CS 3853 Computer Architecture Recitation 8 Answers

Quiz 6a

1. 16K = 2¹⁴, so 14 bits for the page offset. Bits for the page number: 52 - 14 = 38.
2. 32 = 2⁵, so 5 bits for the block offset.
   Number of blocks = 8K/32 = 256. Number of sets = 256/2 = 2⁷, so 7 bits for the index.
3. 43 - 5 - 7 = 31 bits for the tag.

Quiz 6b

1. 32K = 2¹⁵, so 15 bits for the page offset. Bits for the page number: 50 - 15 = 35.
2. 128 = 2⁷, so 7 bits for the block offset.
   Number of blocks = 32K/128 = 256. Number of sets = 256/4 = 2⁶, so 6 bits for the index.
3. 42 - 7 - 6 = 29 bits for the tag.

Problems

1. 1. 64 (virtual address size)
   2. 49 (a - c)
   3. 15 (page size is 32K = 2¹⁵)
   4. 43 (b - e)
   5. 6 (64 = 2⁶, TLB is direct mapped)
   6. 8 (c - g)
   7. 7 (128 = 2⁷, cache block size)
   8. 43 (same as d)
   9. 27 (number of bits in a frame number = m - c)
   10. 27 (same is i)
   11. 1024 (bits in an L1 cache block = 128*8)
   12. 27 (same as i)
   13. 42 (physical address size)
   14. 20 (m - o - p)
   15. 15 (number of blocks in L2 cache is 8MB/128 = 64K, number of sets = 32K = 2¹⁵)
   16. 7 (same as g since both caches have the same block size)
   17. 20 (same as n)
   18. 1024 (bits in an L2 cache block)
2. 1 + .02 × 20 = 1.4
3. Time to transfer 64 bytes: 64/10,000,000,000 = 64 × 10^-10 seconds = 6.4 ns, so the total time is 36.4 ns.
4. 1 + 1.2 × .02 × 30 = 1.72