Demand paging

Question

Answer 1

Let page fault rate be 'p'

EMAT = p*(pfst) + (1 - p)*memory access time       ......(1)

where pfst = page fault service time

Now let's calculate pfst,

pfst = 0.6 * 15 + 0.4 * 5 = 11 msec

$\therefore$ EMAT = p * 11 msec + (1 - p) * 1$\mu$sec

Now, convert every unit into $\mu$sec for simplicity

2 = 11000 p + (1 - p)*1

 p = 0.0001 satisfies best from the given options, which is nothing but 0.01 %

Comments

@srestha
can u pls explain this?

Answer 2

Page fault service time  = 5 ms (if page is empty or not dirty)
                                         15 ms (if page is dirty)

$EMAT = P \times PFST \dotplus (1-P) \times MMAT$
here PFST = page fault service time and MMAT = main memory access time

$2 \mu s = P \times ( 60 \% \times 15ms + 40 \% \times 5ms) \dotplus (1-P) \times 1\mu s$
$1 \mu s = 11000P \mu s - 1P \mu s$

Approximating 10999 to 10000

$P = \frac{1}{10000} \times 100 \%$

$P = 0.01 \%$

I think there is something wrong in the question, considering we are approximating 10999 to 10000. Anyway if you think we can do such kind of approximation then its all fine.