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In a paged memory, the page hit ratio is $0.35$. The time required to service the page fault is $100$ ns. Time required to access a page in primary memory is $10$ ns.

The average time required to access a page is ________ ns
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Page access time=   
(1-page fault)*(Time required to access a page in primary memory) + (page fault) * Miss penalty

We assume page access is Simultaneous !

page fault rate is (1- hit ratio) = 1- 0.35 = 0.65 

Time required to access a page in primary memory is 10 ns as given .

Time to access page in primary memory  ( Tp) =  (1- 0.65)  * 10 = 0.35 * 10 = 3.5 ns.

Time required to access page in secondary memory is the Miss penalty time which is  100 ns as given .
Miss ratio =(1- 0.35)=0.65


Missed pages will be found  in secondary memory.

Time to access page in secondary memory ( Ts)   =  Miss ratio * time for secondary memory
=(1-0.35)*100
=65.0ns


Average time required to access a page is
= Ts+Tp

=65.0ns+3.5ns

=68.5ns

PS :

Read this comment to make everything clear : https://gateoverflow.in/96884/operating-systems-2-29#c125175

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0.35 ( 10 + 10 ) + 0.65( 10 + 100) = 78.5
Answer:

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