Question

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

Answer 1

Answer:

Following are the solution to the given question:

Explanation:

The size of cache  memory $= 2048 (2^{11})$

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

Therefore, $\frac{2048}{16}=128 (2^7)$are blocks.

The collection consists of 4 blocks. Therefore,$\frac{128}{4} =32$

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.