TAOCP 4.2.1 Exercise 8

Section 4.2.1: Single-Precision Calculations

Exercise 8. [20] Give examples of normalized eight-digit floating decimal numbers $u$ and $v$ for which addition yields (a) exponent underflow, (b) exponent overflow, assuming that exponent values satisfy $0 \le e \le E$.

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Solution

Let the floating decimal numbers have $p=8$ digits in the fraction, and let normalized numbers satisfy

$\frac1{10}\le |f|<1,$

unless the number is zero. Exponents are restricted to

$0\le e\le E.$

By Algorithm A, addition first forms the exact sum and then invokes Algorithm N. Exponent underflow occurs in step N6 when normalization requires $e<0$. Exponent overflow occurs in step N6 when normalization requires $e>E$.

(a) Example yielding exponent underflow

Take

v=(0,;-0.99999999\times10^{-1}).$$ Both numbers are normalized eight-digit floating decimal numbers, since their fractions have eight digits and leading digit nonzero. Their sum is $$u+v =0.10000000\times10^{0} -0.099999999\times10^{0} =0.000000001\times10^{0}.

The raw result therefore has

$e=0,\qquad f=0.000000001.$

To normalize, Algorithm N repeatedly executes step N3. The fraction must be shifted left eight places:

$$= 0.10000000\times10^{-8}.$$

Hence the normalized result requires exponent

$e=-8.$

Since $-8<0$, step N6 detects exponent underflow.

Thus

u=(0,0.10000000),\quad v=(0,-0.099999999) }

give exponent underflow.

(b) Example yielding exponent overflow

Take

$$ v=(E,;0.99999999). $$

Both numbers are normalized.

Since the exponents are equal, Algorithm A adds the fractions:

$$ =1.99999998. $$

This exceeds $1$, so step N4 scales right once:

$$= 0.199999998\times10^{E+1}.$$

The normalized result therefore requires exponent

$e=E+1.$

Since $E+1>E$, step N6 detects exponent overflow.

Thus

u=(E,0.99999999),\quad v=(E,0.99999999) }

give exponent overflow. ∎