GATE CS 2015 Set 2 Q.34 (CS Q.24): Memory Hierarchy: Cache, Main Memory and Secondary Storage

Assume that for a certain processor, a read request takes 50 nanoseconds on a cache miss and 5 nanoseconds on a cache hit. Suppose while running a program, it was observed that 80% of the processor's read requests result in a cache hit. The average read access time in nanoseconds is _____.

Your answer

Official answer and solution

Official answer: 14
From: Official GATE 2015 answer key

Solution

Apply the Average Memory Access Time (AMAT) formula: $T_{\text{avg}} = h \cdot T_{\text{hit}} + (1 - h) \cdot T_{\text{miss}}$ .

Parameters given:

Hit time $T_{\text{hit}} = 5\text{ ns}$
Miss time $T_{\text{miss}} = 50\text{ ns}$
Hit ratio $h = 80\% = 0.80$
Miss ratio $1 - h = 20\% = 0.20$

Compute the average access time:

T_{\text{avg}} = h \cdot T_{\text{hit}} + (1 - h) \cdot T_{\text{miss}}

T_{\text{avg}} = (0.80 \times 5\text{ ns}) + (0.20 \times 50\text{ ns})

T_{\text{avg}} = 4\text{ ns} + 10\text{ ns} = 14\text{ ns}.

Final answer

Common trap

Adding the hit time to the miss time when miss time is already specified as the total access latency on a miss.

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

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 2
Question: 34 of 65, printed as CS Q.24
Section: Computer Science and Information Technology

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