Chapter 4, Problem 15E

a)

Explanation of Solution

Lowest and highest addresses with byte addressable memory:

From given 2²⁰ memory with byte addressable length. The format of this memory is $\text{1M×8}$ where 1M is the length of the memory.

• The lowest and highest address of given address would be 0 through 2²⁰-1

b)

Explanation of Solution

Lowest and highest addresses with word addressable memory in 16-bit word:

The following formula used to find the required number of bits.

From given 2²⁰ memory with word addressable length of 16 bits to a word. So, the length of the word calculated is as follows:

$\text{L=}\frac{{\text{2}}^{\text{20}}\text{×8}}{\text{16}}{\text{=2}}^{\text{19}}$

• Hence the length of the word is 2¹⁹

c)

Explanation of Solution

Lowest and highest addresses with word addressable memory in 32-bit word:

The following formula used to find the required number of bits.

From given 2²⁰ memory with word addressable length of 32 bits to a word. So, the length of the word calculated is as follows:

$\text{L=}\frac{{\text{2}}^{20}\text{×8}}{\text{32}}{\text{=2}}^{18}$

• Hence the length of the word is 2¹⁸