GATE CS 2020 Q.31 (CS Q.21): Memory Hierarchy: Cache, Main Memory and Secondary Storage

A direct mapped cache memory of 1 MB has a block size of 256 bytes. The cache has an access time of 3 ns and a hit rate of 94%. During a cache miss, it takes 20 ns to bring the first word of a block from the main memory, while each subsequent word takes 5 ns. The word size is 64 bits. The average memory access time in ns (round off to 1 decimal place) is _______.

Your answer

Official answer and solution

Official answer: 13.5
From: Official GATE 2020 answer key

Solution

Compute the miss penalty by summing the time for the first word and subsequent words of the block, then apply the AMAT formula.

Given parameters:

Cache hit time $T_{\text{cache}} = 3\text{ ns}$
Hit rate $H = 94\% = 0.94 \implies$ Miss rate $M = 1 - 0.94 = 0.06$
Block size = 256 bytes
Word size = 64 bits = 8 bytes
Words per cache block = $\frac{256}{8} = 32\text{ words}$

Miss penalty calculation:

First word retrieval time = $20\text{ ns}$
Remaining $32 - 1 = 31$ words take $5\text{ ns}$ each
Total miss penalty $T_{\text{miss}}$ : $T_{\text{miss}} = 20 + (31 \times 5) = 20 + 155 = 175\text{ ns}$

Average Memory Access Time (AMAT):

\text{AMAT} = T_{\text{cache}} + M \times T_{\text{miss}}

\text{AMAT} = 3 + 0.06 \times 175 = 3 + 10.5 = 13.5\text{ ns}

Final answer

13.5

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

Source

Paper: GATE 2020 Computer Science and Information Technology (CS)
Question: 31 of 65, printed as CS Q.21
Section: Computer Science and Information Technology

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