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how to define an iteration number ?

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Safia 2022-11-19

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评论： Safia 2022-11-19

Capture.PNG

Hi all!

i'm trying to implement a program in matlab attached, i'm wondering how to define the number of iteration j?

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Torsten 2022-11-19

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在 MATLAB Online 中打开

The number of iterations is initialized as 0 :

i = 0;

Within the while loop, for each time the while loop is repeated, the number of iterations i is increased by 1:

i = i+1;

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Personally I wouldn't use j or i. I'd use k as the iterator since we like to recommend not using variables that look like the imaginary variable.

Safia 2022-11-19

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Capture.PNG

@Torsten @Image Analyst thank you for reply,if I have variables depending on the number of iterations and others not,how to differentiate between them for example i will calculate this in attached file,

where the number of iteration denoted by (l), i'm asking how to distinguish beween yn(l) and yn.

because both should exist when i calculate this formula.

Torsten 2022-11-19

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编辑：Torsten 2022-11-19

We don't know what all these functions and variables in the picture mean.

If the old iterates are still needed in further calculations, save them all in a 2d-matrix y(n,l).

If the old iterates are no longer needed, save the y(n) in two one-dimensional arrays yold(n) and ynew(n) and override yold by ynew if a new iterate has been calculated: yold(n) = ynew(n).

Safia 2022-11-19

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@Torsten in each iteration just i need the value of yl (n), it means that each time yl(n) will has new value.

my question is how to write a loop while or for in order to assign a number of iteration just because yl(n) will change.

How can i save yl(n) in one dimensional arrays and each time time it will be updated by new value?

i really need a help to solve this

Torsten 2022-11-19

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编辑：Torsten 2022-11-19

在 MATLAB Online 中打开

Ran in:

I wrote it: by using two variables yold and ynew.

The following code implements the recursion y_(n+1) = y_n + 2 with y(0) = 10.

yold = 10;
i = 0;
while i < 20
    i = i + 1;
    ynew = yold + 2;
    yold = ynew;
end
10 + 20*2
ans = 50
yold
yold = 50

I suggest you learn more about how to handle MATLAB:

https://de.mathworks.com/learn/tutorials/matlab-onramp.html

Your mathematical problem is difficult enough by itself - you should try to improve your programming skills to ease this burden.

Safia 2022-11-19

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编辑：Safia 2022-11-19

在 MATLAB Online 中打开

@Torsten i wrote my code may you could understand more, (l) is th number of iteration , i want to fix at 10.

and the first value of x_new i fix it at 0, iget this error

Index in position 1 exceeds array bounds (must not exceed 1).

x_new(1)=0 ;
l=0;
while l=10
    l = l+1;
    for i=1:m
        for j=1:n
            if N(i,j)==1 & BP(i,j)~=0 
                
                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
end

Torsten 2022-11-19

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编辑：Torsten 2022-11-19

在 MATLAB Online 中打开

Your code will immediately throw a syntax error since

while l=10

will not be accepted by MATLAB.

Maybe you mean

while l <= 10

x_new is only defined for its first element x_new(1). So trying to access x_new(j) for 2 <= j <= n will throw an error.

Since it's not at all obvious what you want to do in the while loop, nobody will be able to help without further explanation.

Safia 2022-11-19

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@Torsten i will use a solver to find in each time the optimal point x through some constraints. This solver will extract as a result a new x which is the (x+1) means the next point, so i want to use this new x to the next iteration.

That's why i'm asking how to fix a number of iteration and in each time update x by the new value resulting by the solver in order to get in the end a vector of all points.

Torsten 2022-11-19

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I don't understand the workflow.

Safia 2022-11-19

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@Torsten i will try to make it easier, i want to use an optimizer to find the optimal location of drone to collect data, at first i will initialize the first point in the first iteration, after doing optimization , a new result will be given which is the next location point denoted x_new, this point will be used in the next iteration,

for example , in the first iteration , i initialize x at 0,the number of iteration is 10

iteration=10;
x=0;
for i=1:iteration
    %doing optimization
    %extract x_new
end

in the next iteration the program will be

x=x_new;
for i=1:iteration
    %doing optimization
    %extract x_new
end

until finish the number of iteration.

How can i do this in a simple loop and each time will update the new value in the next iteration.

Torsten 2022-11-19

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And why do you need two iteration loops with 10 iterations instead of one iteration loop with 20 iterations ?

Safia 2022-11-19

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编辑：Safia 2022-11-19

@Torsten no i don't need two iteration loops, just i need the best way to write my code, i wrote above two in order to explain that i will use the result of the first iteration as input for the second one

Torsten 2022-11-19

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I really don't understand the problem. If you save x in each of the iteration loops (maybe by overwriting a previous x), all would be fine, wouldn't it ?

Safia 2022-11-19