Question

Consider the following memory hierarchy.

Unit Access Latency Local Miss Rate

-----------------------------------------------------------------------

L1 1.5 cycles 80%

L2 15 cycles 65%

Memory 150 cycles 100%

What is the average memory access time?

(Hint: AMAT= hit time + miss rate * miss penalty)

Answer 1

Given

L1 : 1.5 clock cycles and hit rate = 80% so , miss rate is 20%

L2 : 15 clock cycles and hit rate = 65% so , miss rate is 35%

Memory : 150 cycles and 100%

So , miss penalty of L1 = hit time of L2 + miss time of L2* miss penalty of L2 = 15 + 0.35*150 - - - - ( multiplied by 150 because  if L2 misses it has to go to memory )

= 67.5 - - - - (a)

We know average memory access time = hit time + miss rate * miss penalty

= 1.5 + 0.2*(67.5 ) - - - from (a)

= 1.5 + 13.5

= 15 cycles.   

Note : Please do ask any queries regarding the solution , I will be glad to help.