Answer:
Following are the solution to the given question:
Explanation:
The size of cache memory [tex]= 2048 (2^{11})[/tex]
The block size[tex]=16 (2^4)[/tex]
Therefore, [tex]\frac{2048}{16}=128 (2^7)[/tex]are blocks.
The collection consists of 4 blocks. Therefore,[tex]\frac{128}{4} =32[/tex]
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 [tex]32- (4+5)=23[/tex] bits.
