total cache size = (2^(index bits)) * [(associativity)(tag bits + valid bits + data size)]
(Correct me if this is wrong)
The Attempt at a Solution
I came up with 2^11 cache sets, therefore it's 8-way associative because 2^14 / 2^3 = 2^11. And 8=2^3.
Therefore 8 comparators, 10 bits each because the tag is 10 bits.
I'm stuck on the last part where it asks to determine the size of the cache because I'm not sure what the data size is. From what I have so far the equation would be:
2^11 * (8(10+(1+1+1+1)+data size))
If this is correct, can someone tell me how the data size is determined so that I can solve that equation?