IEEE floating point representation

Q: IEEE single precision:

(i) How is 2. represented ??

0 10000000 00000000000000000000000

(ii) What is the next biggest IEEE single precision number larger than 2 ??

0 10000000 00000000000000000000001

(iii) What is the gap between 2 and the rst IEEE single precision number larger than 2?

General Result: Let x = m×2E be a normalized single precision number, with 1 ≤ m < 2. The gap between x and the next largest single precision number is

× 2E .

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Rounding

We use \IEEE Computer Floating Point Numbers" (IEEE CFPNs, or just Floating Point Numbers) to include ±0, subnormal & normalized CFPNs, & ±∞ in a given format, e.g. single. These form a nite set.

If Nmin & Nmax are the minimum & maximum positive normalized such numbers, we say the

binary representation x = m × 2E

of a general real number is \normalized" if

Nmin ≤ |x| ≤ Nmax, and 1 ≤ m < 2.

Q: For any number x which is not a floating point number, what are two obvious choices for the floating point approximation to x ??

x− closest floating point number less than x,

x+ the closest greater than x.

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Rounding, ctd.

Signi cand b0.b1b2, exponent E {−1, 0, 1}:

. . . . . .0 1 x2 3

Rounding in the Toy System.

e.g. if x = 1.7, then x− = 1.5 and x+ = 1.75.

Using IEEE single precision, if

x = (b0.b1b2 . . . b23b24b25 . . .)2 × 2E ,

is positive and normalized, then

x− = (b0.b1b2 . . . b23)2 × 2E ,

is obtained by discarding b24, b25, etc.

An algorithm for x+ is more complicated since it may involve some bit \carries". If x is negative, the situation is reversed: x+ is obtained by dropping bits b24, b25, etc.

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Correctly Rounded Arithmetic

The IEEE standard de nes the

correctly rounded value of x, round(x) . If x is a floating point number, round(x) = x. Otherwise round(x) depends on the rounding mode in e ect:

•Round towards zero:

round(x) is either x− or x+, whichever is between zero and x.

•Round to nearest: round(x) is either x− or x+, whichever is nearer to x. In the case of a tie, the one with its least signi cant bit equal to zero is chosen.

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Correctly Rounded Arithmetic, ctd.

If x is positive, then x− is between zero and x, so round down and round towards zero have the same e ect. Round towards zero simply requires truncating the binary expansion, i.e. discarding bits.

The most useful mode is round to nearest.

Signi cand b0.b1b2, exponent E {−1, 0, 1}:

. . . . . .0 1 x2 3

Rounding in the Toy System

e.g. if x = 1.7, then x− = 1.5 and x+ = 1.75, so with x = 1.7, this gives a \rounded" value of x equal to 1.75

\Round" with no quali cation means \round to nearest".

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Absolute Rounding Error

|round(x)−x| is called the absolute rounding error associated with x, (value depends on mode). For all modes the absolute rounding error is less than the gap between x− and x+.

For general normalized x > 0,

x = (b0.b1b2 . . . b23b24b25 . . .)2 × 2E , b0 = 1.

IEEE single x− = (b0.b1b2 . . . b23)2 × 2E .

So for any mode |round(x) − x| < 2−23 × 2E .

In general for any rounding mode:

Q: (i) For round towards zero, could the absolute rounding error equal × 2E ??

(ii) Does (1) hold if x is subnormal, i.e. E = −126 and b0 = 0 ??

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Relative Rounding Error

The relative rounding error is de ned to be

Since |round(x)−x| < ×2E , & for normalized numbers

Note for any normalized x

round(x) = x(1 + δ), |δ| < .

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An Important Idea

From the de nition of δ we see

round(x) = x(1 + δ),

so the rounded value of an arbitrary normalized number x is equal to x(1 + δ), where, regardless of the rounding mode,

|δ| < .

This is very important, because you can think of the stored value of x as not exact, but as exact within a factor of 1 + . IEEE single precision numbers are good to a factor of about 1 + 10−7, which means that they have about 7 accurate decimal digits.

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Special Case of Round to Nearest

For round to nearest, the absolute rounding error can be no more than half the gap between x− and x+. This means in IEEE single, for all |x| ≤ Nmax:

|round(x) − x| ≤ 2−24 × 2E ,

and in general

|round(x) − x| ≤ 1 × 2E . 2

The previous analysis for round to nearest then gives for normalized x:

round(x) = x(1 + δ),

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