Homework 7 solution

Problem 1

8G × 512 = 2³³ × 2⁹ = 2⁴² = 4T
16G / 256 = 2³⁴ / 2⁸ = 2²⁶ = 64M
4k × 2M = 2¹² × 2²¹ = 2³³ = 8G
log₂(64M) = log₂(2²⁶) = 26

Problem 2

An 2 GB memory has a 32-bit word size. How many words are contained in this memory?
- 2 GB = 2³¹ bytes = 2³¹ × 2³ bits = 2³⁴ bits
- 2³⁴ bits / (2⁵ bits/word) = 2²⁹ words = 512M words
A memory has 8 G words. Each word contains 64 bits. How many bytes are contained in this memory?
- 8 G words = 2²³ words
- 64 bits = 2⁶ bits
- 2²³ words × (2⁶ bits/word) = 2²⁹ bits
- 2²⁹ bits = 2²⁶ bytes = 64 GB
How many bits are required to address the 8 G words of this memory?
- 8 G = 2³³
- log₂(2³³) = 33

Problem 3

Express the following base 10 numbers in 16-bit fixed-point twos-complement representation with 8 integer bits and 8 fraction bits. Some of the numbers can not be expressed exactly, so give the nearest fixed-point representaion.

13.5 is 00001101.10000000 or 0000110110000000 in fixed-point
100.05 is approximately 01100100.00001100 or 0110010000001100 in fixed-point
- 0000110110000001 in fixed-point is 100.046875
100.05 is approximated by 01100100.00001101 or 0110010000001101 in fixed-point

0000110110000000 in fixed-point is 100.05078125

0.3333333333333333333333333333333333333333333333333 is approximately 00000000.01010101 or 0000000001010101 in fixed-point
14 is 00001110.00000000 or 0000111000000000 in fixed-point

Problem 4

Express the following 16-bit fixed-point twos-complement numbers with 8 integer bits and 8 fraction bits as base 10 decimal numbers.

0000100011001100 in fixed-point is 00001000.11001100 or 8 + 204/256 or 8.796875

1100100011001100 in fixed-point is 11001000.11001100 is -56 + 204/256 or -55.203125

Problem 5

What is the 32-bit IEEE 754 representation for 3.5?

In binary fixed point, 3.5 is 11.1
Normalized, 11.1 is 1.11 × 2¹
IEEE 754 sign field is 0 (for positive)
IEEE 754 exponent field is 127+1 or 10000000 in binary
IEEE 754 fraction field is 127+1 or 11000000000000000000000
IEEE 754 representation is 01000000011000000000000000000000 or 0b01000000011000000000000000000000 (Java SE 7) or 0b0100_0000_0110_0000_0000_0000_0000_0000 (Java SE 7) or 0x40600000 (C and Java) or 40600000 (hex)

Problem 6

If 0x42510000 is the 32-bit IEEE 754 representation of a floating point number, what is the floating point number?

0x42510000 is 01000010010100010000000000000000
For IEEE 754, can be divided into 0.10000100.10100010000000000000000
- 0 means number is positive
- 10000100 means exponent field is 132 and binary exponent is 132-127 or 5 so exponent multiplier is 2⁵ or 32
- 10100010000000000000000 means fraction is 1.10100010000000000000000 or 1+1/2+1/8+1/128 or 1+64/128+16/128+1/128 or 1+81/128 or 1.6328125
Result is +32×1.6328125 or 52.25