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If page fault service time is 50 milli second  and memory access time is 100 ns, then what will be EMAT, if the probability of page fault is $p$ ?

a) $500000 + 100p$ ns

b) $100 + 500000p $ ns

c) $10^{-7} - 5 p\times 10^{-2}$ seconds

d) $10^{-7} + 49.9p \times 10^{-3}$ seconds
in Operating System by Active (1k points) | 436 views

3 Answers

+3 votes
Best answer

EMAT = P (Page fault service time )+ (1-P)(Memory access time )

        $ = p(50\times 10^{-3} \times 10^9)$ns + $(1-p)(100)$ns

        $ = p( 50000000)$ns $+ 100$ns$ -100p$ns

        $= p(500000)\times 10^{-7}$sec $+ 10^{-7}$sec$- (10^{-7})p$ sec

        which is nearly equal to ... $10^{-7}$sec$ - 5p \times 10^{-2}$sec 

      Hence ans :Option C)

by Boss (17.1k points)
selected by
0
Why did u eat the memory access time in P (Page fault service time +mem access time)?
0
i just put values in above formula , where is p(ps+ma)??
0
prob of pge fault*(page f serv time+mmem access)+(1-page fault prof)(mem acces)
0

EMAT = P (Page fault service time )+ (1-P)(Memory access time )

        =p(50*$10^{−3}$*$10^{9}$)ns + (1−p)(100) ns

        = 50*$10^{−6}$ P ns + 100ns -100P ns

        =   (50000000)P ns - 100P ns + 100 ns

        = 49999900 P ns +   $10^{−7}$

       = $10^{−7}$  +  49.9P*$10^{−3}$           i.e option (D).

                                                                                                         

+4 votes

Page fault service time includes memory access time too. So need to add it in case of page fault.

$\text{EMAT} =(1-p)10^{-4}+p \times 50 \text{ milliseconds}$

this gives d as ans

by Boss (31.4k points)
+1
This gives D option?
0
Yes it should option d.

Also the sign of 5p*10^-2 is changed in option c..
0
But page fault service time is itself calculated including (memory access time+disc access+write-backs if req.) !

So, p*(page fault service time) + (1-p)(mem access)  is correct.. gives (C)..
0 votes
EMAT = P * 50000000 ns + (1-p) * 100 ns

          = 10^-7 +49.9*10^-3  (option D)
by Junior (621 points)
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