + 200))/(6*(2002*G + 200)^2))^(1/3) + (((G/(4004*G + 400) + (2004*G + 200*G*r + 1001*G^2*r + 200)^3/(27*(2002*G + 200)^3) - ((1003*G + 200*G*r + G^2*r)*(2004*G + 200*G*r + 1001*G^2*r + 200))/(6*(2002*G + 200)^2))^2 + ((1003*G + 200*G*r + G^2*r)/(6006*G + 600) - (2004*G + 200*G*r + 1001*G^2*r + 200)^2/(9*(2002*G + 200)^2))^3)^(1/2) + G/(4004*G + 400) + (2004*G + 200*G*r + 1001*G^2*r + 200)^3/(27*(2002*G + 200)^3) - ((1003*G + 200*G*r + G^2*r)*(2004*G + 200*G*r + 1001*G^2*r + 200))/(6*(2002*G + 200)^2))^(1/3));
>> fx=@(x)f(x(1),x(2));
Also my Confun, which saves as mfile is:
function [c, ceq] = confun(x)
% Nonlinear inequality constraints
c = [(x(1)*x(2))-1];
% Nonlinear equality constraints
ceq = [];
this constrain says that G<(1/r) but when I say the answer which give me matlab, these constrain have been ignored.
Matt J
Matt J 2015-6-17
Since you already know that x(1)=1, what happens when you run as follows:
fx=@(z)f(1,z);
nonlcon=@(z) confun([1,z])
x=fmincon(fx,1,[],[],[],[], 0.99,1.1,nonlcon);
Mohammad
Mohammad 2015-6-17
for simplification the problem i said this actually G and r is not constant
Matt J
Matt J 2015-6-17
But what happens when you do it?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Solver Outputs and Iterative Display 的更多信息

标签

提问:

Mohammad
Mohammad
2015-6-16

评论：

Matt J
Matt J
2015-6-17

Community Treasure Hunt

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

Start Hunting!

Translated by