From Schaum's Comp. Arch (2)

[wood]

Interesting one:

How many bits of storage are required for the tag array of a 32-KB cache with 256-byte cache lines and four-way set-associativity if the cache is write-back but does not require any additional bits of data in the tag array to implement the write-back policy? Assume that the system containing the cache uses 32-bit addresses.

[rafi_dery]

I hope I do not confuse bits with bytes...

number of cache-lines is: 128 (32KB/256B)
tag size is: 32-7-8 = 17bit (32 bit address minus 7bit due to 128 lines minus 8bit due to line-size of 256B).

number of tags = number of lines = 128
size of tag array = 17*128 = 2176b

[wood]

Good try... You forgot to take into account the four-way set-associativity characteristic of the cache.

You're right about needing 8 bits due to the line size. However, you don't need 7 bits for the 128 lines, since you have four entries for each set, and therefore just 5 bits is needed to address 128/4 = 32 sets. So, you have 32 - 13 = 19 bits long (* 128 = 2432 bits).

The official answer throws 2 more bits (making it 21 and thus 2688 bits), arguing that it needs it for the valid and dirty bits. However, the question explicitly mentions not to count those extra bits needed to implement a write-back cache. That's why the answer confused me a bit.

Wood

[rafi_dery]

Sure, I completely forgot the sets matter...
Thanks!

[wood]

But what do you think... is the book correct in adding the two extra bits it supposedly asked not to??