Page fault service problem


#1

Let the page fault service time be 10 ms in a computer with average memory access time being
20ns. If one page fault is generated for every 10^6 memory accesses, what is the effective
access time for the memory ?
A) 21ns
B) 30ns
C) 23ns
D) 35ns


#2

EMAT = P( PAGE FAULT SERVICE TIME( in msec ) + MEMORY ACCESS TIME( in nsec ) ) + ( 1 - p )( MEMORY ACCESS TIME ( in nsec ) )

P = page fault rate = ( 1/10^6 )

So, EMAT = (1/10^6)( 10 msec + 20 nsec ) + (1 - 1/10^6)( 20 nsec)
= (1/10^6)( 10 msec ) + (20 nsec) + (1/10^6)( 20 nsec ) - (1/10^6)( 20 nsec )
= 10 nsec + 20 nsec
= 30 nsec

Answer B.


#3

EMAT =page fault rate * page fault service time + memory access time
=1/10^6 * 10*10^6 + 20
=30ns.

** 1ms = 10^6ns.
Hence answer is B.


#4

Effective Memory Access Time=(probability thet page fault occurs)(page fault service time)+(probability of no page fault)(memory access time)
==>EMAT=p*(page fault service time)+(1-p)*(memory access time)
by give data
EMAT=(1/10^6)(10ms+20 ns)+(1-1/10^6)(20 ns)
=10ns+20ns
=30 nsec
hence the answer is option B.