use fmincon with 0<x1<x2<x3

Hello there,

I would be interested if I can use fmincon in a way that 0<x1<x2<x3 when i have a function f(x) with x=[x1 x2 x3....]

Greetings

Answer 1

[Torsten]

Strict inequality is not possible. If you are satified with <= instead of <, use

-x1      <= 0
 x1 - x2 <= 0
 x2 - x3 <= 0

or in the A,b setting of fmincon

A = [-1 0 0;1 -1 0;0 1 -1]
b = [0;0;0]

[Torsten]

@xone92_ comment moved here:

Perfekt thank you!

[Matt J]

@xone92_ If it worked, then you should Accept-click Torsten's answer.

[Marko]

This thread is "solved" and almost 1year old.

But maybe, somebody need a solution which is strictly "<" instead of "<=".

I suggest this workaround: choose a small number as delta, e.g.:

dx = 2*eps

-x1      <= dx
 x1 - x2 <= dx
 x2 - x3 <= dx

or in the syntax for fmincon:

dx = eps;
 
A = [-1 0 0;1 -1 0;0 1 -1];
b = [dx;dx;dx];

Answer 2

[Steven Lord]

Another possible solution is to redefine your code in terms of d(1), d(2), d(3), etc. Constrain all the elements of the d vector to be greater than eps (or some other small value, whatever difference you want to be the minimum that the elements of x can be separated by) using a lower bound. Inside your objective function compute the x vector as cumsum(d) and use it in your calculations.

d = [1 0.25 3]

d = 1×3

1.0000 0.2500 3.0000

x = cumsum(d)

x = 1×3

1.0000 1.2500 4.2500

If you want to allow some of the consecutive elements of x to be equal, the lower bound for that element in d is 0.

d = [1 0.25 0 3]

d = 1×4

1.0000 0.2500 0 3.0000

x = cumsum(d)

x = 1×4

1.0000 1.2500 1.2500 4.2500