Question

A direct mapped cache is of size 32 KB and has block size 32 Bytes. CPU also generates 32 bit address. Number of bits needed for indexing the cache:
1. 14
2. 15
3. 10
4. 17

Answer 1

Correct Answer - Option 3 : 10

The correct answer is option 3

Data

Memory address = 32 bits

Data Cache = 32 KB = 215 B

Block Size = 32 B = 25 B

Formula:

In a direct Mapped, the main memory can be represented as

Tag

Index

Block offset

32 = tag + lines + Block offset (In bits)

Calculation:

Assume byte-addressable:

\({\rm{number\;of\;blocks\;in\;cache\;memory}} = \frac{{{2^{15}}}}{{{2^5}}} = {2^{10}}\)

So, the number of bits required to index the cache blocks =⌈ log₂ 2¹⁰ ⌉ = 10