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In a demand paging system page fault rate 0.60 and page fault service time 8 milliseconds, memory access time 200 nano seconds. Find out effective access time in nano seconds.
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EMAT= .60(8ms)+.40(200ns).

which is approximately .60*8000000+.40*200=4800080ns

Hi,
Since this is asking about demand paging, we need to take Page fault service time and Main Memory access time into consideration.
So,
EMAT = Page fault rate x (Page fault service time + Main Memory access time)+ (1-page fault rate) x Main memory access time
If you dont understand the formula, think of it this way, the effective memory access time will be the [Probability of page fault happening (page fault rate) multiplied by the time to service the page and then access the memory] + [Probability of page fault NOT happening multiplied by time to access the memory].
$Emat = (0.6 \times ((8\times 10^{6})+ 200 )) + (0.4 \times (200))$

= 4800200ns
Now, you might get a doubt that why did I multiply eight with 10^6, actually that is just conversion from millie to nano seconds.
If you are confused, please note the following conversion:
1sec= 1/10^3 millie seconds

1sec= 1/10^6 micro seconds

1sec= 1/10^9 nano seconds

Hope this helps ! :)