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

# Page fault, the probability of page being dirty?

**prateek111**#1

Memory managemnet

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

**satish**#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+ (p*300)]

finally

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

=> p=0.514

**Harsha_1997**#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