I am trying to answer this question:
Consider a computer with a byte-addressable memory. A 40-bit memory address is divided as follows for cache processing. First, the 8 low-order bits are chopped off to expose the cache-line number. Second, the next 17 low-order bits are inspected to get the cache-container index. Third, the remaining 15 bits are used as the cache tag. Hint: What do the direct-mapped and set-associative placement formulas have in common?
What is the cache size in bytes?
I've figured out that the cache line size is 2^8 bytes and that there are thus 2^32 cache lines in memory. Furthermore I know that there are 2^17 cache sets in the cache. To compute the capacity of the cache, I am using the equation
capacity = line size x #of sets x associativity
However, even using the tag field, I can't see how I can find the associativity of the cache here.