GATE CS 2025 Set 2 Q45 — 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)

View worked solution

Answer: 4:4

To find the average memory access time (AMAT), we analyze the hierarchical access penalty at each level of the cache hierarchy:

L1 Cache: Access time $T_1 = 1$ ns, Hit rate $H_1 = 0.90$ , Miss rate $M_1 = 1 - H_1 = 0.10$ .
L2 Cache: Hit rate $H_2 = 0.80$ , Miss rate $M_2 = 1 - H_2 = 0.20$ .

When an L1 miss occurs, the data must be retrieved from L2 or main memory:

If L2 hits (probability $H_2 = 0.80$ ), the transfer penalty from L2 to L1 is $10$ ns.
If L2 misses (probability $M_2 = 0.20$ ), the data is first transferred from main memory to L2 (penalty $100$ ns) and then from L2 to L1 (penalty $10$ ns). Thus, the total penalty is $100 + 10 = 110$ ns.

The average penalty for an L1 miss is calculated as:

\text{Average L1 Miss Penalty} = (H_2 \times 10) + (M_2 \times 110) = (0.80 \times 10) + (0.20 \times 110) = 8 + 22 = 30\text{ ns}

Now, we compute the overall Average Memory Access Time (AMAT):

\text{AMAT} = T_1 + M_1 \times (\text{Average L1 Miss Penalty}) = 1 + 0.10 \times 30 = 1 + 3 = 4\text{ ns}

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Published 10 Oct 2026

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