What am I doing wrong!?

Hi,

I have the following code, it's part of a larger file that is trying (trying being the crucial word here) to perform optimisation using powell's method:

% testing

clear

a = zeros(1,1);

x0 = [0.2, 0.4, 0.6];

s1 = [0.4, 0.4, 1.6];

dx1 = a*x0;

x1 = x0 + dx1;

func = (x1(1)-x1(2))^2+2*(x1(2)-x1(3))^2+3*(x1(3)-1)^2;

[a] = feval(func, x1(1), x1(2), x1(3));

What I am trying to do is find the value of 'a' that minimises 'func'?

Thanks

Answer 1

[Kevin Holst]

Ah I see the problem now. In your original post you had

s1 = [0.4, 0.4, 1.6];

but nothing using s1 in it. Now that I see what s1 is used for here's the solution:

x0 = [0.2, 0.4, 0.6];
s1 = [0.4, 0.4, 1.6];
x1 = @(n,a) x0(n) + s1(n)*a;
func = @(a)(x1(1,a)-x1(2,a))^2+2*(x1(2,a)-x1(3,a))^2+3*(x1(3,a)-1)^2;
a = fminsearch(func, 0);

a = 0.1364

[Will]

Thank you so much! That's exactly what I wanted. Are you able to explain what having the (n) does? I would understand if it was in a 'for' loop.

Answer 2

[Sean de Wolski]

Syntax issues:

a = zeros(1,1);
x0 = [0.2, 0.4, 0.6];
s1 = [0.4, 0.4, 1.6];
dx1 = a*x0;
x1 = x0 + dx1;
func = @(x)(x(1)-x(2))^2+2*(x(2)-x(3))^2+3*(x(3)-1)^2;
a = fminsearch(func, x1(1:3));

[Will]

Thank you very much!

Is 'feval' still used?

What does @(x) do to the 'func' line?

[Will]

Also, that code works for finding the values of x that minimise func. I am trying to find the value of 'a' which is a scaler multiplier.

Answer 3

[Matt Kindig]

This is not the approach you want to follow. First, you are calling feval, which will not minimize anything, only evaluate a specified. Instead, you will need one of the solvers in the Optimization Toolbox that is designed to minimize a function, such as fminunc. You will pass a function handle into fminunc that defines your function with the variable you wish to change (a) as the input parameter. Something like this should work:

x0 = [0.2, 0.4, 0.6];
s1 = [0.4, 0.4, 1.6];   %note that you don't use this anywhere
func = @(a) ((a+1)*(x0(1)-x0(2))).^2 + 2*((a+1)*(x0(2)-x0(3))).^2 + 3*( (a+1)*(x0(3)-1)).^2;
a = fminunc(func, 0);

[Will]

Thanks for the replies but neither of those suggestions give me the answer I am trying to obtain, probably because I haven't explained it. This is the maths of what I want to achieve.

Original function f1=(x1-x2)^2+2*(x2-x3)^2+3*(x3-1)^2 X1=X0+a1*S1 X1=[0.2 0.4 0.6]+a1*[0.4 0.4 0.6]=[0.2+0.4a1 0.4+0.4a1 0.6+1.6a1]

Substituting the new values of x1,x2,x3 into the original function yields:

fa1=-0.2^2+2*(-0.2-1.2a1)^2+3*(-0.4+1.6a1)^2

Minimisation of this function should return a value of a1=0.1364

Thanks

Answer 4

[Kevin Holst]

I'd suggest using the following code, it's not pretty, but it gets the job done without any toolboxes.

x0 = [0.2, 0.4, 0.6];
x1 = @(n,a) (1+a)*x0(n);
func = @(a)(x1(1,a)-x1(2,a))^2+2*(x1(2,a)-x1(3,a))^2+3*(x1(3,a)-1)^2;
a = fminsearch(func, 0);

[Will]

Thanks for the replies but neither of those suggestions give me the answer I am trying to obtain, probably because I haven't explained it. This is the maths of what I want to achieve.

Original function f1=(x1-x2)^2+2*(x2-x3)^2+3*(x3-1)^2 X1=X0+a1*S1 X1=[0.2 0.4 0.6]+a1*[0.4 0.4 0.6]=[0.2+0.4a1 0.4+0.4a1 0.6+1.6a1]

Substituting the new values of x1,x2,x3 into the original function yields:

fa1=-0.2^2+2*(-0.2-1.2a1)^2+3*(-0.4+1.6a1)^2

Minimisation of this function should return a value of a1=0.1364

Thanks