Chapter 6, Problem 10E

a)

Explanation of Solution

Given:

• The size of the cache is 2²¹; the size of the address bit will be 21.
• The number of address bit is considered to n when the size of the cache is 2ⁿ.
• Cache contains 64 blocks
• Each block contains 4 bytes.

Set field:

The below is used to calculate the number of bits present in the set field.

$\begin{array}{c}\text{Number of set}\text{=}\frac{\text{Number}\text{of}\text{blocks}}{\text{Associativity}}\\ \text{=}\frac{64}{\text{2}}\\ \text{=32}\\ {\text{=2}}^5\end{array}$

Therefore, 5 bits are required for the set field

b)

Explanation of Solution

Given:

• The size of the cache is 2²¹; the size of the address bit will be 21.
• The number of address bit is considered to n when the size of the cache is 2ⁿ.
• Cache contains 64 blocks
• Each block contains 4 bytes.

Set field:

The below is used to calculate the number of bits present in the set field.

$\begin{array}{c}\text{Number of set}\text{=}\frac{\text{Number}\text{of}\text{blocks}}{\text{Associativity}}\\ \text{=}\frac{64}{4}\\ \text{=16}\\ {\text{=2}}^4\end{array}$

Therefore, 4 bits are required for the set field