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I am getting answer as 88.6 time unit. 

using Total page service time = 250 x 0.94 + 0.06 x 450 = 262 time units

Access time = p( page service time + m) + (1-p)(m)  = p * page service time + m   ( p is page fault rate) 

Substituting the values i got access time = 88.6 time unit. 

but the answer is 85.6 time unit. Please point where i am going wrong. 

2 Answers

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The average access time

(1-p)memory access time+p(memory access time+page fault service time +dirty page service time)

=0.7(10)+0.3(10+250+0.06*450)

=7+0.3(260+27)

=7+0.3*287

=93.1 time units
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The question is little ambiguous because it hides various standard properties.
Now Working of virtual memory is quite complicated yet simple .
Here is the reference of page fault summary.
https://en.wikipedia.org/wiki/Page_fault

Now what they assumed in question is if page is dirty then complete procedure takes place in 450 unit of time {Complete this is hiding certain details}
and if now the servicing it means without updating it just fetching detail takes 250 units only.

Now :
Eqn would be :
Avg access time : Page hit *Memory access + page miss[page dirty*dirty service time+page not dirty * page fault service time]
Thus :

0.7*10+0.3[0.06*450+0.94*250]
7+78.6
85.6 time units

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