For loop not terminating when condition is met

I have a for loop that counts up by 1 to a number. This number on each iteration is plugged into a matrix and a column vector of answers is outputted:

% constants:
R1 = 3000 ;
R2 = 8000 ;
R4 = 2000 ;
R5 = 4000 ;
% R3 = 2130 ; <--- this should be the answer for "r" below
v  = 110 ; 

for r = 2000:3000
R3 = r;

A = [0 -R2 0 -R4 0 0 % the "A" matrix
    R1 -R2 R3 0 0 0
    0 0 -R3 -R4 R5 0
    -1 -1 0 0 0 1
    0 1 1 -1 0 0
    1 0 -1  0 -1 0
    0 0 0 1 1 -1];

b = [-v; 0; 0; 0; 0; 0; 0];
x = A\b ; % the answer column vector

if x(4) == 0.0170
    answer = r
    break
end
end

In this code, "x" is a column vector. I want to terminate the for-loop when the "r" index causes the fourth row in x to be equal to 0.0170.

I know that the index value should be 2130, (in other words, I already know the answer) but the code doesn't "break" when this condition is met. Instead,

the for-loop keeps counting to whichever bound I told it to end at (in this case, 3000).

I'm sure I'm missing something simple here but I can't find what it is. Any help would be greatly appreciated!

Answer 1

[Rik]

Since Matlab stores values in a non-decimal representation, there is a lot of potential for round. In general you should be using a tolerance if you want to test equality for a non-integer data type. The required tolerance is an not really arbitrary choice, but depends on the context. If you expect an exact solution you should use something like 2*eps, while if you expect a non-exact solution you should use something that makes sense for you.

To get the exact answer you expected (2130), the precision needs to be a bit of an odd value. Also, a while loop makes more sense if you don't know how many iterations you need.

% constants:
R1 = 3000 ;
R2 = 8000 ;
R4 = 2000 ;
R5 = 4000 ;
% R3 = 2130 ; <--- this should be the answer for "r" below
v  = 110 ;
solution_list=NaN(1,1000);%review the result for debugging
for r = 2000:3000
    R3 = r;
    A = [0 -R2 0 -R4 0 0 % the "A" matrix
        R1 -R2 R3 0 0 0
        0 0 -R3 -R4 R5 0
        -1 -1 0 0 0 1
        0 1 1 -1 0 0
        1 0 -1  0 -1 0
        0 0 0 1 1 -1];
    b = [-v; 0; 0; 0; 0; 0; 0];
    x = A\b ; % the answer column vector
    solution_list(r-1999)=x(4);
    if abs(x(4)-0.0170)<3.68*10^-6
        answer = r
        break
    end
end

[Ryan Fedeli]

Exactly what I was looking for, thank you