Memory Hierarchy Q.31

0. Consider a system with 2 level cache. Access times of Level 1 cache, Level 2 cache and main memory are 1 ns, 10 ns, and 500 ns, respectively. The hit rates of Level 1 and Level 2 caches are 0.8 and 0.9, respectively. What is the average access time of the system ignoring the search time within the cache?

Answer
Comment
Array

Option : C
Explanation : First, the system will look in cache 1. If it is not found in cache 1, then cache 2 and then further in main memory (if not in cache 2 also).
The average access t i me would take into consideration success in cache 1, failure in cache 1 but success in cache 2, failure in both the caches and success in main memory.
Average access time
= [H₁ * T₁] + [(1 – H₁) * H2* T2]+ [(1 – H₁)(1 – H2) * Hm * Tm]
where,
H₁ = Hit rate of level 1 cache = 0.8
T₁ = Access time for level 1 cache = 1 ns
H2 = Hit rate of level 2 cache = 0.9
T2 = Access time for level 2 cache = 10 ns
Hm = Hit rate of Main Memory = 1
Tm = Access time for Main Memory = 500 ns
So, Average Access Time
= (0.8 * 1) + (0.2 * 0.9 * 10) + (0.2 * 0.1 * 1 * 500)
= 0.8 + 1.8 + 10
= 12.6 ns

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