#Operating Systems (Silberschatz, Galvin, Gagne) #VirutalMemory #Page325

Question

In this problem I have a doubt that why effective access time is not taken as 

effective access time = 1 x 200 + p x (8 milliseconds)

because even though pages are accessed after page-fault but all pages must be accessed only through main memory only. Thus I think effective access time should be (p x 8 milliseconds) to bring page from disk to memory,  plus,  200 nanoseconds for accessing all pages in memory

Answer 1

Actually the page fault service time contains three components as given in #Page324

Thanks for your answers

Answer 2

YES you are right

effective access time=(1-p) *memory access time +  p(service time +memory access time )

but generally service time is too large compared to memory access time so we neglect it

 

 

or we can also use this formula

(1-p)*m +p(s+m)===>   m+ps

memory access time + P*service time

Comments

How about adding dirty bit to this. Like, if page fault occurs then there is a probability say D for dirty bit?

Answer 3

Effective memory access time in the case of page fault=Ps*p+m
where Ps=page service time
P=miss rate or frequency of page fault for every memory access.
m=memory access time