Given a computing system with two levels of cache (L1 and L2) and a main…
2025
Given a computing system with two levels of cache (L1 and L2) and a main memory. The first level (L1) cache access time is 1 nanosecond (ns) and the “hit rate” for L1 cache is 90% while the processor is accessing the data from L1 cache. Whereas, for the second level (L2) cache, the “hit rate” is 80% and the “miss penalty” for transferring data from L2 cache to L1 cache is 10 ns. The “miss penalty” for the data to be transferred from main memory to L2 cache is 100 ns.
Then the average memory access time in this system in nanoseconds is ___________ . (rounded off to one decimal place)
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Correct answer: 4
Key formula: Average Memory Access Time (AMAT) = L1 access time + L1 miss rate × L1 miss penalty
Here, the L1 miss penalty equals the time to access L2 plus the additional cost if L2 misses (i.e., fetching from main memory). So:
L1 access time = 1 ns
L1 miss rate = 1 - 0.90 = 0.10
L2 access time (penalty when L2 hits) = 10 ns
L2 miss rate = 1 - 0.80 = 0.20
Main memory penalty (on L2 miss) = 100 ns
Compute L1 miss penalty:
L1 miss penalty = L2 access time + L2 miss rate × main memory penalty = 10 + 0.20 × 100 = 10 + 20 = 30 ns
Now compute AMAT:
AMAT = 1 + 0.10 × 30 = 1 + 3 = 4
Final answer (rounded to one decimal place): 4.0 ns
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