2018 exam, 2 c)

Question

Hi,

In ziele 3 after the first iteration, the sets don't make sense to me.

For i=1 we have t2 = Mem[32]. Then, 32 = 2 * 16 + 0 . So normally, the set would be 16 here, but since that not a valid set, what do we do?

Answer 1

For the 256 bytes, we need 8-bit addresses. The caches have 32 bytes and a block has 2 bytes, so that each cache has 16 blocks. 

The direct mapped cache therefore needs 4 bits for set addresses so that a memory address a[7..0] is split into the tag a[7..5], the set address a[4..1] and the block offset a[0]. The set address a[4..1] is therefore computed as setAdr(adr) := (adr % 32) div 2. We therefore have

• setAdr(30) = 15
• setAdr(31) = 15
• setAdr(32) = 0
• setAdr(33) = 0

Hence, the set address is 0 and not 16 as you assumed. 

For the 4-way set associative cache, we have 4 blocks in a set and therefore each set has 8 bytes, so that the cache has only 4 sets. To that end, we need 2 bits as set addresses, so that a memory address a[7..0] is split into the tag a[7..3], the set address a[2..1] and the block offset a[0]. The set address a[2..1] is therefore computed as setAdr(adr) := (adr % 8) div 2.