fmincon nonlinear inequality constraint including variable with range?

function [c,ceq] = nonlcon(x)

c(1)

The above is the name and content of a nonlinear inequality constraint.

If I want to write a nonlinear inequality constraint c(1) = (complicated function of x)*(n-3) for n between 0 and 9, how can I write it?

I have searched on Google and read many pages but I didn't find the answer.

Thank you very much!

Answer 1

[albara]

In MATLAB, you can define a nonlinear inequality constraint function by writing a separate function file. Based on your given information, you can write a constraint function as follows:

1. Create a new file in MATLAB and name it, for example, nonlcon.m.
2. In the nonlcon.m file, define the nonlinear inequality constraint function:

function [c, ceq] = nonlcon(x)

% Set the value of n, assuming n is an integer between 0 and 9

n = 5; % (You can replace this with the desired value)

% Define your complicated function of x (replace with your actual function)

complicated_function = x(1)^2 + x(2)^2; % (This is an example, replace it)

% Define the nonlinear inequality constraint c(1)

c(1) = complicated_function * (n - 3);

% No nonlinear equality constraints

ceq = [];

end

3- Replace the example complicated_function with your actual function and set the value of n as needed.

4- Save the file.

Now you can use this nonlcon.m file as the nonlinear constraint function in your optimization problem. For example, if you're using fmincon, you can pass @nonlcon as the nonlinear constraint function:

% Define your objective function

fun = @(x) x(1)^2 + x(2)^2;

% Define initial guess, lower bounds, upper bounds, etc.

x0 = [1; 1];

lb = [];

ub = [];

A = [];

b = [];

Aeq = [];

beq = [];

% Call fmincon with nonlcon as the nonlinear constraint function

[x, fval] = fmincon(fun, x0, A, b, Aeq, beq, lb, ub, @nonlcon);

This should solve your optimization problem with the given nonlinear inequality constraint. Remember to replace the example complicated_function and n with the appropriate expressions and values for your specific problem.

Important: There may be some mistakes in this answer Experts can tell if there are any mistakes

[Frank]

You answer is NOT relevant, I have said in the title "including variable with range”. Let me clarify: c(1) = (complicated function of x)*(n-3) for n between 0 and 9. That means ALL the real numbers between 0 and 9, NOT a specific number.

[Torsten]

Is n a solution variable ?

[Frank]

Thank you very much for your answer.

I guess you meant "solution variable" is x.

n is not a solution variable.

n has specific physical meaning, n coupled with solution variable x has to be set < 0 for ALL real 9 > n > 0.

[John D'Errico]

@Frank - you were not even remotely clear in your question. But you can compute the value of n, correct? As a function of x, that is?

Then all you need to do is compute n inside the constraint function, nonlcon. Now you have TWO nonlinear inequality constraints. Since an inequality constraint for fmincon will be of the form C(X) <= 0, then you will have:

C(1) = -n;
C(2) = n - 9;

[Frank]

I apologize if I wasn't clear. Thanks for your answer. Unfortunately, your answer is NOT what I wanted. The value of n is NOT computed, it is given a range from 0 to 9. It is NOT a function of x.

Let me reformulate my question:

If I want the nonlinear constraint

for ALL real 9>n>0, how can I write it?

This is different from the convention nonlinear constraint, which for n has a fixed number. Here I want the nonlinear constraint to satisfy ALL real numbers from 0 to 9.

Having wrote the above done, I just now understand it is an impossible task for Matlab because computer can only understand discreet numbers, computer can't understand continuous numbers in a range. Therefore, I have got the answer to my question: it is an impossible task.

Thank you all for the discussions!

Have a nice day!

[Torsten]

How should C be <=0 if 0 < n < 9 ?

[Frank]

It was a complicated function, so any value is possible, so <=0 is possible. The specific form I wrote was just an example.

[Walter Roberson]

n=0:9;
c(n+1) = complicated_function * (n - 3);

You are not restricted to returning c as a scalar. The test is if all(c<=0) then succeed.

[Torsten]

@Frank

c(1) = (complicated function of x)*(n-3) <= 0

Inserting n = 0, you get (complicated function of x)*(-3) <= 0, thus (complicated function of x) >=0

inserting n = 9, you get (complicated function of x)*6 <= 0, thus (complicated function of x) <=0

Thus your condition can only be satisfied if (complicated function of x) =0, and this an equality condition you should set as

ceq(1) = (complicated function of x).