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Sagot :
Time taken to access the average memory
When direct mapping is utilized, 1. when the block has a 16-byte size.
Given that the memory access time (m) is 20ns and the cache access time (Tc) is 10ns, the cache miss rate is 10%, or 0.1(1-H).
AMAT = HTc + (1-H)(Tc+m) Cache hit rate (H) = 0.9
=HTc + Tc + m -HTc - Hm = Tc + (1-H)m = 10 ns + 0.1 x 20 = 10 ns + 2 = 12 ns
2. direct mapped cache with a 32-byte block size.
Given that the access time to the cache and memory is equal to 10 nanoseconds,
Because of the 25% reduction in the question's cache miss rate, the cache miss rate (1-H) is 0.075. The new miss rate is 75% of 0.1, or 0.075.
Tc + (1-H)m' = 10+0.075 x 25 = 10 + 1.875 = 11.875 ns is what AMAT is.
when there is two-set associativity.
1. when a 16-byte block is in use.
The question indicates that the time it takes to access the new cache goes up by 20% of 10ns.
Access time to the new cache, T'c=12ns
The rate of misses went down by 40%, so the new miss rate is 60% of 0.1. which equals 0.006 (1-H').
AMAT = T'c + (1-H')m.
AMAT is equal to 13.2 ns for 12 ns plus 0.06 x 20.
2. when a cache block is 32 bytes in size.
Access time to memory (m') equals 25 nanoseconds.
By increasing the block size, the miss rate is reduced by 25%.
Therefore, the miss rate is 0.045 = (1-H')/75% of 0.06
AMAT is 12 + 0.45 * 25 = 12 + 1.125, or 13.125 ns.
To learn more about average memory here
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