Question

Number of blocks in main memory is 4M and number of blocks in cache memory is 4K. If a cache tag directory entry contain 1 valid bit and 1 modified bit along with the address tag, then the size of cache tag directory (in KB) is ___________for fully - associative cache mapping.

• A. 11
• B. 13
• C. 12
• D. 14

Answer 1

The correct option is C 12

option (b)

In fully associative mapping


$\text{If no. of blocks in main memory}=4M=2^{22}$

$\text{then main memory block number}=lo{g}_2\left(2^{22}\right)=22-bits$

1 cache tag directory entry size
= 22 + 1 + 1 = 24-bits = 3B

Cache tag directory size
= number of blocks in cache * 1 entry size

= 4K * 3B

=12KB