- #1
hholzer
- 37
- 0
If I have the addressing format:
Tag: 31-12
Index: 11-6
Byte offset: 5-0
And if my cache size is 32KB = 32*2^10 = 2^5*2^10 = 2^15
then this would represent a how many-way set associative
cache? i.e., an x-way associative cache?
Since bits 11-6 are used for the index, then this would
mean that there are 6 bits used for the number of sets:
2^6 sets but how many blocks per set? If I have
determined this correctly, it would be:
2^6 = (2^15)/(x*2^6) or 2^3 = x ?
So it would represent an 8-way set associative cache.
Then, the number of words in a block would be:
2^6/2^2 = 2^4? That doesn't seem right to me.
Tag: 31-12
Index: 11-6
Byte offset: 5-0
And if my cache size is 32KB = 32*2^10 = 2^5*2^10 = 2^15
then this would represent a how many-way set associative
cache? i.e., an x-way associative cache?
Since bits 11-6 are used for the index, then this would
mean that there are 6 bits used for the number of sets:
2^6 sets but how many blocks per set? If I have
determined this correctly, it would be:
2^6 = (2^15)/(x*2^6) or 2^3 = x ?
So it would represent an 8-way set associative cache.
Then, the number of words in a block would be:
2^6/2^2 = 2^4? That doesn't seem right to me.