Notation of floating-point number system

A floating-point number system ${\displaystyle F}$ is characterized by ${\displaystyle 4}$ integers:

• ${\displaystyle \beta }$: base or radix
• ${\displaystyle p}$: precision
• ${\displaystyle [L,U]}$: exponent range, where ${\displaystyle L}$ is the lower bound and ${\displaystyle U}$ is the upper bound

Any ${\displaystyle x\in F}$ has the following form:

$${\displaystyle x=\pm (\underbrace {d_{0}.d_{1}d_{2}\ldots d_{p-1}} _{\text{significand}})_{\beta }\times \beta ^{\overbrace {E} ^{\text{exponent}}}=\pm d_{0}\times \beta ^{E}+d_{1}\times \beta ^{E-1}+\ldots +d_{p-1}\times \beta ^{E-(p-1)}}$$

where ${\displaystyle d_{i}}$ is an integer such that ${\displaystyle 0\leq d_{i}\leq \beta -1}$ for ${\displaystyle i=0,1,\ldots ,p-1}$, and ${\displaystyle E}$ is an integer such that ${\displaystyle L\leq E\leq U}$.

Normalized floating-number system

• A floating-point number system is normalized if the leading digit ${\displaystyle d_{0}}$ is always nonzero unless the number is zero.^[3] Since the significand is ${\displaystyle d_{0}.d_{1}d_{2}\ldots d_{p-1}}$, the significand of a nonzero number in a normalized system satisfies ${\displaystyle 1\leq {\text{significand}}<\beta }$. Thus, the normalized form of a nonzero IEEE floating-point number is ${\displaystyle \pm 1.bb\ldots b\times 2^{E}}$ where ${\displaystyle b\in {0,1}}$. In binary, the leading digit is always ${\displaystyle 1}$ so it is not written out and is called the implicit bit. This gives an extra bit of precision so that the roundoff error caused by representation error is reduced.
• Since floating-point number system ${\displaystyle F}$ is finite and discrete, it cannot represent all real numbers which means infinite real numbers can only be approximated by some finite numbers through rounding rules. The floating-point approximation of a given real number ${\displaystyle x}$ by ${\displaystyle fl(x)}$ can be denoted.
  • The total number of normalized floating-point numbers is
    $${\displaystyle 2(\beta -1)\beta ^{p-1}(U-L+1)+1,}$$

    
    where
    • ${\displaystyle 2}$ counts choice of sign, being positive or negative
    • ${\displaystyle (\beta -1)}$ counts choice of the leading digit
    • ${\displaystyle \beta ^{p-1}}$ counts remaining significand digits
    • ${\displaystyle U-L+1}$ counts choice of exponents
    • ${\displaystyle 1}$ counts the case when the number is ${\displaystyle 0}$.

IEEE standard

In the IEEE standard the base is binary, i.e. ${\displaystyle \beta =2}$, and normalization is used. The IEEE standard stores the sign, exponent, and significand in separate fields of a floating point word, each of which has a fixed width (number of bits). The two most commonly used levels of precision for floating-point numbers are single precision and double precision.

Calculating roundoff error in IEEE standard

Suppose the usage of round-to-nearest and IEEE double precision.

• Example: the decimal number ${\displaystyle (9.4)_{10}=(1001.{\overline {0110}})_{2}}$ can be rearranged into
  $${\displaystyle +1.\underbrace {0010110011001100110011001100110011001100110011001100} _{\text{52 bits}}110\ldots \times 2^{3}}$$

  

Since the 53-rd bit to the right of the binary point is a 1 and is followed by other nonzero bits, the round-to-nearest rule requires rounding up, that is, add 1 bit to the 52-nd bit. Thus, the normalized floating-point representation in IEEE standard of 9.4 is

$${\displaystyle fl(9.4)=1.0010110011001100110011001100110011001100110011001101\times 2^{3}.}$$

• Now the roundoff error can be calculated when representing ${\displaystyle 9.4}$ with ${\displaystyle fl(9.4)}$.

This representation is derived by discarding the infinite tail

$${\displaystyle 0.{\overline {1100}}\times 2^{-52}\times 2^{3}=0.{\overline {0110}}\times 2^{-51}\times 2^{3}=0.4\times 2^{-48}}$$

from the right tail and then added ${\displaystyle 1\times 2^{-52}\times 2^{3}=2^{-49}}$ in the rounding step.

Then ${\displaystyle fl(9.4)=9.4-0.4\times 2^{-48}+2^{-49}=9.4+(0.2)_{10}\times 2^{-49}}$.

Thus, the roundoff error is ${\displaystyle (0.2\times 2^{-49})_{10}}$.

Measuring roundoff error by using machine epsilon

The machine epsilon ${\displaystyle \epsilon _{\text{mach}}}$ can be used to measure the level of roundoff error when using the two rounding rules above. Below are the formulas and corresponding proof.^[3] The first definition of machine epsilon is used here.

Proof

Let ${\displaystyle x=d_{0}.d_{1}d_{2}\ldots d_{p-1}d_{p}\ldots \times \beta ^{n}\in \mathbb {R} }$ where ${\displaystyle n\in [L,U]}$, and let ${\displaystyle fl(x)}$ be the floating-point representation of ${\displaystyle x}$. Since round-by-chop is being used, it is

$${\displaystyle {\begin{aligned}{\frac {|x-fl(x)|}{|x|}}&={\frac {|d_{0}.d_{1}d_{2}\ldots d_{p-1}d_{p}d_{p+1}\ldots \times \beta ^{n}-d_{0}.d_{1}d_{2}\ldots d_{p-1}\times \beta ^{n}|}{|d_{0}.d_{1}d_{2}\ldots \times \beta ^{n}|}}\\&={\frac {|d_{p}.d_{p+1}\ldots \times \beta ^{n-p}|}{|d_{0}.d_{1}d_{2}\ldots \times \beta ^{n}|}}\\&={\frac {|d_{p}.d_{p+1}d_{p+2}\ldots |}{|d_{0}.d_{1}d_{2}\ldots |}}\times \beta ^{-p}\end{aligned}}}$$

In order to determine the maximum of this quantity, the is a need to find the maximum of the numerator and the minimum of the denominator. Since ${\displaystyle d_{0}\neq 0}$ (normalized system), the minimum value of the denominator is ${\displaystyle 1}$. The numerator is bounded above by ${\displaystyle (\beta -1).(\beta -1){\overline {(\beta -1)}}=\beta }$. Thus, ${\displaystyle {\frac {|x-fl(x)|}{|x|}}\leq {\frac {\beta }{1}}\times \beta ^{-p}=\beta ^{1-p}}$. Therefore, ${\displaystyle \epsilon =\beta ^{1-p}}$ for round-by-chop. The proof for round-to-nearest is similar.

• Note that the first definition of machine epsilon is not quite equivalent to the second definition when using the round-to-nearest rule but it is equivalent for round-by-chop.

Addition

Machine addition consists of lining up the decimal points of the two numbers to be added, adding them, and then storing the result again as a floating-point number. The addition itself can be done in higher precision but the result must be rounded back to the specified precision, which may lead to roundoff error.^[3]

• For example, adding ${\displaystyle 1}$ to ${\displaystyle 2^{-53}}$ in IEEE double precision as follows,
  ${\displaystyle {\begin{aligned}1.00\ldots 0\times 2^{0}+1.00\ldots 0\times 2^{-53}&=1.\underbrace {00\ldots 0} _{\text{52 bits}}\times 2^{0}+0.\underbrace {00\ldots 0} _{\text{52 bits}}1\times 2^{0}\\&=1.\underbrace {00\ldots 0} _{\text{52 bits}}1\times 2^{0}.\end{aligned}}}$
  This is saved as ${\displaystyle 1.\underbrace {00\ldots 0} _{\text{52 bits}}\times 2^{0}}$ since round-to-nearest is used in IEEE standard. Therefore, ${\displaystyle 1+2^{-53}}$ is equal to ${\displaystyle 1}$ in IEEE double precision and the roundoff error is ${\displaystyle 2^{-53}}$.

This example shows that roundoff error can be introduced when adding a large number and a small number. The shifting of the decimal points in the significands to make the exponents match causes the loss of some of the less significant digits. The loss of precision may be described as absorption.^[11]

Note that the addition of two floating-point numbers can produce roundoff error when their sum is an order of magnitude greater than that of the larger of the two.

• For example, consider a normalized floating-point number system with base ${\displaystyle 10}$ and precision ${\displaystyle 2}$. Then ${\displaystyle fl(62)=6.2\times 10^{1}}$ and ${\displaystyle fl(41)=4.1\times 10^{1}}$. Note that ${\displaystyle 62+41=103}$ but ${\displaystyle fl(103)=1.0\times 10^{2}}$. There is a roundoff error of ${\displaystyle 103-fl(103)=3}$.

This kind of error can occur alongside an absorption error in a single operation.

Multiplication

In general, the product of two p-digit significands contains up to 2p digits, so the result might not fit in the significand.^[3] Thus roundoff error will be involved in the result.

• For example, consider a normalized floating-point number system with the base ${\displaystyle \beta =10}$ and the significand digits are at most ${\displaystyle 2}$. Then ${\displaystyle fl(77)=7.7\times 10^{1}}$ and ${\displaystyle fl(88)=8.8\times 10^{1}}$. Note that ${\displaystyle 77\times 88=6776}$ but ${\displaystyle fl(6776)=6.7\times 10^{3}}$ since there at most ${\displaystyle 2}$ significand digits. The roundoff error would be ${\displaystyle 6776-fl(6776)=6776-6.7\times 10^{3}=76}$.

Division

In general, the quotient of 2p-digit significands may contain more than p-digits.Thus roundoff error will be involved in the result.

• For example, if the normalized floating-point number system above is still being used, then ${\displaystyle 1/3=0.333\ldots }$ but ${\displaystyle fl(1/3)=fl(0.333\ldots )=3.3\times 10^{-1}}$. So, the tail ${\displaystyle 0.333\ldots -3.3\times 10^{-1}=0.00333\ldots }$ is cut off.

Subtraction

Absorption also applies to subtraction.

• For example, subtracting ${\displaystyle 2^{-60}}$ from ${\displaystyle 1}$ in IEEE double precision as follows,
  $${\displaystyle {\begin{aligned}1.00\ldots 0\times 2^{0}-1.00\ldots 0\times 2^{-60}&=\underbrace {1.00\ldots 0} _{\text{60 bits}}\times 2^{0}-\underbrace {0.00\ldots 01} _{\text{60 bits}}\times 2^{0}\\&=\underbrace {0.11\ldots 1} _{\text{60 bits}}\times 2^{0}.\end{aligned}}}$$

  
  This is saved as ${\displaystyle \underbrace {1.00\ldots 0} _{\text{53 bits}}\times 2^{0}}$ since round-to-nearest is used in IEEE standard. Therefore, ${\displaystyle 1-2^{-60}}$ is equal to ${\displaystyle 1}$ in IEEE double precision and the roundoff error is ${\displaystyle -2^{-60}}$.

The subtracting of two nearly equal numbers is called subtractive cancellation.^[3] When the leading digits are cancelled, the result may be too small to be represented exactly and it will just be represented as ${\displaystyle 0}$.

• For example, let ${\displaystyle |\epsilon |<\epsilon _{\text{mach}}}$ and the second definition of machine epsilon is used here. What is the solution to ${\displaystyle (1+\epsilon )-(1-\epsilon )}$?
  It is known that ${\displaystyle 1+\epsilon }$ and ${\displaystyle 1-\epsilon }$ are nearly equal numbers, and ${\displaystyle (1+\epsilon )-(1-\epsilon )=1+\epsilon -1+\epsilon =2\epsilon }$. However, in the floating-point number system, ${\displaystyle fl((1+\epsilon )-(1-\epsilon ))=fl(1+\epsilon )-fl(1-\epsilon )=1-1=0}$. Although ${\displaystyle 2\epsilon }$ is easily big enough to be represented, both instances of ${\displaystyle \epsilon }$ have been rounded away giving ${\displaystyle 0}$.

Even with a somewhat larger ${\displaystyle \epsilon }$, the result is still significantly unreliable in typical cases. There is not much faith in the accuracy of the value because the most uncertainty in any floating-point number is the digits on the far right.

• For example, ${\displaystyle 1.99999\times 10^{2}-1.99998\times 10^{2}=0.00001\times 10^{2}=1\times 10^{-5}\times 10^{2}=1\times 10^{-3}}$. The result ${\displaystyle 1\times 10^{-3}}$ is clearly representable, but there is not much faith in it.

This is closely related to the phenomenon of catastrophic cancellation, in which the two numbers are known to be approximations.

Unstable algorithms

An algorithm or numerical process is called stable if small changes in the input only produce small changes in the output, and unstable if large changes in the output are produced.^[12] For example, the computation of ${\displaystyle f(x)={\sqrt {1+x}}-1}$ using the "obvious" method is unstable near ${\displaystyle x=0}$ due to the large error introduced in subtracting two similar quantities, whereas the equivalent expression ${\displaystyle \textstyle {f(x)={\frac {x}{{\sqrt {1+x}}+1}}}}$ is stable.^[12]

Ill-conditioned problems

Even if a stable algorithm is used, the solution to a problem may still be inaccurate due to the accumulation of roundoff error when the problem itself is ill-conditioned.

The condition number of a problem is the ratio of the relative change in the solution to the relative change in the input.^[3] A problem is well-conditioned if small relative changes in input result in small relative changes in the solution. Otherwise, the problem is ill-conditioned.^[3] In other words, a problem is ill-conditioned if its condition number is "much larger" than 1.

The condition number is introduced as a measure of the roundoff errors that can result when solving ill-conditioned problems.^[7]