Question

a) Calculate the AMAT for a cache system with one level of cache between the CPU and Main
Memory. Assume that the cache has a hit time of 1 cycle and a miss rate of 11%. Assume
that the main memory requires 300 cycles to access (this is the hit time) and that all instructions
and data can be found in the main memory (there are no misses).
b) Let us modify the cache system from part (a) and add an L2 cache between the L1 cache and
the main memory. If this L2 cache has a hit time of 22 cycles and a miss rate of 4%, then what
is the new AMAT?
c) Calculate the speedup obtained by using the modified cache system from part (b) instead of
the original cache system from part (a).

Answer 1

Answer a

AMAT = Cache Hit Time + Cache Miss Rate * Miss Penalty

= 1 + 0.11*300

= 1 + 33

= 34 Cycles

Answer b

AMAT = L1 Cache Hit Time + L1 Cache Miss Rate * L1 Miss Penalty

= L1 Cache Hit Time + L1 Cache Miss Rate * (L2 Cache Hit Time + L2 Cache Miss Rate * L2 Miss Penalty )

= 1 + 0.11*(22+0.04*300)

= 1 + 0.11*34

= 4.74 Cycles

Answer c

Speedup = 34/4.74 = 7.17 times