Question

Answer the following questions

a) If average page-fault service time is 9 milliseconds, a memory access time is 202 nanoseconds, and the page fault rate is 0.2% then what is the effective access time (in nanoseconds) for a demand paging scheme.

b) Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.

Answer 1

a, Effectice access time = page fault rate *( memory access time + page fault service time )+((1-page fault rate )* Memory acces time ) by puting values

E A T = 0.2*(202+9)+((1-0.2)*202) = 203.2 nano seconds is the anwser.

b. The page fault occur when an access to the page that has nit been brought to the main memory takes pale it leads to the page fault . and when the operating system verifes the memory access aborting the program if it is invalid . it is the interupt that occur when a program request the data which is not currently present in the main memory . the interupt trigger to the operating system to fetch the file form the virtual  memory and load it into the RAM . and the invalid page fault occur when the operating system is not able to find the file in the virtual memory.

the action taken by the operating system is that

1.Assembly routine reolads register iand other information and return to the user space to continue execution .

2.Once the page fault occur the operating system then the system cheks that weather the given address is valid and the weahter there is any no protection access problem.

3.when the disk interupts state that the page has arrived and marked to the reflected palce the state changes to the normal.

4.As soon as the page frame is cleaned , operating system looks for the didk address where the page is to be scheduled .