Problem 5 (20 points) Consider a floating point number representation that is 16 bit wide. The...

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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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