Assume a memory access to main memory on a cache "miss" takes 30 ns and a memory access to the cache (on a cache "hit") takes 3 ns. If 80% of the processor's memory requests results in a cache "hit", what is the average memory access time?

Assume a memory access to main memory on a cache "miss" takes 30 ns and a memory access to the cache (on a cache "hit") takes 3 ns. If 80% of the processor's memory requests results in a cache "hit", what is the average memory access time? Correct Answer 9 ns

The correct answer is option 2.

Concept:

The cache is a random access memory used by the CPU to reduce the average time taken to access memory. AMAT(average memory access time) uses hit time, miss penalty, and miss rate to measure memory performance.

The given data

Memory access to the cache T1 = 3 ns

Cache miss time T₂ = 30 ns

Cache hit H₁ = 80% = 0.8

Cache miss M₁ = 1-H₁ = 20% = 0.2

Formula:

Average memory access time = H₁T₁ + M₁(T₁ +T₂)

AMAT = H1T1 + (1-H₁)(T1 + T2)

Solution:

AMAT=0.8 x 3 + (1 - 0.8)(3 + 30)

AMAT=2.4 + 6.6

AMAT=9 ns

Hence the correct answer is 9 ns.