Answer:
3 bits
Explanation:
Given a 4- way set associative cache that has 64 blocks of 16 words.
Therefore, the number of sets cache has:
![\frac{64}{4} = 16](https://tex.z-dn.net/?f=%5Cfrac%7B64%7D%7B4%7D%20%3D%2016%20)
Now,
Cache data size is 16kB
The number of cache blocks can be calculated as:
![16kB/16 = 1024 bytes/16 byte\times 16 = 256 cache blocks](https://tex.z-dn.net/?f=16kB%2F16%20%3D%201024%20bytes%2F16%20byte%5Ctimes%2016%20%3D%20256%20cache%20blocks)
Now,
Total sets = ![\frac{cache blocks}{associative sets}](https://tex.z-dn.net/?f=%5Cfrac%7Bcache%20blocks%7D%7Bassociative%20sets%7D)
Total sets = ![\frac{256}{4} = 64](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B4%7D%20%3D%2064)
Now,
![2^{n} = 64](https://tex.z-dn.net/?f=2%5E%7Bn%7D%20%3D%2064)
n = 6
For 15 bit address for the architecture, the bits in tag field is given by:
15 - (6 + 6) = 3 bits
Thus the tag field will have 3 bits
Answer:
I think option C is correct
Answer:
B
Explanation:
Key words could also mean special meaning