How write these equation in Matlab for optimization

Dear Friends,

I want to write this optimization problem in matlab.

Anybody here to help me write thi in Matlab please.

the only one decision variable is

Answer 1

[Thiago Henrique Gomes Lobato]

First you don't need the variable w, your problem is to maximize corr(Q,S), which is a unconstrained problem in x. Second, by your problem definition, Q is a single value, and correlation between single values doesn't make much sense, so I believe what you mean by corr is that the values should be equal or Q is a vector and you forgot a summation index. Considering the second hypothesis an example of how to do it could be seen here:

% Random data
S = 1:100;
c = randn(100,4);1
% Function to minimize
f = @(x,S,c)-min(min((corrcoef(S', (x*c')/sum(x)    )) )); % Negative so min = max. Two min's since corrcoef gives a matrix as output
%  Minimizer
[x,fval] = fminsearch(@(x)f(x,S,c),[1,1,1,1])

You can use this code as basis to implement your problem.

[Muhammad Usama]

Q and SA are vectors

[Muhammad Usama]

why we are not using corr?

as corr(Vector A, Vector B) would be a single value

[Thiago Henrique Gomes Lobato]

You can use corr, it will give the same answer, I just forgot about it. You would just need to change the line to

f = @(x,S,c)-corr(S', ((x*c')/sum(x))'    );

[Muhammad Usama]

Actully in f

there was

sum(x*c')/sum(x) 
%instead of
(x*c')/sum(x) 

however when I write sum with numerator it not woks

Also, c would be rectangular matrix, so

sum(x.*c(:,i))/sum(x)

[Thiago Henrique Gomes Lobato]

I replaced sum with a matrix operation. If you want to have the sum it would be:

sum(x.*c,2)'/sum(x)

Which is the same

[Muhammad Usama]

how can we restrict 'x' value to be non-negtive in this ?

[Thiago Henrique Gomes Lobato]

sum(abs(x).*c,2)'/sum(abs(x))

[Muhammad Usama]

But it is giving -ve values of x as well

I want to constraint x >= 0

[Thiago Henrique Gomes Lobato]

Suppose the optimizer finds a value of -2 for a x entry. When it goes to the function the -2 becomes +2 because of the absolute. Which means that if you do

[x,fval] = fminsearch(@(x)f(x,S,c),[1,1,1,1])
x        = abs(x);

You will still get the same cost function, i.e., the abs(x) is exact the same minimum.

[Muhammad Usama]

Thank you Sir.

one more thing, I want to add sum(x) ==1

is it possible?

[Thiago Henrique Gomes Lobato]

You have this already. Note that you divide by sum(x). This is a linear operation, so it would be the same as:

(x/(sum(x)))*c' 

which would give: sum(x) ==1. In other words, if you normalize x by it's sum you'll get the same result as the not normalized one.