GATE CS 2011 Q.20 (CS Q.20): Memory Management and Virtual Memory

Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns. If one page fault is generated for every $10^6$ memory accesses, what is the effective access time for the memory?

21 ns

30 ns

23 ns

35 ns

Official answer and solution

Official answer: B
From: Official GATE 2011 answer key

Solution

Apply the effective memory access time formula with page fault service time.

Let $p$ be the page fault rate. The problem states that 1 page fault occurs every $10^6$ accesses, so $p = \frac{1}{10^6} = 10^{-6}$ .

The memory access time is $T_m = 20\text{ ns}$ and page fault service time is $T_p = 10\text{ ms} = 10 \times 10^6\text{ ns} = 10^7\text{ ns}$ .

Effective access time is:

\text{EAT} = (1 - p) \times T_m + p \times T_p

\text{EAT} = (1 - 10^{-6}) \times 20\text{ ns} + 10^{-6} \times 10^7\text{ ns}

\text{EAT} = 20 - 0.00002 + 10 = 29.99998\text{ ns} \approx 30\text{ ns}

Final answer

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

Source

Paper: GATE 2011 Computer Science and Information Technology (CS)
Question: 20 of 65, printed as CS Q.20
Section: Computer Science and Information Technology

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