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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
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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).

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)..
EMAT = P * 50000000 ns + (1-p) * 100 ns

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