1. Computational Methods
CSE 551
Computational Methods
2018/2019 Fall
Chapter 3
Error Analysis and
Computer Arithmetic

2. Outline
Base Changes
Introduction to Error Analysis
Floating-Point Representations

3. References
W. Cheney, D Kincaid, Numerical Mathematics and Computing, 6ed,
Chapter 1
Chapter 2
Appendix B

4. Introduction general number representation - to bases 2, 8, and 16
bases primarily used in computer arithmetic
The familiar decimal notation for numbers uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
a whole number such as 37294
individual digits represent coefficients of powers of 10:
We begin with a discussion of general number representation but move quickly to bases 2, 8, and 16, as they are the bases primarily used in computer arithmetic.
The familiar decimal notation for numbers uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
When we write a whole number such as 37294, the individual digits represent coefficients
of powers of 10 as follows:
37294 =
= 4 × × × × × 104
Thus, in gener

5. in general, a string of digits represents a number according to the formula
anan− a2a1a0 = a0 × a1 × 101 +· · ·+an−1 × 10n−1 + an × 10n
This takes care of only the positive whole numbers.A number between 0 and 1 is represented
by a string of digits to the right of a decimal point. For example, we see that =
7 × 10−1 + 2 × 10 −2 + 1 × 10 −3 + 5 × 10 −4

6. In general, we have the formula
there can be an infinite string of digits to the right of the decimal point; indeed,
there must be an infinite string to represent some numbers. For example, we note that

7. For a real number of the form
the integer part is the first summation
the fractional part is the second summation
a number represented in base β is signified by
enclosing it in parentheses and adding a subscript β

8. β Base Numbers Other bases used, especially in computers e.g.,
the binary system uses 2 as the base
the octal system uses 8
the hexadecimal system uses 16
In the octal representation of a number
digits - 0, 1, 2, 3, 4, 5, 6, 7
e.g.,
(21467)8 = × × × × 84
= 7 + 8(6 + 8(4 + 8(1 + 8(2))))
= 9015
A number between 0 and 1, expressed in octal
represented with combinations of 8−1, 8−2, and so on.
( )8 = 3 × 8−1 + 6 × 8−2 + 2 × 8−3 + 0 × 8−4 + 7 × 8−5
= 8−5(3 × × × )
= 8−5(7 + 82(2 + 8(6 + 8(3))))
= / 32768 =

9. decimal point - base-10 numbers
If we use another base, say, β, then numbers represented in the β-system look like this:
The digits: 0, 1, , β −2, β −1
If β > 10
necessary to introduce symbols for 10, 11, , β − 1
The separator between the integer and fractional part - called the radix point
decimal point - base-10 numbers

10. Conversion of Integer Parts
formalize the process of converting a number from one base to another
consider separately
the integer and fractional parts of a number
a positive integer N with base γ :
to convert this to the number system with base β Write N in its nested form:

11. replace each of the numbers on the right by its representation in base β
Next, carry out the calculations in β-arithmetic. replacement of the ak ’s and γ by equivalent base-β numbers - a table
how each of the numbers 0, 1, , γ −1 appears
in the β-system
a base-β multiplication table may be required.

12. for hand calculations:
decimal number 3781 to
binary form
the decimal binary equivalences
longhand multiplication in base 2,
for hand calculations:
Write down an equation digits c0, c1, , cm :

13. if N is divided by β, then the remainder in this division is c0, and the quotient is
If this number is divided by β, the remainder is c1, and so on
divide repeatedly by β
saving remainders c0, c1, , cm and quotients.

14. Example
Convert the decimal number 3781 to binary form using the division algorithm.
Solution: divide repeatedly by 2, saving the remainders

15. Here, the symbol ˙↓ is used to remind us that the digits ci are obtained beginning with the
digit next to the binary point. Thus, we have
(3781.)10 = ( )2
and not the other way around: ( )2 = (2615)10

16. Example
Convert the number N = ( )2 to decimal form by nested multiplication.
Solution:

17. Another conversion problem exists in going from an integer in base γ to an integer in base β
when using calculations in base γ
the unknown coefficients
determined by a process of successive division
this arithmetic is carried out in the γ –system
At the end, the numbers ck are in base γ
a table of γ -β equivalents

18. e.g., convert a binary integer into decimal form by repeated division by (1 010)2
equals (10)10
carrying out the operations in binary
A table of binary-decimal equivalents
binary division is easy only for computers
develop alternative procedures

19. Conversion of Fractional Parts
convert a fractional number such as (0.372)10 to binary
a direct yet naive approach:

20. Dividing in binary arithmetic is not straightforward
easier ways conversion.
x in the range 0 < x < 1 and that the digits ck in the representation
are to be determined

21. it is necessary to shift the radix point only when multiplying by base β
the unknown digit c1 can be described as the integer part of βx
denoted by I(βx).
The fractional part, (0.c2c3c )β denoted by F(βx)
The process is repeated in the
following pattern:
the arithmetic is carried out in the decimal system.

22. Example
Use the preceding algorithm to convert the decimal number x = (0.372)10 to binary form.

23. repeatedly multiplying by 2 and removing the integer parts:
(0.372)10 = ( )2

24. Base Conversion 10↔8↔2
Most computers - binary system representation of numbers.
The octal system (base 8) useful in converting from the decimal system to the binary system and vice versa
With base 8, the positional values of the numbers
80 = 1, 81 = 8, 82 = 64, 83 = 512, 84 = 4096,...

25. Example

26. converting between decimal and binary form
convenient - octal representation - intermediate step
Conversion between
octal and decimal
octal and binary – simple
starts at the binary point and proceeds in both directions.
( )2 = ( )8
Conversion of an octal number to binary can be done in a similar manner but in reverse order.
( )8 = ( )2

27. Example
What is ( )10 in octal and binary forms?
Solution: convert the decimal number first to octal and then to binary
For the integer partrepeatedly divide by 8:
2576. = (5020.)8 = ( )2

28. For the fractional part - repeatedly multiply by 8
= (0.266)8 = ( )2
the result =
( )2

29. Base 16
hexadecimal system (base 16)
A, B, C, D, E, and F represent 10, 11, 12, 13, 14, and 15, respectively
table of equivalences:

30. Conversion between binary numbers and hexadecimal numbers
regroup the binary digits to groups of four
( )2 = ( )2
= (2BAD)16
and
( )2
= ( )2
= (7AF2.C9E)16

31. More Examples
convert (0.276)8, (0.C8)16, and (492)10 into different number systems

33. Significant Digits
digits beginning with the leftmost nonzero digit and ending with the rightmost correct digit, including final zeros that are exact.

34. Example
solving for the variable y in this linear system of equations in two variables
x y =
x y =
First, carry only three significant digits of precision in the calculations
Second, repeat with four significant digits throughout
Finally, use ten significant digits.

35. Solution
first task - round all numbers in the original problem to three digits
round all the calculations, keeping only three significant digits
take a multiple α of the first equation and subtract it from the second equation to eliminate the x-term in the second equation
The multiplier is α = 0.208/0.104 ≈ 2.00
in the second equation, - new coefficient of the x-term:: − (2.00)(0.104) ≈ − = 0
new y-term coefficient:
0.425 − (2.00)(0.212) ≈ − = 0.001
righthand side:
0.933 − (2.00)(0.738) = − 1.48 = −0.547
y = −0.547/(0.001) ≈ −547.

36. keep four significant digits:
the multiplier: α = / ≈ 2.009
In the second equation - new coefficient of the x-term:
− (2.009)(0.1036) ≈ − = 0
new coefficient of the y-term:
− (2.009)(0.2122) ≈ − = − new right-hand side:
0.9327−(2.009)(0.7381) ≈ −1.483 ≈ −0.5503
y = −0.5503/(− ) ≈ 343.9
shocked to find that the answer has changed
from −547 to 343.9, which is a huge difference!

37. carry ten significant decimal digits
find that:
even is not accurate
obtain:
y =
The lesson learned:
data thought to be accurate should be carried with full precision and not be rounded off prior to each of the calculations

38. In most computers, the arithmetic operations
a double-length accumulator
twice the precision of the stored quantities
may not avoid a loss of accuracy!
Loss of accuracy
roundoff errors
subtracting nearly equal numbers

39. The point of intersection of the two lines - exact solution
Figure - geometric illustration of what can happen in solving two equations in two unknowns
The point of intersection of the two lines - exact solution
dotted lines - degree of uncertainty
from errors in the measurements or roundoff errors.
sharply defined point v.s. small trapezoidal
area containing many possible solutions.
if the two lines are nearly parallel
area of possible solutions can increase dramatically!
well-conditioned and ill-conditioned systems of linear equations

40. In 2D, wellconditione an ill-conditioned linear systems

41. Errors: Absolute and Relative
α, β - two numbers
one is regarded as an approximation to the other
The error of β as an approximation to α: α − β;
the error – the exact value minus the approximate value
The absolute error of β as an approximation to α:
|α −β|
The relative error of β as an approximation to α:
|α −β|/|α|
in absolute error, the roles of α and β are the same,
in computing the relative error,
relative error is undefined in the case α = 0.

42. relative error is usually more meaningful than the absolute error
e.g., α1 = 1.333, β1 = 1.334
α2 = 0.001, β2 = 0.002
absolute error of βi as an approximation to αi :
the same in both cases - 10−3
However, the relative errors:
(3/4) × 10−3 and 1,
respectively
relative error clearly indicates that
β1 is a good approximation to α1
but that β2 is a poor approximation to α2

43. the exact value - the true value
In summar
the exact value - the true value
A useful way to express the absolute error and relative error - to drop the absolute values:
(relative error)(exact value) = exact value − approximate value
approximate value = (exact value)[1 + (relative error)]
relative error - related to the approximate value rather than to the exact value
the true value may not be known

44. Example
Consider x = rounded to x_head = and y = rounded to y_head = 30.16
What are the number of significant digits, absolute errors, and relative errors?
Interpret the results.

45. Case 1. x_head = 0.35 × 10−2 - two significant digits,
Solution
Case 1. x_head = 0.35 × 10−2 - two significant digits,
absolute error: 0.3 × 10−4
relative error × 10−2
Case 2. y_head = × four significant digits
absolute error: 0.2 × 10−2
relative error 0.66 × 10−4.
the relative error is a better indication of the number of significant digits than the absolute error

46. Accuracy and Precision
Accurate to n decimal places
can trust n digits to the right of the decimal place
Accurate to n significant digits
can trust a total of n digits as being meaningful beginning with the leftmost nonzero digit.

47. a ruler graduated in millimeters to measure lengths
The measurements will be accurate to one millimeter, or m
three decimal places written in meters
A measurement such as m would be accurate to three decimal places
A measurement such as m would be meaningless, since the ruler produces only
three decimal places
and it should be m or m. If the measurement m has five dependable digits
then it is accurate to five significant figures.
a measurement such as m has only two significant figures.

48. using a calculator or computer in a laboratory experiment, one may get a false
sense of having higher precision than is warranted by the data
e.g.,
(1.2) + (3.45) = 4.65
only two significant digits of accuracy
because the second digit in 1.2 may be
the effect of rounding 1.24 down or rounding 1.16 up to two significant figures
Then the left-hand side - as large as
(1.249) + (3.454) = (4.703)
or as small as
(1.16) + (3.449) = (4.609)

49. In Addition and Subtraction
In adding and subtracting numbers
the result is accurate only to the smallest number of significant digits
used in any step of the calculation
In the above example, the term 1.2 has two significant digits;
therefore, the final calculation has an uncertainty in the third digit

50. In multiplication and division of numbers
Rule of Thumb
In multiplication and division of numbers
the results may be even more misleading.
e.g., computations on a calculator:
(1.23)(4.5) = 5.535
(1.23)/(4.5) =
there are four and nine significant digits inhe results
but there are really only two
As a rule of thumb
keep as many significant digits in a sequence of calculations as there are in the least accurate number involved in the computations.

51. Rounding: reduces the number of significant digits in a number
Rounding and Shopping
Rounding: reduces the number of significant digits in a number
The result of rounding:
a number similar in magnitude that is a shorter number having fewer nonzero digits
several slightly different rules for rounding:
The round-to-even method
statistician’s rounding or bankers’ rounding
tends to reduce the total rounding error
with (on average) an equal portion of numbers rounding up as well as rounding down.

52. a number x is chopped to n digits or figures
when all digits that follow the nth digit are discarded and none of the remaining n digits are changed
x is rounded to n digits or figures
when x is replaced by an n-digit number that approximates x with minimum error
The question of whether to round up or down an (n+1)-digit decimal number that ends with a 5
always selecting the rounded n-digit number with an even nth digit
seem strange at first
computers - rounding decimal calculations
standard floating-point arithmetic!

53. Examples
e.g., the results of rounding some three-decimal numbers to two digits:
0.217 ≈ 0.22, ≈ 0.36,
0.475 ≈ 0.48, ≈ 0.59
chopping:
0.217 ≈ 0.21, ≈ 0.36,
0.475 ≈ 0.47, ≈ 0.59
On the compute
has the option - arithmetic operations done with
either chopping or
rounding
The latter is usually preferable

54. Floating-Point Representation
normalized scientific notation
shifting the decimal point and
supplying appropriate powers of 10
e.g.,
= × 102
= × 10−2
= × 107
the number is represented by
a fraction multiplied by 10n
and the leading digit in the fraction is not zero
(except when the number is zero)
write as × 105
not × 106 or × 104

55. Normalized Floating-Point Representation
computer science
normalized scientific notation is also called
normalized floating-point representation
In the decimal system, any real number x (other than zero) represented in normalized floating-point form:
x = ±0.d1d2d × 10n
where
d1  0
n is an integer (positive, negative, or zero)
The numbers d1, d2, are the decimal digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

56. normalized floating-point decimal form as
Stated another way, the real number x, if different from zero, can be represented in
normalized floating-point decimal form as
This representation consists of three parts:
a sign that is either + or −
a number r in the interval [1/10, 1)
an integer power of 10.
The number r is called the normalized mantissa
n the exponent

57. The floating-point representation in the binary system is similar to that in the decimal
system in several ways.
If x  0, it can be written as
The mantissa q would be expressed as a sequence of zeros or ones in the form q =(0.b1b2b )2 where b1  0
necessarily q 1/2 .

58. in Computer
A floating-point number system within a computer is similar to what we have just described,
with one important difference:
Every computer has only a finite word length and a finite total capacity
so only numbers with a finite number of digits can be represented
A number is allotted only
one word of storage single-precision mode
(two or more words in double or extended precision)
the degree of precision is strictly limited
irrational numbers cannot be represented
nor can those rational numbers that do not fit the finite format imposed by the computer
Furthermore, numbers may be
either too large or too small
to be representable
The real numbers - representable in a computer
its machine numbers.

59. Finite v.s. NOnterminating Expension
any number used in calculations with a computer system
conform to the format of numbers in that system
must have a finite expansion
Numbers that have a nonterminating expansion - cannot be accommodated precisely.
Moreover, a number that has a terminating expansion in one base may have a nonterminating expansion in another
e.g.,
1/10 = (0.1)10 = ( )8
= ( )2
The important point:
most real numbers cannot be represented exactly in a computer:

60. The effective number system - a computer
Extreame Example
The effective number system - a computer
is not a continuum
but a rather peculiar discrete set
an extreme example
floating-point numbers
x = ±(0.b1b2b3)2 × 2±k
where
b1, b2, b3, k - value 0 or 1.
List all the floating-point numbers that can be expressed in the form
x = ±(0.b1b2b3)2 × 2±k (k, bi  {0, 1})

61. there is some duplication - nonnegative numbers:
Solution
There are
two choices for the ±
two choices for b1
two choices for b2
two choices for b3
three choices for the exponent
would expect 2 × 2 × 2 × 2 × 3 = 48
different numbers
there is some duplication - nonnegative numbers:

63. Positive machin numbers in Example

64. Altogether there are 31 distinct numbers in the system
The positive numbers obtained are
shown on a line
the numbers are
symmetrically but unevenly
distributed about zero

65. owerflow / underflow
If, in the course of a computation, a number x is produced of the form±q×2m
where m is outside the computer’s permissible range
an overflow or an underflow
has occurred
or that x is outside the range of the computer
Generally, an overflow results
in a fatal error (or exception), and the normal execution of the program stops
An underflow,
usually treated automatically by setting x to zero without any interruption of
the program but with a warning message in most computers.

66. in the Example
In a computer whose floating-point numbers are restricted to the form in Example
any number closer to zero than 1/16 would underflow to zero
any number outside the range −1.75 to would overflow to machine infinity.

67. Hole at Zero
in the example normalized floating-point numbers,
all numbers - with the exception of zero- have the form
x = ±(0.1b2b3)2 × 2±k
This creates a phenomenon known as the hole at zero.
nonnegative machine numbers are now distributed as in the following figure
There is a relatively wide gap between zero and the
smallest positive machine number, which is (0.100)2 × 2−1 = 1/4

68. Normalize machin numbers i Example

69. Floating-Pointg Representation
A computer - operates in floating-point mode except for the limitations imposed by the finite word length
Many binary computers - a word length of 32 bits
The internal representation of numbers and their storage - standard floating-point form

70. single-precision floating-point numbers
all acceptable numbers in a computer using the standard single-precision floating-point arithmetic format
assume - stores these numbers in 32-bit words
This set - finite subset of the real numbers
It consists of:
±0, ±∞, normal and subnormal single-precision floating-point numbers,
but not NotaNumber (NaN) values
most real numbers cannot be represented exactly as floating-point numbers
they have infinite decimal or binary expansions
all irrational numbers and some rational numbers
e.g., π, e, 1/3 , 0.1

71. 32-bitword-length - as much as possible of the normalized floating-point number - ±q × 2m
must be contained in those 32 bits
One way of allocating the 32 bits:
sign of q 1 bit
integer |m| 8 bits
number q 23 bits
Information on the sign of m is contained in the eight bits allocated for the integer |m|.
represent real numbers with |m| as large as 27 − 1 = 127.
The exponent represents numbers from −127 through 128.

72. Single-Precision Floating-Point Form
describe a machine number of the following form in standard single-precision floating-point representation:
(−1)s × 2c−127 × (1.f )2
The leftmost bit - for the sign of the mantissa
s = 0 corresponds to+
s = 1 corresponds to −
The next eight bits - to represent the number c in the exponent of 2c−127
interpreted as an excess-127 code
the last 23 bits represent f from the fractional part of the mantissa in the 1-plus form: (1.f )2

73. normalized representation of a nonzero floating-point number:
the first bit in the mantissa is always 1
this bit does not have to be stored
shifting the binary point to a “1-plus” form (1.f )2
The mantissa - rightmost 23 bits - contains f
with an understood binary point
mantissa (significand) actually corresponds to
24 binary digits
there is a hidden bit.
An important exception is the number ±0

74. procedure for determining the representation of a real number x
If x is zero,
it is represented by a full word of zero bits with the possible exception of the sign bit
For a nonzero x:
first assign the sign bit for x and consider |x|
Then convert both the integer and fractional parts of |x| from decimal to binary
Next one-plus normalize (|x|)2 by
shifting the binary point so that the first bit to the left of the binary point is a 1 and all bits to the left of this 1 are 0
To compensate for this shift of the binary point, adjust the exponent
of 2; that is, multiply by the appropriate power of 2.
The 24-bit one-plus-normalized mantissa in binary is found

75. Now the current exponent of 2 should be set equal to
c − 127 to determine c, which is then converted from decimal to binary.
The sign bit of
the mantissa is combined with (c)2 and (f )2
Finally, write the 32-bit representation of x as eight hexadecimal digits.

76. Partitione floating-poin single-precisio computer word

77. actual exponent of the number is restricted by the inequality
The value of c in the representation of a floating-point number in single precision:
restricted by the inequality
0 < c < ( )2 = 255
The values 0 and 255 are reserved for special cases - ±0 and ±∞, respectively.
actual exponent of the number is restricted by the inequality
−126 <= c − 127 <= 127
the mantissa of each nonzero number is restricted by the inequality
1<=(1.f )2<=( )2=2 − 2−23

78. The largest number representable:
(2 − 2−23)2127 ≈ 2128 ≈ 3.4 × 1038
The smallest positive number:
2−126 ≈ 1.2 × 10−38
The binary machine floating-point number ε = 2−23 is called the machine epsilon
single precision
It is the smallest positive machine number ε such that
1 + ε  1.
2−23 ≈ 1.2 × 10−7,
in a simple computation, approximately six
significant decimal digits of accuracy may be obtained in single precision
Recall that 23 bits are allocated for the mantissa.

79. Double-Precision Floating-Point Form
each double-precision floating-point number is stored in two computer words in memory
52 bits allocated for the mantissa
The double precision machine epsilon - 2−52 ≈ 2.2 × 10−16 - approximately 15 significant decimal digits of precision
11 bits allowed for the exponent
which is biased by 1023
represents numbers from −1022 through 1023
A machine number in standard doubleprecision floating-point form:
(−1)s × 2c−1023 × (1.f )2

80. The leftmost bit - for the sign of the mantissa The next eleven bits
s = 0: + and s = 1: −
The next eleven bits
to represent the exponent c corresponding to 2c−1023
52 bits represent f
from the fractional part of the mantissa in the one-plus form: (1.f )2
The value of c - floating-point number in double precision:
restricted by the inequality
0 < c < ( )2 = 2047
the values at the ends of this interval are reserved for special cases.
actual exponent - restricted by the inequality
−1022 <= c − 1023 <= 1023

81. the mantissa of each nonzero number – restricted:
= 2 − 2−52
2−52 ≈ 1.2 × 10−16 in a simple computation approximately 15 significant decimal digits of accuracy may be obtained in double precision.
52 bits are allocated for the mantissa
The largest double-precision machine number:
(2 − 2−52)21023 ≈ ≈ 1.8 × 10308
The smallest double-precision positive machine
number
2−1022 ≈ 2.2 × 10−308

82. In single precision, 31 bits - available for an integer
Single precision on a 64-bit computer -comparable to double precision on a 32-bit computer
whereas double precision on a 64-bit computer gives four times the precision available on a 32-bit computer
In single precision, 31 bits - available for an integer
only 1 bit is needed for the sign
range for integers from−(231−1) to (231−1) =
In double precision, 63 bits - for integers
in the range −(263 − 1) to (263−1)
integer arithmetic, accurate calculations can result in only approximately
nine digits in single precision
18 digits in double precision!
For high accuracy, most computations should be done by using double-precision floating-point arithmetic.

83. Ezample
Determine the single-precision machine representation of the decimal number−
in both single precision and double precision.

84. Solution Converting the integer part to binary
(52.)10 =(64.) 8 =( )2
converting the fractional part
( )10 = (.17)8 = ( )2
( )10 = ( )2
= ( )2 × 25
one-plus form in base 2
( )2 - stored mantissa.
the exponent: (5)10
since c−127 = 5
(132)10 = (204)8 = ( )2 stored exponent
the single-precision machine representation
of − :
[ ]2 =
[ ]2 = [C250F000]16

85. In double precision, for the exponent (5)10,
(2004)8 = ( )2, - stored exponent.
double-precision machine representation of − :
[ · · · 00]2 =
[ · · · 0000]2
= [C04A1E ]16
[· · ·]k - the bit pattern of the machine word(s) that represents floating-point numbers,
displayed in base-k

86. Example
Determine the decimal numbers that correspond to these machine words:
[45DE4000]16 [BA390000]16

87. Solution
The first number in binary:
[ ]2
The stored exponent:
( )2 = (213)8 = (139)10,
139 − 127 = 12
The mantissa – positive:
( )2 × 212 = ( )2
= (15710.)8
= 0 × × × × × 84
= 8(1 + 8(7 + 8(5 + 8(1))))
= 7112

88. the second word in binary:
[ ]2
The exponential part of the word:
( )2 = (164)8 = 116
the exponent: 116 − 127 = −11
The mantissa - negative:
−( )2 × 2−11 = −( )2
= −( )8
= −2 × 8−4 − 7 × 8−5 − 1 × 8−6
= −8−6(1 + 8(7 + 8(2)))
= −185 / ≈ − × 10−4

89. Computer Errors in Representing Numbers
errors - occur
represent a real number x in the computer
a model computer bitword length
for x = , x = 2−32591
exponents - far exceed the limitations of the machine
overflow and underflow - respectively
relative error in replacing x by the closest machine number - very large
Such numbers are outside the range of a 32-bit word-length computer.

90. positive real number x - normalized floating-point form
x = q × 2m ,1/2 <= q < 1, −126 <= m <= 127
The process of replacing x by its nearest machine number - correct rounding
the error involved - roundoff error
q - expressed in normalized binary notation:
x = (0.1b2b3b b24b25b )2 × 2m
one nearby machine number
by rounding down or by simply dropping the excess bits b25b , only 23 bits allocated stored mantissa
This machine number: x− = (0.1b2b3b b24)2 × 2m
lies to the left of x on the real-number axis.

91. Another machine number, x+, to the right of x on the real axis
obtained by rounding up
adding one unit to b24 in the expression for x−:
x+ = ](0.1b2b3b b24)2 + 2−24 ]× 2m
The closer of these machine numbers is the one chosen to represent x
|x − x−| <= (½)|x+ − x−| = 2−25+m
the relative error is bounded:

92. where u = 2−24 is the unit roundoff error for a 32-bit binary computer with standard floating-point arithmetic
Recall that the machine epsilon is ε = 2−23
so u = 1/2 ε.
Moreover:
u = 2−k , where k is the number of binary digits used in the mantissa, including the hidden bit (k = 24 in single precision and k = 53 in double precision)

93. On the other hand, if x lies closer to x+ than to x−, then
|x − x+| <= (1/2) |x+ − x-|
and the same analysis shows that the relative error is no greater than 2−24 = u
So in the case of rounding to the nearest machine number
the relative error is bounded by u

94. In Case3 of Choping
when all excess digits or bits are discarded, the process is called chopping
If a 32-bit word-length computer has been designed to chop numbers, the relative error bound
would be twice as large as above:
2u = 2−23 = ε.

95. A possibl relationsh between x−, x+, and x

96. Notation fl(x) and Backward Error Analysis
errors produced - elementary arithmetic operations
to illustrate - a five-place decimal machine
add numbers
in normalized floatingpoint
form
x = × 104 y = × 10−1
Many computers perform arithmetic operations in a double-length work area,
assume that our computer - a ten-place accumulator
First, the exponent of the smaller number is adjusted - both exponents are the same.
Then the numbers are added in the accumulator
the rounded result is placed in a computer word:
x = × 104
y = × 104
x + y = × 104

97. The nearest machine number is z = 0.37219 × 104
the relative error:
would be regarded as acceptable on a machine of such low precision

98. introduce the notation fl(x)
denote the floating-point machine number corresponds to the real number x
the function fl depends on the particular computer involved
for the hypothetical five-decimal-digit machine:
fl( × 104) = × 104
For a 32-bit word-length computer,
x - any real number within the range of the computer

99. assume that correct rounding is used
This inequality can expressed in the form
fl(x) = x(1 + δ) |δ| <= 2−24
two inequalities equivalent
let δ = [fl(x) − x]/x
by |δ| <= 2−24
solving for fl(x) - fl(x) = x(1 + δ)

100. details in the addition 1 + ε
if ε >= 2−23, then fl(1 + ε) > 1,
if ε < 2−23, then fl(1 + ε) = 1
Consequently, if machine epsilon is
the smallest positive machine number ε such that
fl(1 + ε) > 1 then ε = 2−23.

101. the symbol arithmetic operations +, −, ×, or ÷.
a 32-bit word-length computer designed
two machine numbers x and y
computer produce fl(x  y) instead of x  y
imagine that x  y
first correctly formed
then normalized
finally rounded to become a machine number.
the relative error not exceed 2−24
fl(x  y) = (x  y)(1 + δ)

102. Special cases:
δ - variable −2−24 <= δ <= 2 24
The assumptions about - a model 32-bit word-length computer - not quite true
x  y - overflow or underflow

103. the equation above can be written in a variety of ways,
alternative interpretations of roundoff
e.g.,
fl(x + y) = x(1 + δ) + y(1 + δ)
result of adding x and y not in general x + y
true sum of x(1 + δ) and y(1 + δ)
x(1 + δ) - result of slightly perturbing x.
machine version of x + y -fl(x + y)
exact sum of a slightly perturbed x and a slightly perturbed y

104. Backward v.s. Direct Error Analysis
backward error analysis
attempts to determine
what perturbation of the original data would cause the computer results to be the exact results for a perturbed problem
direct error analysis
how computed answers differ from exact answers based on the same data

105. Example
If x, y, and z are machine numbers in a 32-bit word-length computer
what upper bound can be given for the relative roundoff error in computing z(x + y)?

106. Solution In the computer x + y will be done first
This arithmetic operation produces
the machine number fl(x + y),
differs from x + y roundoff
there is a δ1 such that
fl(x + y) = (x + y)(1 + δ1) |δ1| <= 2−24
z - a machine number
When multiplies the machine number fl(x + y),
the machine number fl[z fl(x + y)]
differs from its exact counterpart,
for some δ2
fl[z fl(x + y)] = z fl(x + y)(1 + δ2) |δ2| <= 2-24

107. |
δ1δ2| <= 2−48 ignore it
δ = δ1 + δ2
|δ| = |δ1 + δ2| <= |δ1| + |δ2| <= 2− −24 = 2 −23

108. Example
Critique the following attempt to estimate the relative roundoff error in computing the sum
of two real numbers, x and y. In a 32-bit word-length computer, the calculation yields
Therefore, the relative error is bounded as follows:
Why is this calculation not correct?

109. the quantities δ that occur in such calculations are not, in general, equal to each other
The correct calculation is

110. the relative roundoff error:

111. cannot be bounded
the second term has a denominator that can be zero or close to zero
if x and y are machine numbers, then δ1 and δ2 are zero
and a useful bound results—namely, δ3.
But we do not need this calculation to know that! It has been assumed that when machine numbers are combined with any of the four arithmetic operations, the relative roundoff error will not exceed 2−24 in magnitude