Find optimized x1, x2, x3 for a given y.

Hi,

I have a function like below:

y=x1^2+2*x1*x2+x3^2;

I want to find the values for x1, x2 and x3 for y=2.

How can I do that? I tried with genetic algorithm using 'ga' function but looks like that's not the right one. Can anyone please help me how should I approach or is there any built in function to get that? Thanks in advance.

Answer 1

[Walter Roberson]

y = 2;
fun = @(x) (x(1).^2 + 2.*x(1).*x(2) + x(3).^2 - y).^2
bestx = ga(fun, 3);
x1 = bestx(1); x2 = bestx(2); x3 = bestx(3);

Note: there are an infinite number of solutions. It does not start to get interesting until you put on constraints.

[Easir Papon]

Thank you so much.

I think there should not be the square sign you used at the end of the function.

And if my x2 is bounded by 2<x2<5 and x1=2, and I want to find optimized result for y=2. That means I am looking for the values of x2 and x3. How should I write it?

Thanks again.

[Walter Roberson]

No, the ^2 I used is needed. ga is a minimizer. If you were to use

fun = @(x) x(1).^2 + 2.*x(1).*x(2) + x(3).^2 - y;

then it would happily minimize to negative infinity by making x(2) negative (if it was unconstrained). You are not asking the question "which x gives x(1).^2 + 2.*x(1).*x(2) + x(3).^2 most less than y", you are asking the question "which x gives x(1).^2 + 2.*x(1).*x(2) + x(3).^2 equal to y". The way to use a minimizer to get one (real) value equal to another is to minimize the square of the difference: the difference will be 0 when they are the same, and the square of the difference increases as they get further apart.

"And if my x2 is bounded by 2<x2<5 and x1=2, and I want to find optimized result for y=2. That means I am looking for the values of x2 and x3"

y = 2;
fun = @(x) (x(1).^2 + 2.*x(1).*x(2) + x(3).^2 - y).^2;
nvars = 3;
A = []; b = [];
Aeq = []; beq = [];
lb = [2 2*(1+eps) -inf];
ub = [2 5*(1-eps) inf];
bestx = ga(fun, nvars, A, b, Aeq, beq, lb, ub);
x1 = bestx(1); x2 = bestx(2); x3 = bestx(3);

For such a simple system you could also consider fmincon:

y = 2;
fun = @(x) (x(1).^2 + 2.*x(1).*x(2) + x(3).^2 - y).^2;
x0 = [2 3.5 20];
A = []; b = [];
Aeq = []; beq = [];
lb = [2 2*(1+eps) -inf];
ub = [2 5*(1-eps) inf];
bestx = fmincon(fun, x0, A, b, Aeq, beq, lb, ub);
x1 = bestx(1); x2 = bestx(2); x3 = bestx(3);

[Easir Papon]

Thank you so much. I got the understanding now. I don't know how to thank you.

Answer 2

[Akira Agata]

Another solution would be to use fsolve function, like:

fun = @(x) x(1)^2+2*x(1)*x(2)+x(3)^2 - 2;
options = optimoptions(@fsolve,'Algorithm','Levenberg-Marquardt');

% Find the solution from the initial point (x1,x2,x3) = (1,2,3)
fsolve(fun,[1,2,3],options)