Given,
Total cache size = 16 KB = $2^{14}$ bytes
word length = 32 bits, i.e. 4 bytes or $2^2$ bytes.
Since, the word length is given, I am assuming memory to be word-addressable, and not byte-addressable.
So, we have $2^{12}$ words cache size.
1 block is 8 words, so block offset = $\log 8 = 3$ bits.
cache size(in terms of blocks) = $2^{12} / 8$ = $2^9$ blocks.
Given the cache is 2-way set associative. So,
2 blocks - 1 set
$2^9$ blocks - $2^8$ sets.
so, we need 8 bits to identify the set.
The physical address space is 4GB,i.e. $2^{32}$ bytes, or in terms of words, $2^{30}$ words.
So, the physical address will be of 30 bits.
Out of 30, 3 bits are used for offset, and 8 bits for the set field.
Therefore, tag bits are 30 - 3 - 8
= 19 bits.
Option (B) 19, 8
