Question

Problem 5 (20 points) Consider a floating point number representation that is 16 bit wide. The leftmost bit is the sign bit, and the next 5 bits from the left make up an exponent (which has a bias of 15). The remainder 10 bits give the magnitude of the number. This representation assumes a hidden 1. Consider the number -1.3215 x 10-1 How doe its rine and acrac cmpare wit a he same number, this time b) How does its range and accuracy compare with that of the same number, this time represented in IEEE 754 format? Justify your answer. c) Calculate the product of the number by itself by hand. d) Briefly describe the steps involved in a computer calculated result of this product

Answer 1

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ANS:

THIS IS ARTHEMATIC FOR COMPUTERS

a. bit representation - 0(sign) 11111 ( Exponent) 1111111111 mantissa

b .Accuracy in floating point representation is governed by number of significand bits, whereas range is limited by exponent.

A bias of (2n-1 – 1), where n is # of bits used in exponent, is added to the exponent (e) to get biased exponent (E), therefor here it would be 2^5-1=15;

Largest number will be 1.111111111×2^(+15)

Smallest = 1.00000×2^(-15)

whereas IEEE 754 has
Largest = 1. 1 1 1 … x 2^ +127 = 2 x 10 ^+38 (aprrox for 32 bit representation)
Smallest = 1.000 … x 2 ^–128 = 1 x 10 ^-38 (aprrox for 32 bit representation)

c & d:

let A = -1.3215 x 10^-1
Step 1. Exponent of A x A = -1 + (-1) = -2
Step 2. Multiply significands
-1.3215 x -1.3215 = 1.74636225
Step 3. Normalize the product
1.74636225 = 1.74636225 x 10^0
Step 4. Round off
A x A = 1.7463 x 10^0
Step 5. Decide the sign of A x B (- x - = +)
So, A x B = + 1.7463 x 10^0

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