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GATE CSE 2025 Session 2 Q55: Average memory access time with two cache levels

Question

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)

Try it: numerical answer: type the value

Numerical answer. No negative marking.
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Answer: 4 (official GATE 2025 final answer key)

Concept used: AMAT = hit time + miss rate × miss penalty, applied level by level.

First step: The L1 miss penalty is itself 10+0.2×10010 + 0.2 \times 100 ns.

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Source: official GATE 2025 Session 2 CS question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key

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