Question

You have a K-Way set associative cache with following parameters

Cache Data Size (C) 2048
Block Size (b) 16
Blocks/Ways per set (K) 4

After partitioning the address, which is 32 bits big, into Tag, Set, and Offset, how many bits will be in each field?

Answer 1

Following are the solution to the given question:

Step-by-step explanation:

The size of cache memory
`= 2048 (2^(11))`

The block size
`=16 (2^4)`

Therefore,
`(2048)/(16)=128 (2^7)`are blocks.

The collection consists of 4 blocks. Therefore,
`(128)/(4) =32`

are sets. Because the size of its cache block is 16, the block offset comprises 4 bits. There would be 5 bits inset offset since there are 32 caches. An Address contains 32 bits, while the space of the tags contains
`32- (4+5)=23` bits.