GATE CS 2025 Set 2 Q.55: Memory Hierarchy: Cache, Main Memory and Secondary Storage

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)

Your answer

Official answer and solution

Official answer: 4
From: Official GATE 2025 answer key

Solution

Apply the hierarchical Average Memory Access Time (AMAT) formula $\text{AMAT} = T_{\text{L1}} + (1 - H_{\text{L1}})\left(M_{\text{L2}\rightarrow\text{L1}} + (1 - H_{\text{L2}}) M_{\text{Mem}\rightarrow\text{L2}}\right)$ .

Given:

L1 access time $T_{\text{L1}} = 1\text{ ns}$ , L1 hit rate $H_{\text{L1}} = 0.90 \implies$ L1 miss rate $= 1 - 0.90 = 0.10$ .
L2 hit rate $H_{\text{L2}} = 0.80 \implies$ L2 miss rate $= 1 - 0.80 = 0.20$ .
Miss penalty from L2 to L1 $= 10\text{ ns}$ .
Miss penalty from main memory to L2 $= 100\text{ ns}$ .

Therefore, the average memory access time is:

\text{AMAT} = 1 + 0.10 \times \left(10 + 0.20 \times 100\right) = 1 + 0.10 \times (10 + 20) = 1 + 3.0 = 4.0\text{ ns}

Final answer

4.0

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
Question: 55 of 65
Section: Computer Science and Information Technology

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