How do I use GlobalSearch to solve the minimal problem of a customized function with two outputs

I code a customized function U ,it has a input d which is a 5-dimensional vector, a output 'flag' ,and a output 'P' witch is the parameter I want to find it's minimal value:

function [P flag]=U(d)
...
end

The question is how can I to use GlobalSearch to find the minimal value of P.

Thanks for helping me....

Answer 1

[infinity]

Hello

I will give an example of find minimum value of objective function

First, we can create the objective function, named by U(x), for example

function res = U(x)

res = sqrt(x(1)^2 + x(2)^2 + x(3)^2 + ...
    x(3)^2 + x(4)^2 + x(5)^2);

Then, we can use the globalsearch as follows

problem = createOptimProblem('fmincon','objective', ...
          @(x) U(x), 'x0', [1 1 1 1 1], 'lb', [-3 -3 -3 -3 -3], ...
          'ub', [3 3 3 3 3]);
      
gs = GlobalSearch;

[x, f] = run(gs, problem) 

where x0 = [1, 1, ..., 1] is the initial guess, lb is the lower bound of the varibale, ub is the upper bound of the variable and x is the optimal solution, f is the minimum value of the objective function corresponding to x.

Best regards,

Trung

[mathy]

Thanks for your answer. I noticed that in your answer the function U only has one output parameter 'res',but what I asked is what if the function U had two output parameters,take this code as example:

function [P flag]=U(d)
if (d(1)+d(2))>0
    P=d(1)+d(2);
    flag=1;
else
    P=d(1)*d(2);
    flag=0;
end
end

To find the minimal P,how do I code the globalsearch correctly? Thanks for your answer again!

[infinity]

Hello,

Sorry for misunderstad your question. I have tried to solve your example function and I have got a reasonable result. For example, here is my code for your objective function:

clear

x0 = [1 1];
lb = [-3 -3];
ub = [3 3];
problem = createOptimProblem('fmincon','objective', ...
          @(x) U(x), 'x0', x0, 'lb', lb, ...
          'ub', ub);  
      
gs = GlobalSearch;

[x, f] = run(gs, problem) 

It will give the results like

x =

   -3.0000    3.0000


f =

   -9.0000

Maybe in your problem, you will get problem with the optimal solution. Can you tell me some what happen with your working objective function.

Best regards,

Trung

[mathy]

Thank you! You answer helps me a lot!

The objective function I working on is a little complicated than the example I gave,but in a way they are the same for the number of output parameters they have is two and one of the output parameters is a binary number

But there is a small question that I still want to make sure:

Can the GlobalSearch automatically identify which the output parameter that I want to find it's minimal value is？

[infinity]

Hello,

It is great that the solution can help you.

For your question, I could give you some infomation, say

If we define the objective function like

function [P flag]=U(d)
if (d(1)+d(2))>0
    P=d(1)+d(2);
    flag=1;
else
    P=d(1)*d(2);
    flag=0;
end
end

then we can obtain the value of this function by many ways, for example

x = [0, 4];
a = U(x)

we will get "a = 4" which is the value of "P" in the output of the function. It will automatically give us the value of the first output of the function. If we want to get the second output only, usually it will be like

x = [0, 4];
[~,b] = U(x)

and "b = 1" which is the value of "flag" of the function.

So, in this specific function, we force "P" as the first output and when we implement globalsearch function, it will find the minumum of "P".

If we change the order of the outputs in the function, we will get different results. For example, if we change the function like

function [flag P]=U(d)

then,

x = [0, 4];
a = U(x)

we will get the result "a = 1"

If you use the modified function [flag P] to optimize, the globalsearch will give you wrong result. Becarefullly with this. Also, "flag" is variable only 0 and 1.

Hope this can help you.

Best regards,

Trung

[mathy]

Your answers do help me a lot! Thank you again for your patience and your answers!

[infinity]

I am glad to know your problem has been solved.

You are wellcome!