CS 3853 Computer Architecture Recitation 8 Answers

Problems

1. 7 bits for the offset, 9 bits for the index, and the rest (16) for the tag.
2. Then index is now 8 bits and the tag is 17 bits.
3. There is no index (0 bits), the offset is still 7 bits and the tag is 32-7 = 25 bits.
4. 64K/32 = 2K = 2¹¹, so 11 bits are needed.
5. None, a write-through cache does not need a dirty bit.
6. The first has a time per instruction of 1.21 × .4 = .484 ns and the second has a time per instruction of 1.32 × .38 = .5016, so the first is better.
7. Time = .5 + .05(1.3 + .01 × 20) = .575 ns