Quiz submission record for quiz6-2-1 at Wed Jul 28 10:20:08 2004:

Your Answer for Question 1:
53 = 32'b000000000000000000000000000111001

8 words = 32 bytes = 2^5, so five bits in offset.  if it is 16KB, then 2^14/2^5 = 2^9, so 9 bits in the index.  This puts the contents at this memory address at the 27th byte of the second block of the chache with a remaining tag of zero

Your Answer for Question 2:
this way it can keep the j entirely in a lower level and hence faster cache while it's doing the iterations on it.  Because accesses to i are spaced out by the loop of j, putting it in a slower cache doesn't affect performance as much.


Your unique submission ID is quiz6-2-1-cs61c-bg-1091035208-1650.