Nt1330 Unit 4 Case Study
Let’s take an average page fault service time of 25 ms and a memory access time of 100 ns, then the effective access time in ns is effective access time = (1-p)*(100) +p(25 ms)
=(1-p)*100+p*25,000,000)
=100+24,999,900*p
We saw effective access time is directly proportional to the page fault rate . suppose one access out of 1,000 cause a page fault, effective access time is 25 ms the computer might be slow down by a factor of 250 because of demand paging. Suppose we want less than 10 % degradation,
We require:
110>100+25,000,000*p
10>25,000,000*p
P<0.0000004
that is to keep the slowdown due to paging to a reasonable level, we can allow only less than one memory access out 2,500,000 to page fault. in a demand paging system keep the page
=(1-p)*100+p*25,000,000)
=100+24,999,900*p
We saw effective access time is directly proportional to the page fault rate . suppose one access out of 1,000 cause a page fault, effective access time is 25 ms the computer might be slow down by a factor of 250 because of demand paging. Suppose we want less than 10 % degradation,
We require:
110>100+25,000,000*p
10>25,000,000*p
P<0.0000004
that is to keep the slowdown due to paging to a reasonable level, we can allow only less than one memory access out 2,500,000 to page fault. in a demand paging system keep the page