How to deal with rounding error in this example?

I want to add 0.2 to -2000 until I reach 2000. Apparently, some rounding errors happen in the code below. What is the solution to avoid that and the result will be exact, so -99.6?

A = (-2000: 0.2 : 2000)
A(9503)

>> -99.599999999999909

Answer 1

[Alan Stevens]

How about

i = 1:20001;
A = -2000 + (i-1)*0.2;

[Kamia Kavi]

It does not work. For A(9503), it outputs -99.599999999999909 when the output should be -99.6

[Alan Stevens]

You've ht one of the problems of dealing with floating point numbers I'm afraid!

[Star Strider]

@Kamia Kavi — See: Floating-Point Numbers

[Walter Roberson]

The only way to get the exact same bit representation is the literal -99.6 is to perform the operation in integers until the last moment -- though scaling the addend before doing addition and then scaling back gets close .

Remember, though, that 1/10th is simply not exactly representable in any finite-length binary floating point system. There is no integers N, M such that N / (2^M) is 1/10 exactly . In order for N / (2^M) == 1/10 exactly then (10*N)/(2^M) would have to equal 1, and you would have the task of finding an integer multiple of 10 that is an exact power of 2. Which is not going to happen since 10 = 2*5 so you would need to find an integer multiple of 5 that is an exact power of 2.

i = 1:20001;
A = -2000 + (i-1)*0.2;
A9503 = A(9503);
B9503 = -2000 + 9502*0.2

B9503 = -99.6000

C9503 = -2000 + 9502/5

C9503 = -99.6000

D9503 = (-2000*5 + 9502)/5

D9503 = -99.6000

E9503 = (-2000/.2 + 9502) * 0.2

E9503 = -99.6000

fprintf('A(9503) = %.999g\n', A9503);

A(9503) = -99.59999999999990905052982270717620849609375

fprintf('B(9503) = %.999g\n', B9503);

B(9503) = -99.59999999999990905052982270717620849609375

fprintf('C(9503) = %.999g\n', C9503);

C(9503) = -99.59999999999990905052982270717620849609375

fprintf('D(9503) = %.999g\n', D9503);

D(9503) = -99.599999999999994315658113919198513031005859375

fprintf('E(9503) = %.999g\n', E9503);

E(9503) = -99.6000000000000085265128291212022304534912109375

fprintf('-99.6   = %.999g\n', -99.6)

-99.6   = -99.599999999999994315658113919198513031005859375

differenceA = -99.6 - A9503

differenceA = -8.5265e-14

differenceB = -99.6 - B9503

differenceB = -8.5265e-14

differenceC = -99.6 - C9503

differenceC = -8.5265e-14

differenceD = -99.6 - D9503

differenceD = 0

differenceE = -99.6 - E9503

differenceE = 1.4211e-14

[differenceA, differenceB, differenceC, differenceD, differenceE] / eps(-99.6)

ans = 1×5

-6 -6 -6 0 1