Minimum of a function

Question

Ray Louise 2020-4-26

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编辑： John D'Errico 2020-4-26

I am having a hard time figuring out how to call the file-functions. I want to make a nonlinear contraint , and then another file function to call the initial approximations x0(1) и х0(2), then call the the equtaion below

f = @(х) (х(1) - 2).^2 + (х(2) - 4)

Below are my limits:

c=[-x(1);

-x(2);

2*x(1).^2+x(2).^2-34;

2*x(1)+3*x(2).^2-18];

So far I got to here

function f = fun( x )

f = @(х) (х(1) - 2).^2 + (х(2) - 4)

end

function [ с,сequ ] = mycon( x )

с=[-x(1);

x(2);

-x(3);

2*x(1)+2+(x)+3*x(3).^2-34; %% I'm not sure if this is even written correctly.

сequ=[]; end

This speaks nothing to me and I've been trying every option under the moon to make it run but it doesn't work. How can I even write down the last limit as a MATLAB function? I need a slight boost, please. I read the fmincon documentation but this function is giving me a hard time.

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Answer 1

Thiago Henrique Gomes Lobato 2020-4-26

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https://ww2.mathworks.cn/matlabcentral/answers/520818-minimum-of-a-function#answer_428448

编辑：Thiago Henrique Gomes Lobato 2020-4-26

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You had some typos in your constrain function and also you don't need to create a function file for your function since you already use an annonymous one. Use this and it should work:

f = @(x) (x(1) - 2).^2 + (x(2) - 4);
x0 = [1,1]; % Here you put your x0 function you said you had
nonlcon = @mycon;
x = fmincon(f,x0,[],[],[],[],[],[],nonlcon)
% Another file
function [ c,cequ ] = mycon( x ) 
 c=[-x(1);
-x(2); 
2*x(1).^2+x(2).^2-34;
2*x(1)+3*x(2).^2-18]; %% I'm not sure if this is even written correctly.
cequ=[];
end 
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
<stopping criteria details>
x =
    2.0000    0.0000

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John D'Errico 2020-4-26

编辑：John D'Errico 2020-4-26

在 MATLAB Online 中打开

I would only add that the first two inequalities are just bound constraints on x(1) and x(2). so they could be put in directly as bound constraints, rather than needing a nonlinear inequality constraint. If there were 3 variables, then LB would be just

LB = [0,0,-inf];

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