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

  1. $21$ ns
  2. $30$ ns
  3. $23$ ns
  4. $35$ ns
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4 Answers

Best answer
72 votes
72 votes

Open slides $12-13$ to check : 

http://web.cs.ucla.edu/~ani/classes/cs111.08w/Notes/Lecture%2016.pdf$$\begin{align*} \text{EMAT} &= \frac{1}{10^6} \times 10\ \text{ms} + \left(1-\frac{1}{10^6} \right ) \times 20\ \text{ns} \\ &= 29.99998\ \text{ns} \\ &\approx 30\ \text{ns} \end{align*}$$Answer = option B

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25 votes
25 votes

Effective memory access time = Memory access time + page fault rate *page fault service time

so here

$\text{EMAT} = 20 ns  + \frac{ 1}{10^6} \times 10 \times 10^6 ns$

$= 20 + 10 =30 ns$                  

Ans is B.

5 votes
5 votes

Page Service = 10 ms

P = 1 / 10^6 

M = 20ns

EMAT = (Page service) P + M

EMAT = (10 ms) * (1 / 10^6) + 20 ns

=10ns + 20ns

so, EMAT = 30ns

Ans) Option B.

0 votes
0 votes

Here the explanation

Answer:

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