Set Associative Cache: 8-Way Determination

[hholzer]

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.

 

[Future Bruno]

Can you help me confirm this?No, the number of words in a block is determined by the byte offset (bits 5-0), which indicates the number of bytes. So if the byte offset has 6 bits, that means there are 2^6 = 64 bytes per block, or 16 words per block.