Consider a system with memory access time 50nsec and demand paging with the page
ID: 3816187 • Letter: C
Question
Consider a system with memory access time 50nsec and demand paging with the page table held in MMU registers. It takes 4msec to service a page fault if an empty page is available and 8msec if replaced page is modified. a How long it takes to service a page fault if the replaced page is not modified? b. Assuming that the page to be replaced is modified 5% of time, an empty page is available 80% of time and the page to be replaced is not modified 15%, what is the maximum acceptable page fault rate for an effective access time of no more than 60nsec.Explanation / Answer
Given memory access time = 50nsec
Time to service a page= 4msec = 4000000 ns =
Time for replaced page modified = 8msec = 8000000 ns
A) If the replaced page is not modified:
EAT = ((1-P) * 50ns)+ p 4000000
= 50-50P + 4000000* p
= 50+3999950*p
B) Given that 80% of time for empty page fault and replaced is modified 5% and not modified is 15%
0.06us = ((1-P) * 0.05 us) + (0.8P * 4,000 us) + (0.05P * 8,000 us)+ (0.15p*8000)
0.06 = -0.05P+0.05 + 3200P + 1600P