How to use fmincon with vectors? (error message: not enough input arguments)

Hello,

I am trying to use fmincon to find values that maximizes the function. (2 variables)

The document on the website provides sample that provides scalar and I want to return vectors instead of it.

Below is the sample code I wrote to understand how fmincon works.

test = [1 2 3 4];
fun = @(c,d)100*(c-d^2)^2 + (test(:)-c).^2;
fun(2,3)
lb = [0,0.2,0,0.2 ; 0,0.2,0,0.2];
ub = [0.5,0.8,0.5,0.8 ; 0.5,0.8,0.5,0.8];
A = [];
b = [];
Aeq = [];
beq = [];
x0 = [1/4,1/4,1/4,1/4 ; 1/4,1/4,1/4,1/4 ];

[x,val] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)

So basically, I want to have (c,d) that maximizes fun for each value in test.

But when I run this code, it says 'Not enough input arguments'.

Seems like I made some mistake but couldn't find it.

Any help? Thanks in advance.

Answer 1

[Matt J]

fun = @(x) -100*(x(1,:)-x(2,:).^2).^2 + (test-x(1,:)).^2;

[Chang seok Ma]

Thanks for the answer but then I got an error message as "Supplied objective function must return a scalar value"

[Matt J]

I guess you solved it (since you Accept-clicked the answer)?

[Chang seok Ma]

Sorry, I thought it will solve it but I realized that it doesn't solve the problem.

test = [1 2 3 4];
fun = @(x) -100*(x(1,:)-x(2,:).^2).^2 + (test-x(1,:)).^2;
lb = [0,0.2,0,0.2 ; 0,0.2,0,0.2];
ub = [0.5,0.8,0.5,0.8 ; 0.5,0.8,0.5,0.8];
A = [];
b = [];
Aeq = [];
beq = [];
x0 = [1/4,1/4,1/4,1/4 ; 1/4,1/4,1/4,1/4 ];
[x,val] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)

This is the code I have right now which gives me an error message "Supplied objective function must return a scalar value."

[Matt J]

fun = @(x) -100*sum( (x(1,:)-x(2,:).^2).^2 + (test-x(1,:)).^2  ,'all');

[Chang seok Ma]

Thanks now I can see some results and I think it is right but I just want to make sure I understand this correctly.

Here is how I thought.

Since the length of test is 4, (x(1,:)-x(2,:).^2).^2 + (test-x(1,:)).^2 will give me some (1,4) vector(A)

and I thought I need to choose x(1) and x(2) that will maximize each value in (1,4) vector. (x(1,1) and x(2,1) will maximize A(1)...)

By putting sum, now the problem becomes choosing x(1,1) x(2,1) x(1,2) x(2,2) x(3,1) x(3,2) x(4,1) x(4,2) that maximizes A(1) + A(2) + A(3) + A(4).

But this is also identical to choose x(1) and x(2) that will maximize each value in (1,4)

Am I understanding correctly?

[Matt J]

Yes. Although because the problem is additively separable, it might be better just to use a loop and solve 4 separate 2-dimensional problems. The finite differencing operations will be less wasteful that way.

[Chang seok Ma]

So if the problem is not additively separable, is there other way I could solve this problem without loop?

[Matt J]

If the problem weren't separable, using a loop would not be an option. You would be forced to solve it as an 8D minimization problem.