Minimizing a function using fmincon with special constraints and intervals

Question

HAT 2023-1-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1889562-minimizing-a-function-using-fmincon-with-special-constraints-and-intervals

编辑： HAT 2023-1-6

采纳的回答： Jon

在 MATLAB Online 中打开

I would like to minimize a function using fmincon as follows:

The function to be minimized is: fun = @(x)x(1)^2 + 3*x(2)/x(3) - x(4)^2 ;

The constraints are:

1/9 <=x(i)/x(j)<=9, for i,j=1,2,3,4 by considering i<j.
x(1) + x(2) + x(3) + x(4) = 1;
x(i) > 0 for i=1,2,3,4.

My challenge is how to apply the first constraint (1) in fmincon. Without the first constraint, I was able to solve it using only the second and the third constraints as follows:

clear; clc
% function to be minimized
fun = @(x)x(1)^2 + 3*x(2)/x(3) - x(4)^2 ;
%initial value
x0=ones(1,4); 
%bound limits
lb=0.0001*ones(1,4); % lower bound as x(i)>0
ub=[];   %no upper bound
% normalize x inside fmincon as x(1)+...+x(4) = 1
Aeq = ones(1,4);  
beq = 1;
% using Matlab's fmincon
x= fmincon(fun,x0,[],[],Aeq,beq,lb,ub)
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.
x = 1×4
    0.0001    0.0001    0.0123    0.9875

I was trying to convert the first constraint into a set of 12 inequalites below. But still I couldn't manage to implement it inside the fmincon. I wonder if you could provide me help. Thank you in advance!

% writing inequalities for: 1/9 <=x(i)/x(j)<=9, for i,j=1,2,3,4 by
% considering i<j.
% (1) for x(i)/x(j)<=9:
x(1)-9*x(2)<=0;
x(1)-9*x(3)<=0;
x(1)-9*x(4)<=0;
x(2)-9*x(3)<=0;
x(2)-9*x(4)<=0;
x(3)-9*x(4)<=0;
%(2) for 1/9<=x(i)/x(j), i,j=1,2,3,4:
1/9*x(2)-x(1)<=0;
1/9*x(3)-x(1)<=0;
1/9*x(4)-x(1)<=0;
1/9*x(3)-x(2)<=0;
1/9*x(4)-x(2)<=0;
1/9*x(4)-x(3)<=0;

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Answer 1

Jon 2023-1-6

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https://ww2.mathworks.cn/matlabcentral/answers/1889562-minimizing-a-function-using-fmincon-with-special-constraints-and-intervals#answer_1142937

在 MATLAB Online 中打开

Here is one approach

% function to be minimized
fun = @(x)x(1)^2 + 3*x(2)/x(3) - x(4)^2 ;
%initial value
x0=ones(1,4); 
%bound limits
lb=0.0001*ones(1,4); % lower bound as x(i)>0
ub=[];   %no upper bound
% normalize x inside fmincon as x(1)+...+x(4) = 1
Aeq = ones(1,4);  
beq = 1;
% using Matlab's fmincon
x = fmincon(fun,x0,[],[],Aeq,beq,lb,ub,@mycon)
function [c,ceq] = mycon(x)
% compute nonlinear inequality constraints
c = zeros(12,1); % preallocate
k = 0;
for i = 1:3
    for j = i+1:4
        k = k+1;
        c(k) = 1/9 - x(i)/x(j);
        c(k+6) = x(i)/x(j) - 9;
    end
end
% compute nonlinear equality constraints at x.
ceq = []; % no nonlinear equality constraints
end

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Jon 2023-1-6

在 MATLAB Online 中打开

So this would be better, as it is probably "easier" for the solver to use a linear constraint than a nonlinear one,

note that I made the initial guess [1/4,1/4,1/4,1/4] so that it satisfied the equality constraint.

% function to be minimized
fun = @(x)x(1)^2 + 3*x(2)/x(3) - x(4)^2 ;
%initial value
x0=ones(1,4)/4; 
%bound limits
lb=0.0001*ones(1,4); % lower bound as x(i)>0
ub=[];   %no upper bound
% normalize x inside fmincon as x(1)+...+x(4) = 1
Aeq = ones(1,4);  
beq = 1;
% define inequality constraints
A = zeros(12,4);
b = zeros(12,1);
k = 0;
for i = 1:3
    for j = i+1:4
        k = k + 1;
        A(k,i) = 1;
        A(k,j) = -9;
        A(k+6,i) = -1;
        A(k+6,j) = 1/9;
    end
end
% using Matlab's fmincon
[x,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)
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.
x = 1×4
    0.0500    0.0500    0.4500    0.4500
fval = 0.1333

Jon 2023-1-6

Thanks @John D'Errico for reinforcing that point. This is an interesting case, because at first glance the constraints are non-linear, but with the additional constraint that the x's are all positive you can multiply through (without the possibility of a reversal in the direction of the inequality) to clear the denominator and obtain the linear constraints.

HAT 2023-1-6