Modifying the objective function in order to meet the optimization variable boundaries

First Try

if (A <= 1e-04) & (A >= 4e-04)
    fun = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2;
else
    fun = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2 + 1e30 * ( (A - 4e-04).^2 + (A - 1e-04).^2 );

or second try

for i = 1:size(A,2)
    if (A(i) >= 1e-04) & (A(i) <= 4e-04)
        fun(i) = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2;
    else
        fun(i) = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2 - 1e30 * ( (A(i) - 4e-04).^2 + (A(i) - 1e-04).^2 );
    end
end

Hello,

I wrote above a piece of a code where fun is an objective function that I aim to minimize. And A, an array of size (1,13), is the optimization variable. f is also an array of size(1,13) and my aim is to tune its first two values f(1) and f(2) where the values of f depend on A. The ultimate goal is to optimize A such that f(1) and f(2) obtain the specific values 500 and 1800 respectively, and this has to be done by minimizing the value of objective fucntion where its return value is a scalar.

I needed to impose my conditions that if A lies outisde the specified boundaries then the value of the objective function function will decrease rapidly depending on the distance of this variable from the boundaries.

The displayed error in Matlab is "Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-13." of the first try.

The second gave unlogical result.

I'm not knowing how to fix it.

Thank you.

Answer 1

[Zahraa]

My professor at university helped, and this is how it should work:

fun = (1- (f(1))/500)^2 + (1- (f(2))/1000)^2;

for i = 1:length(A)

if (A(i) <= 0.5e-04)

fun = fun + 1e10 * (A(i) - 0.5e-04).^2 ;

elseif (A(i) >= 6e-04)

fun = fun + 1e10 * (A(i) - 6e-04).^2 ;

end

end

[Matt J]

That's the same as my answer. You don't need the loop.

[Torsten]

@Zahraa

Why don't you simply use lower and upper bounds as 0.5e-4 and 6e-4 for the A(i) in your optimization ?

Answer 2

[Torsten]

if (A <= 1e-04) & (A >= 4e-04)

How can A be <= 1e-4 and >= 4e-4 at the same time ?

And if-statements have to be done elementwise:

for i = 1:size(A,2)
    if A(i) ...
        f(i) = ...
    else
        f(i) = ...
    end
end

[Zahraa]

I modified it this way:

for i = 1:size(A,2)

if (A(i) >= 1e-04) & (A(i) <= 4e-04)

fun(i) = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2;

else

fun(i) = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2 - 1e30 * ( (A(i) - 4e-04).^2 + (A(i) - 1e-04).^2 );

end

end

But still it gave this error: Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-13.

[Torsten]

What is your objective function ? It must return a scalar value that you want to minimize by varying the inputs f(1) and f(2), I guess.

In your case I don't understand how A is related to the objective function.

[Zahraa]

Thank you for your question, I have edited my main question.

[Zahraa]

Also this didn't work

fun = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2;

for i = 1:length(A)

if (A(i) <= 1e-04)

fun = fun + 1e10 * (A(i) - 1e-04).^2 ;

elseif (A(i) >= 4e-04)

fun = fun + 1e10 * (A(i) - 1e-04).^2 ;

end

end

[Torsten]

f is also an array of size(1,13) and my aim is to tune its first two values f(1) and f(2) where the values of f depend on A.

In your objective function, compute f(1) and f(2) from your optimization variables A and return (f(1)-500)^2 + (f(2)-1800)^2.

Answer 3

[Matt J]

If the boundaries are given by simple lower and upper bound vectors lb and ub, you can do,

function fun=objective(A,lb,ub)

    arguments
       A  (1,:)
       lb (1,:)
       ub (1,:)
    end
  

    Athresh=max(min(A,ub),lb);
    
    distance=norm(A-Athresh);
    
    f=...some function of A....
    
    fun = (1- (f(1))/500)^2 + (1- (f(2))/1800)^2 + 1e30*distance^2;
    
end