Average access time in two level cache system
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29-09-2020 - |
Question
In a two-level cache system, the level one cache has a hit time of 1 ns (inside the CPU), hit rate of 90%, and a miss penalty of 20 ns. The level two cache has a hit rate of 95% and a miss penalty of 220 ns. What is the average memory access time?
What is a two-level cache system and how to calculate the time required? Since the hit time of level two cache is missing...
Solution
First Let's define all given things
- L1 Cache hit: 90% & Time: 1ns
L2 and RAM has hit rate 10% out of which 95% which is 9.5% in total
- L2 Cache hit: 9.5% & Time 20ns
RAM has hit rate 5% of 10% = 0.5%
- RAM hit: 0.5% & Time 220ns
Formula Avg Memory Access Time
AMAT = L1_hit * L1_T + L2_hit * L2_T + RAM_hit * RAM_T
AMAT = 0.9*1 + 9.5*20 + 0.5*220
AMAT = 300.9ns
What is 2 level cache system:
First cache called L1 is reside on CPU is too fast. When CPU needs data, it checks in L1 cache but if it is not there it will go to L2. L2 cache is sometimes on CPU or outside CPU it depends on architecture of CPU. It does similarly and here also it missed then it fetch data from RAM which is expensive process.