MATLAB: Array elements not equal despite only difference being their respective end points

Question

I have two time arrays t1400 and t1500 defined as follows: The only difference between these two arrays is that t1500 has 100000 more cycles than t1400.

t1400=0:1e-3:1400;
t1500=0:1e-3:1500;

I would have assumed that the first 1400001 elements of both arrays would be absolutely equal but alas this is not the case as can be seen from the output of the following code:

equal=ones(length(t1400),1)*36;
for n=1:length(t1400)
equal(n)=(t1400(n)==t1500(n));
end
for n=1:length(equal)
if equal(n)==0
break;
end
end
n
n =
        700003

The 700003rd being the first of many positions of the two arrays that are ever so slightly unequal and it's very slight:

>> format long
>> t1400(n)
ans =
       7.000020000000000e+02
>> t1500(n)
ans =
       7.000020000000001e+02

Unfortunately the system I am modelling is incredibly sensitive and by changing the end time of the time array can lead to very different results. I have accepted this as the case but I was just wondering whether someone could explain to me as to why Matlab gives different values depending upon the end time? Thanks in advance 🙂

Answer 1

I received an answer from Tech Support which clarifies this issue:

This is a very interesting question. This is an expected behaviour, to understand why this happens we need to understand how the colon operator works.

v = a:d:b is not constructed using repeated addition (as one would think) as that might accumulate round-off errors.

In fact, this kind of round-off error is inherent to software or hardware systems that perform calculations using floating point numbers. Since floating point numbers have to be represented within a limited number of bits, the result of some calculations is rounded off in order to fit the result into the finite representation (sign, exponent, mantissa) for floating-point numbers.

An example of a number that cannot be represented exactly in binary floating point is the number 0.1, which, while finite in decimal representation, would require an infinite number of bits to be exactly represented in floating-point notation. For instance, executing the following in MATLAB:

0.1+0.1+0.1 == 0.3

will not return true due to such round-off errors.

To counteract such error accumulation, the algorithm of the COLON operator dictates that:

1. The first half of the output vector (in this example ‘v’) is calculated by adding integer multiples of the step ‘d’ to the left-hand endpoint ‘a’. 2. The second half is calculated by subtracting multiples of the step ‘d’ from the right-hand endpoint.

The COLON operator ensures symmetry of the output vector about its midpoint and also that the round off error is minimized at both endpoints. For example, even though 0.01 is not exactly represented in binary,

v = -1 : 0.01 : 1

consists of 201 floating points numbers centered symmetrically about zero.

For more detailed information on how the COLON operator works, refer to the attached function (COLONOP).

function v = colonop(a,d,b)
% COLONOP  Demonstrate how the built-in a:d:b is constructed.
%




%   v = colonop(a,b) constructs v = a:1:b.
%   v = colonop(a,d,b) constructs v = a:d:b.
%
%   v = a:d:b is not constructed using repeated addition.  If the
%   textual representation of d in the source code cannot be
%   exactly represented in binary floating point, then repeated
%   addition will appear to have accumlated roundoff error.  In
%   some cases, d may be so small that the floating point number
%   nearest a+d is actually a.  Here are two imporant examples.
%
%   v = 1-eps : eps/4 : 1+eps is the nine floating point numbers
%   closest to v = 1 + (-4:1:4)*eps/4.  Since the spacing of the
%   floating point numbers between 1-eps and 1 is eps/2 and the
%   spacing between 1 and 1+eps is eps,
%   v = [1-eps 1-eps 1-eps/2 1 1 1 1 1+eps 1+eps].
%
%   Even though 0.01 is not exactly represented in binary,
%   v = -1 : 0.01 : 1 consists of 201 floating points numbers
%   centered symmetrically about zero.
%
%   Ideally, in exact arithmetic, for b > a and d > 0,
%   v = a:d:b should be the vector of length n+1 generated by
%   v = a + (0:n)*d where n = floor((b-a)/d).
%   In floating point arithmetic, the delicate computatations
%   are the value of n, the value of the right hand end point,
%   c = a+n*d, and symmetry about the mid-point.
if nargin < 3
   b = d;
   d = 1;
end
tol = 2.0*eps*max(abs(a),abs(b));
sig = sign(d);
% Exceptional cases.
if ~isfinite(a) || ~isfinite(d) || ~isfinite(b)
  v = NaN;
  return
elseif d == 0 || a < b && d < 0 || b < a && d > 0
  % Result is empty.
  v = zeros(1,0);
  return
end
% n = number of intervals = length(v) - 1.
if a == floor(a) && d == 1
  % Consecutive integers.
  n = floor(b) - a;
elseif a == floor(a) && d == floor(d)
  % Integers with spacing > 1.
  q = floor(a/d);
  r = a - q*d;
  n = floor((b-r)/d) - q;
else
  % General case.
  n =  round((b-a)/d);
  if sig*(a+n*d - b) > tol
     n = n - 1;
  end
end
% c = right hand end point.
c = a + n*d;
if sig*(c-b) > -tol
  c = b;
end
% v should be symmetric about the mid-point.
v = zeros(1,n+1);
k = 0:floor(n/2);
v(1+k) = a + k*d;
v(n+1-k) = c - k*d;
if mod(n,2) == 0
  v(n/2+1) = (a+c)/2;
end