Assume that we have a demand paging The page table is stored

Assume that we have a demand paging. The page table is stored in registers (no penalty for accessing the page table). It takes 10 milliseconds to srrvice a page fault if an empty frame is available or if the victim page is not dirty. However, it takes 20 milliseconds if the victim page is dirty.

a. Assume that the victim page is dirty 70% of the time, what is the average page-fault service time?

b. Memory-acess time is 100 nanoseconds. What is the maximum acceptable page-faut rate for an effective access time of no more than 200 nanoseconds?

Solution

Let p be the page fault. Then (1 p) is the probability that a memory access costs 100 nsec. The

probability that a page fault costs 20 msec is 0.7 p and the probability that a page fault costs 8 msec is

0.3 p. Since 1 nsec = 1000000 msec,

(1 p) 100 + 0.7 p 20000000 + 0.3 p 8000000 = 200

(14000000 + 2400000 100)p = 100

p = 100/(16400100) = 6.1 106 = .0000061 = .00061%