1. Introduction To Computer Science
CS500
Introduction To Computer Science
Lecture 3

2. Floating-Point Representation
Numbers used in scientific calculations are designed by a sign, by the magnitude of the number, and by the position of the radix point.
The position of the radix point is required to represent fractions, integers, or mixed integer-fraction numbers.
There are two ways of specifying the position of the radix point which are:
Fixed-point representation (which we have studied up till now) .
Floating-point representation.

3. Floating-Point Representation (2)
In Floating-point representation, there are two ways for positioning the radix point.
They are:
1- Putting the radix point at the extreme left of the number.
2- Putting the radix point at the extreme right of the number.
According to the first way, any number can be expressed as a fraction (mantissa) and a positive exponent. (e.g., = * 10+3)
According to the second way, any number can be expressed as an integer and a negative exponent. (e.g., = * 10-3)

4. Floating-Point Representation (3)
In decimal system, the floating-point notation; which is also called “scientific notation”; of any real number can be represented as:
(N)10= .F × 10E
(N)10 : is the real number in decimal.
F : is a fraction(mantissa).
E : is an exponent.
Example 1: The real number can be expressed using floating-point notation
notation as:

5. Floating-Point Representation (4)
There are 4 different formats for representing floating-point binary numbers:
Signed-Magnitude Method
2’s Complement Method
Excess Method
IEEE Standard Method

6. Signed-Magnitude Method
Example 2: Using 14-bit signed-magnitude method, the binary number can be represented with a 6-bit exponent and 8-bit fraction as:
Therefore, the floating-point representation is:

7. Signed-Magnitude Method (2)
Example 3: Using 14-bit signed-magnitude method, the binary number can be represented with a 8-bit fraction and 6-bit exponent as:
= - ( ) x 2-2
Therefore, the floating-point representation is:

8. Signed-Magnitude Method (3)
Example 4: Using 14-bit signed-magnitude method with a 8-bit fraction and 6-bit exponent, calculate the following:
(a) The allowed decimal range for signed floating-point binary numbers.
(b) The forbidden decimal range for signed floating-point binary numbers.
Solution:
(a) The allowed decimal range for signed floating-point binary numbers is the range extended form the lowest negative decimal value to the highest positive decimal value that can be represented by the method. .

9. Signed-Magnitude Method (4)

10. Signed-Magnitude Method (5)
(b) The forbidden decimal range for signed floating-point binary numbers is the range extended form the highest negative decimal value to the lowest positive decimal value that can be represented by the method. .

11. Signed-Magnitude Method (6)
In the allowed decimal range for signed floating-point binary number, the lowest negative decimal value is the negative image of the highest positive decimal value.
In the forbidden decimal range for signed floating-point binary number, the highest negative decimal value is the negative image of the lowest positive decimal value. .

12. 2’s Complement Method
In this method, only the exponent part is expressed using 2’s complement notation.
There is only 1-bit sign exists for the mantissa part (with 0 for + and 1 for -), i.e., if the exponent is (–ve), it is converted using the two’s complement.
If 32-bit 2’s complement representation of floating-point binary number with 1-bit sign, 7-bit exponent and 24-bit fraction, then its format will be:
Example 5: Using 14-bit 2’s complement method, the binary number can be represented with 1-bit sign, 6-bit exponent and 7-bit fraction: x 2-2
Therefore, the floating-point representation is:
where the exponent (-2) is the 2’s complement of (+2)
Sign bit for fraction

13. 2’s Complement Method (2)
Example 6: Using 32-bit 2’s complement representation of floating-point binary number with 1-bit sign, 7-bit exponent and 24-bit fraction, find the following:
(a) The allowed decimal range for signed floating-point binary numbers.
(b) The forbidden decimal range for signed floating-point binary numbers.
Solution: (a) The allowed decimal range: .

14. 2’s Complement Method (3)
(b) The forbidden decimal range: .

15. Excess Method
In this method, only the exponent part is expressed using excess notation. There is only 1-bit sign exists for the fraction part (with 0 for + and 1 for -) and the exponent is converted using the excess notation.
Example 7: Using 14-bit excess method with 1-bit sign, 4-bit exponent and 9-bit fraction, the binary number can be represented as:
Because the exponent takes 4-bit, then it’s in excess-8 notation. Thus, 8 + (+4) = 12 → 1100
Therefore, the excess –based floating-point representation is:

16. Excess Method (2)
Example 8: Using 14-bit excess method with 1-bit sign, 3-bit exponent and 10-bit fraction, the binary number be represented as:
Because the exponent takes 3-bit, then it’s in excess-4 notation. Thus, 4 + (-3) = +1 → 001.
Therefore, the excess-based floating-point representation is:
Example 9: Using 10-bit excess method with 1-bit sign, 4-bit exponent and 5-bit fraction, find the following:
(a) The allowed decimal range for signed floating-point binary numbers.
(b)The forbidden decimal range for signed floating-point binary numbers.

17. Excess Method (3) Solution: (a) The allowed decimal range:
(b)The forbidden decimal range for signed floating-point binary numbers.

18. IEEE Standard Method
In this method, the single precision floating-point number is expressed using 32-bits
The exponent part is expressed using 8-bits in excess-127 notation.
The fraction part is considered to be 1.f , with 1 is hidden and appears only during conversion to decimal. E.g., if fraction is , then fraction would be
If the exponent is (+4), the exponent will be (+4) = 131 ( )

19. IEEE Standard Method (2)
Example 10: Code the decimal value as a binary IEEE floating-point format
Solution:
+ ( ) = + ( ) = + (.101) X 2-9= + (1.01) X 2-10
(-10) exponent in excess-127 = = 117 =
Since the fraction part is considered to be 1.f , with 1 is hidden and
appears only during conversion to decimal, so f = .01
2 X
.00255
.0051
.0102
.0204
.0408
.0816
.1632
.3264
.6528
.3056 1
.6112
.2224
1-bit
Sign of
Fraction
(f)
8-bits Exponent
(e)
23-bit Fraction
the binary pattern of floating-point format of ( ) =
( ) = (3AA00000)16

20. IEEE Standard Method (3)
Example 11: Decode the binary IEEE bit pattern (C3D1E000)16 to its decimal value
Solution:
(C3D1E000)16 =
Since the fraction part is considered to be 1.f , with 1 is hidden and appears only during
conversion to decimal, so f =
e = ( ) in excess-127 to decimal, we will subtract the notation = 135 – 127 = +8
(f) X 2e = - ( ) X 2+8 = - ( ) =
- ( ) =
1-bit Sign of Fraction (f)
8-bits Exponent (e)
23-bit Fraction (f) 1

21. IEEE Standard Method (4)

22. Floating-Point Representation Questions
Example 12: Assuming a binary pattern of length 14-bits is used to store signed numbers as floating point values with 1-sign bit, 5-exponent bits, and 8-fraction bits.
Find the following:
Decode the bit pattern into its equivalent decimal value
Code the decimal value as a binary pattern of floating-point format
If the following floating-point representations are used:
Signed-Magnitude Method
2’s Complement Method
Excess Method

23. Floating-Point Representation Questions (2)
Solution:
Decode the bit pattern into its equivalent decimal value
Signed-Magnitude is used
f = ( )
e = + (00101) = +5
(f) X 2e = - ( ) X 2+5 = - ( ) =
1-bit
Sign of
Exponent
(e)
5-bits Exponent
Fraction
(f)
7-bit Fraction
00101 1

24. Floating-Point Representation Questions (3)
Solution:
Decode the bit pattern into its equivalent decimal value
2’s Complement is used
f = ( )
e = (00101) = +5
(f) X 2e = + ( ) X 2+5 = + ( ) =
1-bit
Sign of
Fraction
(f)
5-bits Exponent
(e)
8-bit Fraction
00101

25. Floating-Point Representation Questions (4)
Solution:
Decode the bit pattern into its equivalent decimal value
Excess is used
f = ( )
e = (00101) in excess-16  to decimal, we will subtract the notation = 5 – 16 = -11
(f) X 2e = + ( ) X 2-11 =
1-bit
Sign of
Fraction
(f)
5-bits Exponent
(e)
8-bit Fraction
00101

26. Floating-Point Representation Questions (5)
Solution:
Code the decimal value as a binary pattern of floating-point format
Signed-Magnitude is used
= - ( ) = X 2+4
4 in 5-bits magnitude only = 00100
1-bit
Sign of
Exponent
(e)
5-bits Exponent
Fraction
(f)
7-bit Fraction
00100 1
the binary pattern of floating-point format of (-12.75) =

27. Floating-Point Representation Questions (6)
Solution:
Code the decimal value as a binary pattern of floating-point format
2’s Complement is used
= - ( ) = X 2+4
(+4) = (00100) in 5-bits 2’s Complement
1-bit
Sign of
Fraction
(f)
5-bits Exponent
(e)
8-bit Fraction 1
00100
the binary pattern of floating-point format of (-12.75) =

28. Floating-Point Representation Questions (7)
Solution:
Code the decimal value as a binary pattern of floating-point format
Excess is used
= - ( ) = X 2+4
(+4) in binary 5-bits excess-16 = (+4) + 16 = 20 = (10100)
1-bit
Sign of
Fraction
(f)
5-bits Exponent
(e)
8-bit Fraction 1
10100
the binary pattern of floating-point format of (-12.75) =