Operating System Demand Paging

Question

In a demand-paged system, it takes 100 nanoseconds to access memory. The page table has 8 entries and is held in registers. It takes 10 milliseconds to service a page fault if an empty frame is available or if the victim frame is not dirty. If the victim frame is dirty, the page fault service time is 20ms. Assuming that the victim frame is dirty 80% of the time, find the maximum page fault rate for which the effective memory access time remains within 200 nanoseconds

Comments

http://gauss.ececs.uc.edu/Courses/c4029/exams/Fall2012/c4029_fall_1_12_sol.pdf  

Q:5

Answer 1

11ms

Comments

Given,

Effective memory access Time = (Page Fault Rate x Average access time to service page fault) + [(1-Page Fault Rate) x (Main memory access time)]

 

Average access time to service page fault = (Page Rate which is modified) x (Modified service time)  +  (Page Rate which is empty) x (Page Fault service which is empty frame)

= 0.8(2 * 10^7 ns) + 0.2(10^7 ns)

= 18 * 10^6 ns

 

then 200 ns = [(1-p)*100ns ] +  [p * 18* 10^6ns]

p= 0.00055%