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35 votes
35 votes

Suppose the time to service a page fault is on the average $10$ milliseconds, while a memory access takes $1$ microsecond. Then a $99.99\%$ hit ratio results in average memory access time of

  1. $1.9999$ milliseconds
  2. $1$ millisecond
  3. $9.999$ microseconds
  4. $1.9999$ microseconds 
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5 Answers

Best answer
55 votes
55 votes
Since nothing is told about page tables, we can assume page table access time is included in memory access time.

So, average memory access time

$= .9999 \times 1 + 0.0001 \times 10,000$
$= 0.9999 + 1$
$= 1.9999$ microseconds

Correct Answer: $D$
edited by
8 votes
8 votes

p=miss rate, 1−p=hit rate, ma=memory access time, ps=page fault time (or) service time

EMAT = p(ps+ma)+(1−p)×ma = p(ps)+p(ma)+ma−p(ma) = p(ps)+ma


Given Values:  p = $\frac{0.001}{100}$ = $10^{-4}$,  ps = 10×$10^{-3}$ sec,  ma=$10^{-6}$ sec

EMAT = $10^{-4}$x(10×$10^{-3}$)+$10^{-6}$ sec

EMAT = $10^{-4}$×($10^{-2}$)+$10^{-6}$ sec

EMAT = $10^{-6}$+$10^{-6}$ sec

EMAT = 1+1 μsec

EMAT = 2 μsec
 

So correct answer is D

2 votes
2 votes
CAN ANYONE TELL ME WHERE AM I WRONG

Step 1: memory read to access the page table (1 micro second)

case 1) frame number found -> second memory access to access the desired page from memory (1 microsecond)

case 2) frame number not found -> page fault -> service the page fault (10000 microseconds)

Step 2:   (.9999*(first memory access time + second memory access time)) +

             (.0001*(first memory access time + time to service page fault))

             = (.9999*2)+(.0001*10001)

             =2.9999 micro seconds
Answer:

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