Numbers Representation
&
Error Analysis
&
Accuracy & Precision
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We count in Base 10
(Decimal)
Number
Representation
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Computer’s perception …

101
100
99
98
97
96
95
14
13
12
11
10
24
23
22
21
20
19
18
17
16
15
9
8
7
6
5
4
3
2
1
0
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

Computers are made of a series of
switches
Each switch has two states: ON or OFF
Each state can be represented by a
number:
Ran out of symbols (0-9), so increment the digit on the left by one unit.
1 for “ON” and 0 for “OFF”
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1
Computers count in Base 2
(Binary)
Binary Numbers (Bits)

Bits can be represented as:









1 or 0
On or Off
Up or Down
Open or Closed
Yes or No
Black or White
Thick or Thin
Long or Short
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Counting in Binary is the same, but with
only two symbols


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On (1)
Off (0)
10000
1111
1110
1101
1100
1011
1010
1001
1000
111
101
100
110
11
10
1
0
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7
Equivalence of Binary and
Decimal


Every Binary number has a corresponding Decimal
value (and vice versa)
8
Converting Binary to Decimal

In general, the "position values" in a binary
number are the powers of two.
Examples:

Binary Number
1
10
11
…
1010111
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Decimal Equivalent
1
2
3
…
87
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




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The first position value is 20 , i.e. one
The 2nd position value is 21 , i.e. two
The 2nd position value is 22 , i.e. four
The 2nd position value is 23 , i.e. eight
The 2nd position value is 24 , i.e. sixteen
etc.
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2
Converting Binary to Decimal
Converting Binary to Decimal
1 0 1 0 1 1 0 0
0 1 0 1 0 0 0 1
128
128
64
32
16
8
128 + 0 + 32 + 0 + 8
4
2
64
16
8
4
2
1
1
+ 4 + 0 + 0
0 + 64 + 0 + 16 + 0
128 + 32 + 8 + 4 = 172
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32
+ 0 + 0 + 1
64 + 16 + 1 = 81
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12
Representing Fractions
in Binary

Converting a decimal# to a
binary#
Consider ordinary Base-10 notation: what does
e.g. 1205.91410 mean?
1
2
0
5
.
9
1
4
103
102
101
100
.
10-1
10-2
10-3
• Binary works the same way; e.g., what is 101.012?
1
22
0
21
1
20
.
.
0
2-1
1
2-2
= 4 + 1 + 1/4 = 5.025
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3
13
Converting a decimal# to a
binary#
Floating
Point
Number
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Floating-Point Numbers



History: origins of computers in finance
(banking) and military (physics).
Financial calculations are (usually) fixedpoint: the decimal place is fixed in one
position, two digits from right: $129.95, $0.59,
etc.
In physical calculations, we need the decimal
place to “float”: 3.14159, 0.333,
6.0221415 × 1023 , etc.
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The Radix


In the decimal system, a decimal point
(radix point) separates the whole
numbers from the fractional part
Examples:
37.25 ( whole=37, fraction = 25)
123.567
10.12345678
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4
Exponential Notation




Normalized Notation
Format: Mantissa X 10Exponent, or a X 10n
Abbreviated a e n or a E n; for example,
6.0221415e23
Mantissa (a.k.a. significand, a.k.a. fractional part)
is a floating-point number; exponent is an integer.
So someone has to decide the precision:

IEEE 754 standard: with 64-bit numbers, use 52 bits for
mantissa, 11 for exponent, 1 for sign (+ or − )
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
So a normalized number in exponential notation
has the decimal place immediately before the first
nonzero digit:
.314159e1,

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-0.9876 x 10-3
Location of
decimal point
Exponent
Base
Integer
part
mantissa
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!
a.be
Binary normalization
What is 12.5 in floating-point representation?
1.
Convert 12.5 to binary fixed point
2.
Normalize the number by moving the radix point,
producing the mantissa
1100.12 = 0.11001 * 24
Integer
part
mantissa
exponen
Baset of the number
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used
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12.510 = 1100.12
Sign of
exponent
Mantissa
etc.
Significant digits of a normalized floating-point
number are all digits except leading zeros.
Floating Point Number (Computer)
Sign of
mantissa
.60221415e24,
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!
a.be
exponen
Baset of the number
system used
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5
Normalization … cont. …
Every binary number, except the one
corresponding to the number zero, can be
normalized by choosing the exponent so that the
radix point falls to the right of the leftmost 1 bit.
37.2510 = 100101.012 = 1.0010101 x 25
7.62510 = 111.1012 = 1.11101 x
0.312510 = 0.01012 = 1.01 x
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22
So what Happened ?
After normalizing, the numbers now have different
mantissas and exponents.
37.2510 = 100101.012 = 1.0010101 x 25
7.62510 = 111.1012 = 1.11101 x 22
0.312510 = 0.01012 = 1.01 x 2-2
2-2
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IEEE Floating Point Representation

From Binary Floating
Point To Machine
representation
Floating point numbers can be
represented by binary codes by dividing
them into three parts:
the sign, the exponent, and the
mantissa.
0
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1
8
9
32
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6
IEEE Floating Point Representation

The first, or leftmost, field of our floating point
representation will be the sign bit:
 0 for a positive number,
 1 for a negative number.
IEEE Floating Point Representation


The second field of the floating point number will be the
exponent.
Since we must be able to represent both positive and negative
exponents, we will use a convention which uses a value known
as a bias of 127 to determine the representation of the
exponent.



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IEEE Floating Point Representation

The mantissa is the set of 0’s and 1’s to
the left of the radix point of the
normalized (when the digit to the left of
the radix point is 1) binary number.


ex:1.00101 X 23
The biased exponent, the value actually stored, will range
from 0 through 255. This is the range of values that can be
represented by 8-bit, unsigned binary numbers.
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Converting decimal floating point values to stored
IEEE standard values.
Example:
40.15625.
Find the IEEE FP representation of
Step 1.
Compute the binary equivalent of the whole part and
the fractional part. ( convert 40 and .15625. to their
binary equivalents)
The mantissa is stored in a 23 bit field,
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An exponent of 5 is therefore stored as 127 + 5 or 132;
an exponent of -5 is stored as 127 + (-5) OR 122.
40.1562510 = 101000.001012
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7
Converting decimal floating point values to stored
IEEE standard values.
Step 2. Normalize the number by
moving the decimal point to the right of
the leftmost one.
Converting decimal floating point values to stored
IEEE standard values.
Step 3. Convert the exponent to a
biased exponent
127 + 5 = 132
101000.00101 = 1.0100000101 x 25
==>
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13210 = 100001002
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Covert 40.15625 to IEEE 32-bit format
40.15625
Step 1
Find binary of whole number
Converting decimal floating point values to stored
IEEE standard values.
Step 4. Store the results from above
Sign
0
Exponent (from step 3)
10000100
40-32-8
Binary value of 40
32 16 8 4 2 1
1 0 1 0 0 0
Step 1b
Find binary of fraction number
.15625 -.1250 - .03125
Binary value of .15625
.5 .25 .125 .0625 .03125
0
0
1
0
1
Step 2a
Move decimal (Radix point) to the
right of the left most 1 to come up
with the exponent
.
101000.00101
Step 2b
Normalize by multipling the number by 2
and the exponent from step 3
Step 3a
convert the exponent to a biased exponent
by adding the bias of 127 to the exponent
Mantissa( from step 2)
01000001010 .. 0
Equals
101000.00101 x 25 = 1.0100000101
127 + 5 = 13210
Step 3b
Convert the biased exponent to its binary
value
Binary value of 132
132-128-4
128 32 16
1 0 0
8
0
4 2 1
1 0 0
Step 4a
Piece together values.
Step 4c. Left pad exponent and right
pad mantissa to determine the binary
equivalent of IEEE
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Sign
Exponent
0
10000100
0
010000100
Mantissa
0100000101
01000001010…0
Sign
Exponent
0
010000100
Mantissa
01000001010…0
Step 4b
Move to mantissa
without the 1.
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8
Converting decimal floating point values to stored
IEEE standard values.
One more example
Step 2.
Normalize the number by moving the
decimal point to the right of the leftmost
one.
Find the IEEE FP representation of –24.75
Step 1.
-24.7510 = -11000.112
-11000.11 = -1.100011 x 24
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Converting decimal floating point values to stored
IEEE standard values.
Step 3. Convert the exponent to a biased exponent
127 + 4 = 131
==> 13110 = 100000112
Step 4. Store the results from above
Sign
Exponent
mantissa
1
10000011
1000110..0
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Converting from IEEE format to the decimal floating
point values.


Do the steps in reverse order
In reversing the normalization step
move the radix point the number of
digits equal to the exponent. if exponent
is +ve move to the right, if –ve move to
the left.
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9
Converting from IEEE format to the decimal floating
point values.
Converting from IEEE format to the decimal floating
point values.
Ex: Convert the following 32 bit binary numbers to their
decimal floating point equivalents.
Step 1 Extract
exponent)
a.
Sign
Exponent
Mantissa
1
01111101
010..0
exponent
biased exponent = 01111101 = 125
exponent: 125 - 127=
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(unbias
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-2
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Converting from IEEE format to the decimal floating
point values.
Converting from IEEE format to the decimal floating
point values.
Step 2 Write Normalized number
Step 3: Write the binary number (denormalize value
from step2)
1 . ____________
mantissa
x 2
-1. 01 x 2 –2
Exponent
----
-0.01012
Step 4: Convert binary number to FP equivalent ( add
column values)
-0.01012 = - ( 0.25 + 0.0625) = -0.3125
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10
Converting from IEEE format to the decimal floating
point values.
Ex: Convert the following 32 bit binary
numbers to their decimal floating point
equivalents.
Sign
0
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Exponent
10000011
Converting from IEEE format to the decimal floating
point values.
Step 1 Extract
exponent)
(unbias
biased exponent = 10000011 = 131
Mantissa
1101010..0
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exponent
exponent: 131 - 127=
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4
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Converting from IEEE format to the decimal floating
point values.
Converting from IEEE format to the decimal floating
point values.
Step 2 Write Normalized number
Step 3 Write the binary number (denormailze value
from step 2)
mantissa
1 . ____________
x 2
1. 110101 x 2 4
Exponent
----
11101.012
Step 4 Convert binary number to FP equivalent ( add
column values)
11101.012 = 16 + 8 + 4 + 1 + 0.25
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= 29.2510
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11
Proof your work
Convert
0 10000100 01000001010…0 back to IEEE
Sign Exponent
0 010000100
Step 1a
Determine the decimal of the binary
number
Step 1b
Unbias the number by Subtracting 127
from the decimal number to determine
the exponent
Practise example
Mantissa
01000001010…0
128 32 16 8 4 2 1
1 0 0 0 1 0 0
128+4
Binary value = 132
Step 2
Denormalize by multipling the number by 2 and
the exponent from step 2.
Set up format 1 mantissa x 2 exponent

132 – 127 = 5
.
1.0100000101 x 25
Step 3a
Move decimal (Radix point) to the right of
the left most 1 to come up with the exponent
What decimal value is represented by
the following 32-bit floating point
number?
1 10000010 11110110000000000000000
Equals
101000 . 001012
Step 3b
Convert binary number to FP equvalent
Step 4a
Find whole number of exponent
32 16 8 4 2 1
1 0 1 0 0 0
.5 .25 .125 .0625 .03125
0 0
1
0
1
32+8 = 40
.1250 + .03125 = .15625
Step 4b
Find fractional numberof
the mantissa

Answer: -15.6875
Step 4c
Add together values. Make sure to include
the sign if it is a negative value
40.15625
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Another Practise Example
Convert -12.510 to binary:
1.
Convert 12.5 to fixed point  01100.1
2.
Normalize  0.11001 * 24
3.
Convert exponent base to binary: 4  0100
4.
2s complement the mantissa by flipping bits
and adding 1:
011001  100111
5.
Final number  1 00111 0100
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Double and single
Precision
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12
Single Precision Format
Double Precision Format
32 bits (4x8bits = 4 bytes … Real*4)
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64 bits (8x8bits = 8 bytes … Real*8)
Mantissa (23 bits)
Mantissa (52 bits)
Exponent (8 bits)
Exponent (11 bits)
Sign of mantissa (1 bit)
Sign of mantissa (1 bit)
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51
Smallest value for a binary #


The smallest value for a binary number
of any number of bits is zero.
52
Smallest value for a binary #

The smallest value for a binary number with any number of
bits is zero (i.e. when all the bits are zeros)
# of bits
1 bit:
2 bits:
3 bits:
4 bits:
5 bits:
6 bits:
7 bits:
8 bits:
etc.
This is the case when all bits are zero:
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smallest binary #
decimal value
0
00
000
0000
00000
000000
0000000
00000000
0
0
0
0
0
0
0
0
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13
53
Largest value for a binary #


54
Largest numbers
The largest value for a binary number
with a specific number of bits (i.e. digits)
is when all of the bits are one.

The following are the largest values for binary
numbers with a specific number of bits:
General rule: for a binary number with
n bits, the largest possible value is :
2n - 1
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largest binary #
decimal value
1 bit:
2 bits:
3 bits:
4 bits:
5 bits:
6 bits:
7 bits:
8 bits:
etc.
1
11
111
1111
11111
111111
1111111
11111111
1
3
7
15
31
63
127
255
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Maximums & minimums
mantissa
# of bits
Normal&Abnormal IEEE Singles
excess
floating
Number name
Normal
s, e, and f
0 < e < 255
Value of single
(-1)sx2e-127x1.f
Subnormal
e=0, f = 0
(-1)sx2-126x 0.f
Signed zero
(+/- 0)
+
e=0, f=0
(-1)sx0.0
s=0,e=255,f=0
+ INF
s=1,e=255,f=0
- INF
s=u,e=255,f=0
NaN
"
- "
Not a Number
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55
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56
!
!
14
Normal&Abnormal IEEE Doubles
Number name
Normal
s, e, and f
0 < e < 2047
Value of single
(-1)sx2e-1023x1.f
Subnormal
e=0, f = 0
(-1)sx2-1022x0.f
Exceptions
Signed zero (+/- e=0, f=0
(-1)sx0.0
0)
+
s=0,e=2047,f=0 + INF
-
"
"
Not a Number
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s=1,e=2047,f=0 - INF
s=u,e=2047,f=0
NaN
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57
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!
!
Overflow and Underflow
Overflow and Underflow


Overflow is an error condition that occurs when
there are not enough bits to express a value in a
computer.

Possible for the number to be too large or too
small for representation
Underflow is an error condition that occurs when
the result of a computation is too small for the
computer to represent.
0.00001 x 10-50 = 10-55
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15
There are discrete points
on the number lines that
can be represented by our
computer.
Error Analysis
How about the space
between ?
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Possible Errors

truncation error


overflow error


round off errors using floating-point numbers
because not all real numbers can be represented
accurately
attempting to represent a number that is greater
than the upper bound for the given number of bits
underflow error

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attempting to represent a number that is less than
the lower bound for the given number of bits
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64
16
Rounding in binaries
Round-off Error


ROUNDING
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Phys381


Round-off error is the problem of not having
enough bits (or digits) to store an entire floatingpoint number and approximating the result to the
nearest number that can be represented.
Numbers such as π, e, or 7 also cannot be expressed by a
fixed number of significant figures.
Computers use a base-2 representation, they cannot precisely
represent certain exact base-10 numbers.
E.g., 1/3 = 0.333 ≠ 0.3 ≠ 0.33 ≠ etc.
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Implications
First, familiar example
0.1000x101 + 0.0000(4)x101 = 0.1000x101
(a+b)+c =
a+(b+c) =
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0.0000(8)x101 = 0.0001x101
17
Adding decimals
Adding in
Binary
system
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Adding Binaries
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Adding Binaries
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18
Adding Binaries
Precision
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Precision

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Precision
Consider the addition of two 32-bit
words (single precision): 7 + 1.0 x 10-7.


Consider the addition of two 32-bit
words (single precision): 7 + 1.0 x 10-7.
These numbers are stored as:
7= 0
10-7 =
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0
10000010
11100000000000000000000
01100010
11010110101111111001010
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19
Precision


Precision
Because the mantissa of the numbers are
different, the exponent of the smaller number is
made larger while the mantissa is made
smaller until they are same(shifting bits to the
right while inserting zeros).
Until:



As a result when the calculation is carried out
the computer will get: 7 + 1.0 x 10-7 = 7
Effectively the 10-7 term has been ignored.
The machine precision can be defined as
the maximum positive value that can be
added to 1 so that its value is not changed.
Another example:
10-7 = 0

01100010
00000000000000000000000(0001101…)
NB: The bits in the brackets are lost.
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(1% 2
2& # " = 0.6666666 " 0.6666667 = "0.0000001 ! 0
' 3$ 3
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Numerical Errors: fractions
Phys381
One more Example
Example:
>b
= 1/3
>b
= 0.333333
> b*3 - 1 = 0
or
>b
=4/3 - 1
>b
= 0.33333
> b*3 - 1 = -2.2204e-16
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????
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80
20
Implications
Solution

When you program:

You should write
these instead:
x 2 +1 +1
f ( x) = x 2 + 1 ! 1
f ( x) = ( x 2 + 1 ! 1)
g ( x) = ln( x) ! 1
x
g ( x) = ln( x) ! ln(e) = ln( )
e
x 2 +1 +1
=
x2
x 2 +1 +1
Every FP operation introduces error, but the
subtraction of nearly equal numbers is the worst
and should be avoided whenever possible
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81
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Avoiding Round-off Error
Avoiding Round-off Error

Some values (1/3) will always result in round-off;
however, we can help reduce error:

E.g. consider computing (x/y)z on a machine with
three significant digits of precision, where
x = 2.41


When adding numbers whose magnitudes are drastically
different, accumulate smaller number before combining
them with larger ones.
Otherwise they get absorbed !
When multiplying and dividing, perform all multiplications
in the numerator before dividing by the denominator.
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y = 9.75
z = 1.54
(x / y)*z = (2.41 / 9.75)*(1.54) = (0.247)*(1.54) = 0.380
But (x / y)*z = (x*z) / y:
(x*z) / y = ((2.41) )*(1.54)) / 9.75 = 3.71 / 9.75 = 0.381
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21
Truncation Errors
and the Taylor Series
Truncation/Chopping errors
Example:
π=3.14159265358 to be stored on a base-10
system carrying 7 significant digits.
π=3.141592
chopping error εt=0.00000065
If rounded
π=3.141593
εt=0.00000035
 Some machines use chopping, because
rounding adds to the computational overhead.
Since number of significant figures is large
enough, resulting chopping error is negligible.
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

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85


Non-elementary functions such as trigonometric,
exponential, and others are expressed in an
approximate fashion using Taylor series when
their values, derivatives, and integrals are
computed.
Any smooth function can be approximated as a
polynomial. Taylor series provides a means to
predict the value of a function at one point in
terms of the function value and its derivatives at
another point.
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Truncation error is decreased by addition of terms to the Taylor series.
If h (discretization) is sufficiently small, only a few terms may be
required to obtain an approximation close enough to the actual value
for practical purposes.
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86
Example
To get the cos(x) for small x:
cos x = 1 !
If x=0.5
cos(0.5)
x2 x4 x6
+ ! +L
2! 4! 6!
=1-0.125+0.0026041-0.0000127+ …
=0.877582
From the supporting theory, for this series, the error is no
greater than the first omitted term.
!
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87
x8
8!
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for
x = 0.5 = 0.0000001
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88
22
Truncation Errors: example
Absolute and Relative Error

If correct is the answer and result is the result
obtained, then
absolute error = | correct - result |
x = tan(pi/6)
y = sin(pi/6)/cos(pi/6)
relative error = (absolute error)
but,
| correct |
x - y = 1.1102e-16
????
= (absolute error)
X 100%
| correct |
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Example
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Total Error
Total error = Round-off Error + Truncation Error
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The total numerical error is the summation of
the truncation and roundoff errors.
The truncation error generally increases as
the step size increases, while the roundoff
error decreases as the step size increases this leads to a point of diminishing returns for
step size.
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Decreasing the step h, reduces round-off error
but increase truncation error
Optimization issue: step size h
ROUND-OFF ERROR
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Error for one step h = dx
TRUNCATION ERROR
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Increasing the number of terms of series reduces
truncation error but increase round-off error
Optimization issue: step h
Optimization issue: # terms
ROUND-OFF ERROR
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TRUNCATION ERROR
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Optimization issue: # terms
Error propagation
Imagine computing the following in a Do loop !!
See Upcoming Assignment #2
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Looping and Error Propagation
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Accumulating values in a loop (repeatedly) can
produce error
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Patriot Missile Failure
A loop is a segment of code [computer instructions]
that is executed repeatedly.
E.g., Patriot Missile failure in 1991 Gulf War
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
Patriot’s internal clock measured time in 1/10 of a
second.
Each tick of the clock added 1/10 s to current time.
But 1/10 cannot be represented in a finite number
of bits (just as 1/3 cannot be represented in a finite
number of decimal digits)
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Patriot Missile Failure

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Each one-tenth increment produced an error of
about .000000095 seconds.
After ten hours of running Patriot’s computer:
10 hr
60 min
1 hr
60 sec .000000095 sec
1 min 1 sec
Numerical Stability !
Numerical stability and well-posed problems
= 0.34 sec
An algorithm is called numerically stable if an error, whatever its cause,
does not grow to be much larger during the calculation.
This happens if the problem is well-conditioned, meaning that the
solution changes by only a small amount if the problem data are
changed by a small amount.
• Scud travels 1,676 meters / sec. So error =
1676 meters 0.34 sec
1 sec
= 570 meters
To the contrary, if a problem is ill-conditioned,
then any small error in the data will grow to be a large error.
• Result Patriot missed Scud, 28 Americans died
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Accuracy versus precision
Accuracy
vs
Precision
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Accuracy. How close is a computed or
measured value to the true value
Precision (or reproducibility). How close is
a computed or measured value to
previously computed or measured values.
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?
Accuracy versus precision
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