Page fault, the probability of page being dirty?


#1

A demand paging system takes 100 time units to service a page fault and 300 time units to replace a dirty page. Memory access time is 1 time unit. The probability of a page fault is p. In case of a page fault, the probability of page being dirty is also p. It is observed that the average access time is 3 time units. Then the value of p is
(A) 0.194 (B) 0.233 © 0.514 (D) 0.981


Memory managemnet
#2

Given that
Page fault service time =100
page fault service time =300 (for dirty page)

probability of page fault =p

probability of being dirty is =p

so total page fault service time= p(300)+ (1-p)(100) = 200p+100

now given Average Access Time =3 so

Estimated(Average) Access Time = p(page fault service time +memory access time)+(1-p) memory access time

3= p*page fault service time + memory access time
3 =p(200p+100)+1= 200p2+100p+1
p=0.0194


#3

probability tht pagefault doesnt occur is (1-p)
so (1-p)*1
Probability tht pagefault occur is p
within tht comes the probability of dirty page occuring or nt wich s again p
so p[(1-p)100+ (p300)]

finally
(1-p)*1 + p[(1-p)100+ (p300)] = 3
=> p=0.514


#4

The answer is option A

explanation:

p(p * 300 + (1-p) * 100) + (1-p) * 1 = 3

p(300p + 100 - 100p) + 1-p=3

200p^2 + 99p - 2=0

After solving this equation : −b+(√b^2−4ac)/2a, we get

p ≈ 0.0194