Minimizing a function using fmincon with special constraints and intervals

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;

Answer 1

[Jon]

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

[Jon]

But actually all of your constraints are linear, so it would be better to formulate your linear inequality constraints in matrix form and use these

[Jon]

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]

If you do try my original solution, using the non-linear constraints, I suggest modifying the initial guess so that it satisfies, the equality constraint, e.g. x = [1/4,1/4,1/4,1/4]. I noticed that using your initial guess, the solution seems to terminate prematurely, with a function value that is larger than actual optimum. Again, this reinforces that it is probably better to use the linear constraints, as then, even using the initial guess of [1 1 1 1] it reaches the same solution as using [1/4 1/4 1/4 1/4], so it is less sensitive to a correct initial guess.

[John D'Errico]

Linear constraints are ALWAYS preferable to nonlinear constraints. ALWAYS.

[Jon]

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]

Thank you all very much. It helps a lot.