Computer Arithmetic
module 2
By
Soumya Das
Asst prof. dept. of cse
GCE Kalahandi
Number Representation
Sign Magnitude
Sign magnitude is a very simple representation of negative numbers. In sign magnitude the first bit is
dedicated to represent the sign and hence it is called sign bit.
Sign bit ‘1’ represents negative sign.
Sign bit ‘0’ represents positive sign.
In sign magnitude representation of a n – bit number, the first bit will represent sign and rest n-1 bits
represent magnitude of number.
For example,
+25 = 011001
Where 11001 = 25
And 0 for ‘+’
-25 = 111001
Where 11001 = 25
And 1 for ‘-‘.
Range of number represented by sign magnitude method = -(2n-1-1) to +(2n-1-1) (for n bit number)
2’s complement method
To represent a negative number in this form, first
we need to take the 1’s complement of the
number represented in simple positive binary
form and then add 1 to it.
For example:
(-8)10 = (1000)2
1’s complement of 1000 = 0111
Adding 1 to it, 0111 + 1 = 1000
So, (-8)10 = (1000)2
Range of number represented by 2’s
complement = (-2n-1 to 2n-1 – 1)
Ripple carry adders
for each adder block, the two bits that are to
be added are available instantly. However,
each adder block waits for the carry to arrive
from its previous block. So, it is not possible to
generate the sum and carry of any block until
the input carry is known. The block waits for
the block to produce its carry. So there will be
a considerable time delay which is carry
propagation delay.
Carry Look-ahead Adder
A carry look-ahead adder reduces the
propagation delay by introducing more
complex hardware. In this design, the
ripple carry design is suitably
transformed such that the carry logic
over fixed groups of bits of the adder is
reduced to two-level logic.
Advantages and Disadvantages of
Carry Look-Ahead Adder :
Advantages –
The propagation delay is reduced.
It provides the fastest addition logic.
Disadvantages –
The Carry Look-ahead adder circuit gets
complicated as the number of variables increase.
The circuit is costlier as it involves more number of
hardware.
Booth algorithm
• Booth algorithm gives a procedure for multiplying binary integers in
signed 2’s complement representation in efficient way, i.e., less number
of additions/subtractions required. It operates on the fact that strings of
0’s in the multiplier require no addition but just shifting and a string of 1’s
in the multiplier from bit weight 2^k to weight 2^m can be treated as
2^(k+1 ) to 2^m.
• As in all multiplication schemes, booth algorithm requires examination of
the multiplier bits and shifting of the partial product. Prior to the shifting,
the multiplicand may be added to the partial product, subtracted from the
partial product, or left unchanged according to following rules:
• The multiplicand is subtracted from the partial product upon encountering
the first least significant 1 in a string of 1’s in the multiplier
• The multiplicand is added to the partial product upon encountering the
first 0 (provided that there was a previous ‘1’) in a string of 0’s in the
multiplier.
• The partial product does not change when the multiplier bit is identical to
the previous multiplier bit.
Booth algorithm contd.
• Example – A numerical example of
booth’s algorithm is shown below for n = 4.
It shows the step by step multiplication of -
5 and -7.
• MD = -5 = 1011, MD = 1011, MD'+1 = 0101 MR
= -7 = 1001
The explanation of first step is as
follows:
OPERATION AC MR QN+1 SC
0000 1001 0 4
AC + MD’ + 1 0101 1001 0
ASHR 0010 1100 1 3
AC + MR 1101 1100 1
ASHR 1110 1110 0 2
ASHR 1111 0111 0 1
AC + MD’ + 1 0010 0011 1 0
Product is calculated as follows:
Product = AC MR Product = 0010 0011=35
Division algorithm(Restoring)
• A division algorithm provides a quotient and a
remainder when we divide two number. They
are generally of two type slow algorithm and
fast algorithm. Slow division algorithm are
restoring, non-restoring, non-performing
restoring, SRT algorithm and under fast comes
Newton–Raphson and Goldschmidt. Restoring
term is due to fact that value of register A is
restored after each iteration.
Division algorithm flow chart
Examples:
Perform Division Restoring Algorithm Dividend = 11 Divisor = 3
N M A Q OPERATION
4 00011 00000 1011 initialize
00011 00001 011_ shift left AQ
00011 11110 011_ A=A-M
Q[0]=0 And
00011 00001 0110
restore A
3 00011 00010 110_ shift left AQ
00011 11111 110_ A=A-M
00011 00010 1100 Q[0]=0
2 00011 00101 100_ shift left AQ
00011 00010 100_ A=A-M
00011 00010 1001 Q[0]=1
1 00011 00101 001_ shift left AQ
00011 00010 001_ A=A-M
00011 00010 0011 Q[0]=1
Non-Restoring division
• Non-Restoring division is less complex than
the restoring one because simpler operation
are involved i.e. addition and subtraction, also
now restoring step is performed. In the
method, rely on the sign bit of the register
which initially contain zero named as A.
Division algorithm(Non Restoring)
Examples: Perform Non_Restoring Division for
Unsigned Integer Dividend =11 Divisor =3 -M =11101
N M A Q ACTION
4 00011 00000 1011 Start
00001 011_ Left shift AQ
11110 011_ A=A-M
3 11110 0110 Q[0]=0
11100 110_ Left shift AQ
11111 110_ A=A+M
2 11111 1100 Q[0]=0
11111 100_ Left Shift AQ
00010 100_ A=A+M
1 00010 1001 Q[0]=1
00101 001_ Left Shift AQ
00010 001_ A=A-M
0 00010 0011 Q[0]=1
Quotient = 3 (Q) Remainder = 2 (A)
Floating point representation of
numbers
• 32-bit representation floating point numbers IEEE
standard
Normalization
• Floating point numbers are usually normalized
• Exponent is adjusted so that leading bit (MSB) of
mantissa is 1
• Since it is always 1 there is no need to store it
• Scientific notation where numbers are normalized to
give a single digit before the decimal point like in
decimal system e.g. 3.123 x 103
Floating point arithmatic
• A finite number can also represented by four
integers components, a sign (s), a base (b), a
significand (m), and an exponent (e). Then the
numerical value of the number is evaluated as
• (-1)s x m x be ________ Where m < |b|
• Depending on base and the number of bits used
to encode various components, the IEEE
754 standard defines five basic formats. Among
the five formats, the binary32 and the binary64
formats are single precision and double precision
formats respectively in which the base is 2.
IEEE Standard 754 Floating Point
Numbers
• The IEEE Standard for Floating-Point Arithmetic (IEEE 754) is a technical standard for floating-point
computation which was established in 1985 by the Institute of Electrical and Electronics Engineers
(IEEE). The standard addressed many problems found in the diverse floating point implementations
that made them difficult to use reliably and reduced their portability. IEEE Standard 754 floating
point is the most common representation today for real numbers on computers, including Intel-
based PC’s, Macs, and most Unix platforms.
• There are several ways to represent floating point number but IEEE 754 is the most efficient in most
cases. IEEE 754 has 3 basic components:
• The Sign of Mantissa –
This is as simple as the name. 0 represents a positive number while 1 represents a negative
number.
• The Biased exponent –
The exponent field needs to represent both positive and negative exponents. A bias is added to the
actual exponent in order to get the stored exponent.
• The Normalised Mantissa –
The mantissa is part of a number in scientific notation or a floating-point number, consisting of its
significant digits. Here we have only 2 digits, i.e. O and 1. So a normalised mantissa is one with only
one 1 to the left of the decimal.
• IEEE 754 numbers are divided into two based on the above three components: single precision
and double precision.
Table – 1 Precision Representation
Precision Base Sign Exponent Significand
Single 2 1 8 23+1
precision
Double 2 1 11 52+1
precision
Example
85.125 85 = 1010101 0.125 = 001 85.125 = 1010101.001
=1.010101001 x 2^6 sign = 0
1. Single precision: biased exponent 127+6=133 133 =
10000101 Normalised mantisa = 010101001 we will add 0's
to complete the 23 bits The IEEE 754 Single precision is: = 0
10000101 01010100100000000000000 This can be written
in hexadecimal form 42AA4000
2. Double precision: biased exponent 1023+6=1029 1029 =
10000000101 Normalised mantisa = 010101001 we will add
0's to complete the 52 bits The IEEE 754 Double precision
is: = 0 10000000101
010101001000000000000000000000000000000000000000
0000 This can be written in hexadecimal form
4055480000000000
• Special Values: IEEE has reserved some values that can ambiguity.
• Zero –
Zero is a special value denoted with an exponent and mantissa of 0. -0 and +0 are
distinct values, though they both are equal.
• Denormalised –
If the exponent is all zeros, but the mantissa is not then the value is a
denormalized number. This means this number does not have an assumed leading
one before the binary point.
• Infinity –
The values +infinity and -infinity are denoted with an exponent of all ones and a
mantissa of all zeros. The sign bit distinguishes between negative infinity and
positive infinity. Operations with infinite values are well defined in IEEE.
• Not A Number (NAN) –
The value NAN is used to represent a value that is an error. This is represented
when exponent field is all ones with a zero sign bit or a mantissa that it not 1
followed by zeros. This is a special value that might be used to denote a variable
that doesn’t yet hold a value.
EXPONENT MANTISA VALUE
0 0 exact 0
255 0 Infinity
0 not 0 denormalised
255 not 0 Not a number (NAN)
APPROXIMATE
DENORMALIZED NORMALIZED
DECIMAL
± approximately
± 2-149 to (1 – 2- ± 2-126 to (2 – 2- 10-44.85 to
Single Precision 23)×2-126 23)×2127 approximately
1038.53
± approximately
± 2-1074 to (1 – 2- ± 2-1022 to (2 – 2- 10-323.3 to
Double Precision 52)×2-1022 52)×21023 approximately
10308.
Floating point addition and subtraction
floating point addition
For example, we have to add 1.1 * 103 and 50.
We cannot add these numbers directly. First, we need to align
the exponent and then, we can add significand.
After aligning exponent, we get 50 = 0.05 * 103
Now adding significand, 0.05 + 1.1 = 1.15
So, finally we get (1.1 * 103 + 50) = 1.15 * 103
Here, notice that we shifted 50 and made it 0.05 to add these
numbers.
Now let us take example of floating point number addition
We follow these steps to add two numbers:
1. Align the significand
2. Add the significands
3. Normalize the result
Let the two numbers be
x = 9.75 y = 0.5625
Converting them into 32-bit floating point representation,
9.75’s representation in 32-bit format = 0 10000010
00111000000000000000000
0.5625’s representation in 32-bit format = 0 01111110
00100000000000000000000
Now we get the difference of exponents to know how much
shifting is required.
(10000010 – 01111110)2 = (4)10
Now, we shift the mantissa of lesser number right side by 4 units.
Mantissa of 0.5625 = 1.00100000000000000000000
(note that 1 before decimal point is understood in 32-bit
representation)
Shifting right by 4 units, we get 0.00010010000000000000000
Mantissa of 9.75 = 1. 00111000000000000000000
Adding mantissa of both
0. 00010010000000000000000
+ 1. 00111000000000000000000
1. 01001010000000000000000
In final answer, we take exponent of bigger number
So, final answer consist of :
Sign bit = 0
Exponent of bigger number = 10000010
Mantissa = 01001010000000000000000
32 bit representation of answer = x + y = 0
10000010 01001010000000000000000
FLOATING POINT SUBTRACTION
Subtraction is similar to addition with some differences like we subtract
mantissa unlike addition and in sign bit we put the sign of greater number.
Let the two numbers be
x = 9.75
y = – 0.5625
Converting them into 32-bit floating point representation
9.75’s representation in 32-bit format = 0 10000010
00111000000000000000000
– 0.5625’s representation in 32-bit format = 1 01111110
00100000000000000000000
Now, we find the difference of exponents to know how much shifting is required.
(10000010 – 01111110)2 = (4)10
Now, we shift the mantissa of lesser number right side by 4 units.
Mantissa of – 0.5625 = 1.00100000000000000000000
(note that 1 before decimal point is understood in 32-bit representation)
Shifting right by 4 units, 0.00010010000000000000000
Mantissa of 9.75= 1. 00111000000000000000000
Subtracting mantissa of both
0. 00010010000000000000000
– 1. 00111000000000000000000
1. 00100110000000000000000
Sign bit of bigger number = 0
So, finally the answer = x – y = 0 10000010 00100110000000000000000
