CS-343 Assignment 10 Solutions

• Byte addressable
• 8 bytes per word
• 8 GB primary memory
• 2-way set associative cache
• 8 words per cache line
• 2 MB cache capacity
• cache access time is 500 psec
• miss penalty time is 5 nsec

Questions

1. Number of blocks of primary memory

   total bytes ÷ bytes per block = total blocks
   (23 × 230) ÷ (23 × 23) = 233 ÷ 26 = 227 blocks.

2. Number of cache lines

   cache capacity ÷ bytes per line = number of lines
   (21 × 220) ÷ (23 × 23) = (221 ÷ 26) = 215 lines.

3. Number of cache sets

   Number of lines ÷ lines per set = number of sets
   215 ÷ 21 = 214 = 16,384 sets.

4. Primary memory address size

   log2(number of addressable bytes) = 33 bits

5. Size of byte offset field

   log2(bytes per word) = 3 bits

6. Size of block offset field

   log2(words per block|line) = 3 bits

7. Size of index field

   log2(number of sets = 14 bits

8. Size of tag field

   Address width - (index + block offset + byte offset) = 13 bits

9. p(hit) if the CPU were to access memory totally randomly

   Proportion of main memory that fits in cache = 221 233 = 22−12 = 0.0002441.

10. Average access time if the CPU were to access memory totally randomly

    p(hit) × 500 psec + p(miss) × (500 psec + 5 nsec) = 0.0002441 × 500 + .9997559 × 5500 = 5.4987794 nsec
    This is correct: the hit time has to be added to the miss penalty on a miss. Other semesters, the “miss time” = was used, which would have been given as 5.5 nsec for this example. You will not have to worry about which number to use on the final: both values will be given to avoid confusion.

11. Average access time if p(hit) is 0.97

    0.97 × 500 + 0.03 × 5,500 = 650 psec.