Test: Floating Point Number- 2 - Computer Science Engineering (CSE) MCQ

Rounding to plus and minus infinity are useful in implementing a technique known as interval arithmetic. Interval arithmetic provides an efficient method for monitoring and controlling errors in floating point computations by producing two values for each result. The two values correspond to the lower and upper endpoints of an interval that contains the true result. The width of the interval, which is the difference between the upper and lower endpoints, indicates the accuracy of the result of the endpoints of an interval are not representable then the interval endpoints are rounded down and up respectively.

The decimal number is 0.239 x 2¹³
We have to find hexadecimal representation without normalization.
Biased exponent = 13 + 64 = 77
Representing 77 in binary
(77)₁₀ = (1001101)₂ 
Representing mantissa in binary

(0.239)₁₀ = 0.00111101000101 
Floating point representation is as follows:

The decimal number is 0.239 x 2¹³
We have to find hexadecimal representation with normalization.
Binary representation of 0.239 is 0.00111101000101 
∴ N is (0.00111101000101) x 2¹³
After normalization we have 1.11101000101 x 2¹⁰
Then, biased exponent - 10 + 64 = 74 Representating biased exponent 74 in binary (74)₁₀ = (1001010)₂
• Floating point representation of number is as follows-:

No overflow

Binary subtraction is like decimal subtraction:
0 - 0 = 0, 1 - 1 = 0, 1 - 0 = 1,
0 - 1 = 1 with 1 borrow.

Sign bit is 1 implies number is negative. Exponent bits: 10000011 Exponent is added with 127 bias in IEEE single precision format.
So, Actual exponent
= 10000011 - 127 
= 131 - 127 = 4
Mantissa bits: 101000000000000000000000
In IEEE format, an implied 1 is before mantissa.
Hence the Number is -1.101*2⁴
= - (11010)₂ = - 26

An exponent of 0 together with a fraction of 0 with positive sign represents +0 which is a special value.

Apply Booth’s algorithm:

4 additions and 4 substractions total 8 operations.

2’s complement of P is given as (F87B)₁₆
2’s complement of 8 x P = 2³ x (F87B)₁₆

Range of 2’s complement no is
(-2^n-1 to 2^n-1 - 1)
So minimum no represents by 8 bit is
- 2^{8 - 1} = - 128
maximum no represented by 8 bit is
2^{8 - 1} - 1 = 127

Let n = 5 bit
f = 2 bit
So i = 5 - 2 = 3 bit
Min value : 000.00
⇒ 0
Max value : 111.11
⇒ 7.75 
[2ⁱ - 2^-f]
⇒ [2³ - 2^-2]
⇒ 8 - 0.25 = 7.75