Loss is still larger than 30?

Loss is still larger than 30?

i need these columns a(:,2:2:24) to be close to these ones a(:,1:2:24) and a(:,2:2:24) should be smaller than a(:,1:2:24)

Loss = zeros(27, 12);  % Adjust the size according to the number of pairs

for ii = 1:27
    colIndex = 1;  % Initialize column index for Loss matrix
    for i = 1:2:24
        % Calculate initial loss
        Loss(ii, colIndex) = (a(ii, i) - a(ii, i+1)) * 100 / a(ii, i);

        % Check condition and adjust if necessary
        if Loss(ii, colIndex) < 0 && a(ii, i) < a(ii, i+1)
            a(ii, i) = a(ii, i+1) + 0.00004;
        elseif Loss(ii, colIndex) > 30
            a(ii, i+1) = a(ii, i) - 0.00004;
        end

        % Recalculate loss after adjustment
        Loss(ii, colIndex) = (a(ii, i) - a(ii, i+1)) * 100 / a(ii, i);
        
        colIndex = colIndex + 1;  % Increment column index for Loss matrix
    end
end

% Display the updated matrix 'a'
disp(a);

% Display the Loss matrix
disp(Loss);

[John D'Errico]

How can we possibly help you, if you do not supply the matrix a? Even then, if we are given no clue what this code is supposed to do, it is just random code. Since it does not do what you seem to expect it to do, then it might not even be correct code.

I've tried the tea reading toolbox, to see if it would tell me what a is, but it was of no value. Admittedly, I may have used the wrong kind of tea. And since I dropped my crystal ball, it now has a large crack in it. The new one is on order, but you know how mail order works these days.

[arian hoseini]

sorry i forgot to add my data

Answer 1

[Steven Lord]

Let's take a look at a few lines of your code. I've commented them out since I want to also run some code and these code snippets would not execute.

%{
        elseif Loss(ii, colIndex) > 30
            a(ii, i+1) = a(ii, i) - 0.00004;
        end
%}

So if this else block executes we know that a(ii, i+1) - a(ii, i) is effectively -0.00004. So what does that mean for your recalculation?

%{    
        % Recalculate loss after adjustment
        Loss(ii, colIndex) = (a(ii, i) - a(ii, i+1)) * 100 / a(ii, i);
%}

a(ii, i) - a(ii, i+1) is effectively 0.00004 (from above, multiplying both sides of "a(ii, i+1) - a(ii, i) = -0.00004" by -1.) That times 100 is 0.004. Can 0.004/a(ii, i) be greater than 30? Let's take one value greater than 30 and see if we can find a value for the value that makes 0.004/x equal to it.

syms x
digits(5)
vpasolve(0.004/x == 31)

ans = 

0.00012903

Is there something about your code that rules out a(ii, i) being equal to 0.0012903? [Of course, a simpler way to compute that would be]

0.004/31

ans = 1.2903e-04

Can other values of x satisfy 0.004/x > 30?

fplot(@(x) 0.004./x, [1e-5 1e-3])
yline(30, ':')

Any x where the solid line is above the dotted line would make that correction greater than 30.

x = 0.004/100 % makes the correction 100

x = 4.0000e-05

figure
fplot(@(x) 0.004./x, [1e-5 1e-3])
yline(30, ':')
line(x, 100, Marker='o')

That marker looks pretty squarely on the blue curve to my eye.

Depending on what you're trying to do you should consider using a while statement (to keep adjusting the elements of the a vector while the condition is not satisfied instead of if the condition is not satisfied.)

[arian hoseini]

i dont understand what u tryring to say ...i need these columns a(:,2:2:24) to be close to these ones a(:,1:2:24) and a(:,2:2:24) should be smaller than a(:,1:2:24)

[Steven Lord]

What exactly do you want to happen if even after correcting the values in the a vector once, the value of Loss computed using those corrected values is still greater than 30? Right now your code continues on with the Loss value being greater than 30; you need to decide how you want to handle that scenario.

• Do you want to set Loss to 30 even if it would be greater for those particular values in a? For this you'd want to use the min function.
• Do you want to perform an additional correction to the values in a? If so, describe that additional correction in detail.
• Do you want the code to throw an error? If so use error or assert
• Or do you want it to do something else (and if so, what? Be specific.)?

[arian hoseini]

I want to set loss to under 30 and after that i need to change those that has upper 30 loss and i need these columns a(:,2:2:24) to be close to these ones a(:,1:2:24) and a(:,2:2:24) should be smaller than a(:,1:2:24) cause a(:,1:2:24) are the real parameters

[Steven Lord]

Setting the corresponding elements of a to be 1 and 1+eps would satisfy the requirements you stated in your comments. But I'm guessing that isn't going to satisfy the requirements you haven't stated. What are those unstated requirements?

a1 = 1+eps;
a2 = 1;
Loss = (a2-a1)*100/a2

Loss = -2.2204e-14

a1 is close to a2. a2 is smaller than a1.

isSmaller = a2 < a1

isSmaller = logical
   1

The corresponding Loss value is less than 30 (both in value and in absolute value.)

isLessThan30 = (Loss < 30) & (abs(Loss) < 30)

isLessThan30 = logical
   1

Are they an acceptable solution to your problem? Why or why not?