CS 1557

Assignment #6 - Solution

 

8.2 XOR truth table

Ci-1 Ci

0 0 | 0 clearly not overflow

0 1 | 1 overflow

1 0 | 1 overflow

1 1 | 0 not overflow

 

yes

  

8.6

true, for integer arithmetic, the remainder of J/K is truncated.

 

 

8.8

Strengths

Weaknesses

8.10

total bits = P

1 sign bit,

biased Exponent bits = ciel(log (X+1))

significand bits = P-1- ciel(log (X+1))

smallest exponent is b-q

largest exponent is b X-q

a) Smallest: 2-(P-1- ciel(log (X+1))). b-q

Largest: (1 – 2-(P-1-logX)).b(x-q)

b) normalized significand.

smallest: 0.5 . b-q

largest: (1 – 2-(P-logX). b (X- q)

 

 

8.11

Note: all numbers are normalized

base 2, bias 127

a)

-5

1 10000001 01000000000000000000000

b)

-6

1 10000001 10000000000000000000000

c)

-1.5

1 01111111 10000000000000000000000

d)

384

0 10000111 10000000000000000000000

e)

1/16

0 01111011 00000000000000000000000

f)

-1/32

1 01111010 00000000000000000000000

8.13 a) 25

b) 86