how to use the result of current iteration as input for next iteration ?

hello everyone!

this is my code in matlab, i want to put the result of x_new for the first iteration after doing optimization as input for the next iteration,

for example i have 10 iterations , at first i assign x_new at 0, and after each iteration it will be updated by the new value.

Do you have a suggestion?

x_new=0,           %initialize x_new
for l=1: 10     %number of iteration
    for i=1:m
        for j=1:n
            
            R_lb(i,j)=BP(i,j)*(log2(1+(Puissance(i,j)*(g0/n0(i,j)))/(sqrt(x_new(j)-Pos_c(i,j))^2+H^2))-(Puissance(i,j)*log2(e)*((sqrt(x(j)-Pos_c(i,j))^2)-(sqrt(x_new(j)-Pos_c(i,j))^2))*(g0/n0(i,j)))/((sqrt(x_new(j)-Pos_c(i,j))^2+H^2)*((sqrt(x_new(j)-Pos_c(i,j))^2+H^2)+(Puissance(i,j)*(g0/n0(i,j))))));
            
        end 
    end 
end

Answer 1

[Walter Roberson]

for l=1: 10     %number of iteration
    for i=1:m
        xnew = zeros(1,n+1);
        for j=1:n
            R_lb(i,j)=BP(i,j)*(log2(1+(Puissance(i,j)*(g0/n0(i,j)))/(sqrt(x_new(j)-Pos_c(i,j))^2+H^2))-(Puissance(i,j)*log2(e)*((sqrt(x(j)-Pos_c(i,j))^2)-(sqrt(x_new(j)-Pos_c(i,j))^2))*(g0/n0(i,j)))/((sqrt(x_new(j)-Pos_c(i,j))^2+H^2)*((sqrt(x_new(j)-Pos_c(i,j))^2+H^2)+(Puissance(i,j)*(g0/n0(i,j))))));
            xnew(j+1) = SomethingAppropriate
        end 
    end 
end

[Safia]

@Walter Roberson thank you for your reply! but how to deal with this error "Index exceeds the number of array elements (1)."

[Walter Roberson]

You need to figure out which array is being referred to, by examining all of the mentioned variables and examining the index at the time of the error. You did not post full code so we cannot test this on your behalf.

I notice that you refer to x(j) . Is there a relationship between x and xnew ?

[Safia]

@Walter Roberson xnew is x updated, x is depending only on(n) but x_new is the result updated after each iteration,it is depending on n and l

[Walter Roberson]

If xnew is the updated x, then you do not need to index it in your calculations: any time you needed to refer to indexed xnew you would just refer to indexed x instead.

An example flow would be

x(1) = some_value
for j = 1 : n
    xnew = some calculation that uses x(j) to create a new x
    some_output(j+1) = some calculation involving xnew and maybe x(j)
    x(j+1) = xnew
end

[Safia]

@Walter Roberson it seems that you didn't understand my problem, so i will try to explain by this equation attached , it is the same as above but with different notation (just focus on "q" which is in my case "x").

if you notice that i have q(n) and qj(n), the different between them is one is depending on calculation that i did before (i already calculated q(n)), but qj(n) is the new depending on the result of each iteration, because the output after doing calculation will give new q which will be used as input for the next iteration, but q(n) will not change.

at first i should initialize the iteration number(j) and give the value for qj(n)

second, i will do my calculation and a new q*j(n) will be given

third, i will replace the qj+1(n) with q*j(n) , it means i will use the new result of current iteration denoted by q*j(n) for the next iteration (j+1)

and each time i will update.

[Walter Roberson]

We cannot help you debug code that we cannot see.