Problem 3: Data Representation (20 points 16 points) Consider a 5-bit floating-p

Question

Problem 3: Data Representation (20 points 16 points) Consider a 5-bit floating-point based on the IEEE foating point format with one sign bit, two exponent bits, and two fraction bits. S-1 (1 point) What is the bias in this representation (3 points) Provide the following information for the largest normal positive number in this representation? sign bit exponent bits mantissa bits (3 points) Provide the following information for the largest denormal positive number in this representation? sign bit O exponent bits mantissa bits: (3 points) Provide e following information for the smallest normal positive number in this representation? sign bit exponent bits: mantissa bits (3 points) Provide the following information for the smallest denormal positive number in this representation? sign bit exponent bits mantissa bits (3 points) Provide the following information for the number One in this representation? sign bit exponent bits mantissa bits:

Explanation / Answer

Given to consider
5 bit floating point representation based on the IEEE floating point format with
One sign bit
Two exponent bits
Two fraction bits

* Bias in this representation

--> 2^(k-1) - 1 (Here, k is number of exponent bits)
--> 2^(2-1) - 1
--> 2^1 - 1
--> 2 - 1 = 1

* Largest normal positive number

s = 0 (positive) , exp!=00 (minimum) and exp!=11,
So, largest two bit binary number is "10" exp=(10) = 2^1 = 2.
E= exp-bias = 2 - 1 = 1 , so fraction = (01)
M = 1 + fraction * 2^(-2) = 1 + 0.01 = 1.01

largest Normalized positive Value,

V = (-1)^s * M * 2^E
V = (-1)^0 * 1.01 * 2^1
V = (1.01) * 2^1   
V = (10.1) base 2

* largest denormal positive value

s = 0, exp =00, fraction = 01
M = fraction * 2^(-2) = 0.01
E = 1 - bias = 1 - 1 = 0

Largest denormalized value

V = M * 2^E
V = 0.01 * 2^0 = 0.01

* Smallest normal Positive value

s=0: positive value
V = (-1)^s * M * 2^E
exp!=00 (minimum) and exp!=11 (max),
So, smallest two bit binary number is "01" exp=(01) = 2^0 = 1.
E= exp-bias = 1 - 1 = 0 ,
M = 1 + fraction * 2(-n) = 1+0 =1.0

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