# Minimizing a function using fmincon with special constraints and intervals

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HAT on 6 Jan 2023
Edited: HAT on 6 Jan 2023
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. 1/9 <=x(i)/x(j)<=9, for i,j=1,2,3,4 by considering i<j.
2. x(1) + x(2) + x(3) + x(4) = 1;
3. 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;

Jon on 6 Jan 2023
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
HAT on 6 Jan 2023
Edited: HAT on 6 Jan 2023
Thank you all very much. It helps a lot.