Precision of Floating Point

Question

So, I know a little bit about how floating point are represented, but not enough to be sure of my answer.

The general question: for a given precision (for my purposes, the number of accurate decimal places in base 10), what range of numbers can be represented for 16-, 32-, and 64-bit IEEE-754 numbers?

Specifically, I'm only interested in the range of 16-bit and 32-bit numbers accurate to +/-0.5 (the ones place) or +/- 0.0005 (the thousandths place).

Answer 1

If I understand your question correctly, it depends on your language.
For C#, check out the MSDN ref. Float has a 7 digit precision and double 15-16 digit precision.

Answer 2

First off, neither IEEE-754-2008 nor -1985 have 16-bit floats; but it is a proposed addition with a 5-bit exponent and 10-bit fraction. IEE-754 uses a dedicated sign bit, so the positive and negative range is the same. Also, the fraction has an implied 1 in front, so you get an extra bit.

If you want accuracy to the ones place, as in you can represent each integer, the answer is fairly simple: The exponent shifts the decimal point to the right-end of the fraction. So, a 10-bit fraction gets you ±2¹¹.

If you want one bit after the decimal point, you give up one bit before it, so you have ±2¹⁰.

Single-precision has a 23-bit fraction, so you'd have ±2²⁴ integers.

How many bits of precision you need after the decimal point depends entirely on the calculations you're doing, and how many you're doing.

• 2¹⁰ = 1,024
• 2¹¹ = 2,048
• 2²³ = 8,388,608
• 2²⁴ = 16,777,216
• 2⁵³ = 9,007,199,254,740,992 (double-precision)
• 2¹¹³ = 10,384,593,717,069,655,257,060,992,658,440,192 (quad-precision)

See also

Wikipedia: Double-precision
Wikipedia: Half-precision

Answer 3

The precision quoted form Peter R's link to the MSDN ref is probably a good rule of thumb, but of course reality is more complicated.

The fact that the "point" in "floating point" is a binary point and not decimal point has a way of defeating our intuitions. The classic example is 0.1, which needs a precision of only one digit in decimal but isn't representable exactly in binary at all.

If you have a weekend to kill, have a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic. You'll probably be particularly interested in the sections on Precision and Binary to Decimal Conversion.

Answer 4

I'm drawing this answer from the MATLAB documentation for the function EPS, but it should apply universally to IEEE-754 floating point numbers.

For a given floating point number X, if

2^E <= abs(X) < 2^(E+1)

then the distance from X to the next largest representable floating point number (epsilon) is:

epsilon = 2^(E-52)    % For a 64-bit float (double precision)
epsilon = 2^(E-23)    % For a 32-bit float (single precision)
epsilon = 2^(E-10)    % For a 16-bit float (half precision)

The above equations allow us to compute the following:

• For half precision...

  If you want an accuracy of +/-0.5 (or 2^-1), the maximum size that the number can be is 2^10. Any larger than this and the distance between floating point numbers is greater than 0.5.

  If you want an accuracy of +/-0.0005 (about 2^-11), the maximum size that the number can be is 2. Any larger than this and the distance between floating point numbers is greater than 0.0005.

• For single precision...

  If you want an accuracy of +/-0.5 (or 2^-1), the maximum size that the number can be is 2^23. Any larger than this and the distance between floating point numbers is greater than 0.5.

  If you want an accuracy of +/-0.0005 (about 2^-11), the maximum size that the number can be is 2^13. Any larger than this and the distance between floating point numbers is greater than 0.0005.

• For double precision...

  If you want an accuracy of +/-0.5 (or 2^-1), the maximum size that the number can be is 2^52. Any larger than this and the distance between floating point numbers is greater than 0.5.

  If you want an accuracy of +/-0.0005 (about 2^-11), the maximum size that the number can be is 2^42. Any larger than this and the distance between floating point numbers is greater than 0.0005.

Answer 5

It took me quite a while to figure out that when using doubles in Java, I wasn't losing significant precision in calculations. floating point actually has a very good ability to represent numbers to quite reasonable precision. The precision I was losing was immediately upon converting decimal numbers typed by users to the binary floating point representation that is natively supported. I've recently started converting all my numbers to BigDecimal. BigDecimal is much more work to deal with in the code than floats or doubles, since it's not one of the primitive types. But on the other hand, I'll be able to exactly represent the numbers that users type in.

Answer 6

For floating-point integers (I'll give my answer in terms of IEEE double-precision), every integer between 1 and 2^53 is exactly representable. Beyond 2^53, integers that are exactly representable are spaced apart by increasing powers of two. For example:

• Every 2nd integer between 2^53 + 2 and 2^54 can be represented exactly.
• Every 4th integer between 2^54 + 4 and 2^55 can be represented exactly.
• Every 8th integer between 2^55 + 8 and 2^56 can be represented exactly.
• Every 16th integer between 2^56 + 16 and 2^57 can be represented exactly.
• Every 32nd integer between 2^57 + 32 and 2^58 can be represented exactly.
• Every 64th integer between 2^58 + 64 and 2^59 can be represented exactly.
• Every 128th integer between 2^59 + 128 and 2^60 can be represented exactly.
• Every 256th integer between 2^60 + 256 and 2^61 can be represented exactly.
• Every 512th integer between 2^61 + 512 and 2^62 can be represented exactly. . . .

Integers that are not exactly representable are rounded to the nearest representable integer, so the worst case rounding is 1/2 the spacing between representable integers.

So, for example, integers between 1 and 2^53 are accurate to within 1/2 unit in the ones place, and integers between 2^61 + 512 and 2^62 are accurate to within 1/2 unit in the thousands place.