Consider a direct mapped cache of size 32 kbyte with block

Home » Computer Organization and Architecture » Computer organization and architecture miscellaneous » Question

Consider a direct mapped cache of size 32 kbyte with block size 32 byte. The CPU generates 32 bit address. The number of bits needed for cache indexing and the number of tag bits are respectively

1. 10, 17
2. 10, 22
3. 15, 17
4. 5, 17

Correct Option: A

As given
Cache size = 32 kbyte = 32 × 2¹⁰ byte
= 2¹⁵ byte = 15 bit
Therefore, size of tag = 32 – 15 = 17 bit
Since, 32 byte of block is used by cache, so there actually 1 k blocks and for indexing that we need 10 bit.