Consider the following
page trace: 4,3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5
Percentage of page fault
that would occur if FIFO page replacement algorithm is used with number of
frames for the JOB m = 4 will be
[Paper II June 2012]
(A)
8
(B) 9
(C)
10
(D) 12
Answer is C
Read detailed explanation with page fault and FIFO page replacement algorithm.
What is page fault?
A page fault is the situation arises when running
program tries to access the required page in Physical Memory (RAM) but the page is not
available in RAM this is called page fault. when page is not available in RAM then operating system tries to find page data from virtual
memory and load the page into RAM.
An invalid page fault or page fault error occurs when
the operating system cannot find the data in virtual memory. This usually
happens when the virtual memory area, that maps virtual addresses to real
addresses, becomes corrupt.
The term page fault is a bit misleading as it implies that
something went seriously wrong. Although page faults are undesirable – as they
result in slow accesses to the hard disk – they are quite common in any
operating system that uses virtual memory.
Now, we need
to actually solve the problem. The easiest way to do this is to break the
problem down into the number of pages. See what happens each time a page is
referenced by the program, and at each step see whether a page fault is
generated or not. Of course, we want to keep track of what pages are currently
in the physical memory (the RAM).
The first four page accesses will result in page faults because the
frames are initially empty. After that, if the program tries to access a page
that’s already in one of the frames then there’s no problem. But if the page
that the program is trying to access is not already in one of the frames then
that results in a page fault. In this case, we have to determine which page we
want to take out (or ‘swap’) from the RAM, and for that we use the FIFO
algorithm.
Below
is the explanation of page 4,
3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5 and page frame size is 4.
Page
|
4
|
3
|
2
|
1
|
4
|
3
|
5
|
4
|
3
|
2
|
1
|
5
|
Sr. No.
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
1
|
4
|
4
|
4
|
4
|
4
|
4
|
5
|
5
|
5
|
5
|
1
|
1
|
2
|
3
|
3
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3
|
3
|
3
|
3
|
4
|
4
|
4
|
4
|
5
|
|
3
|
2
|
2
|
2
|
2
|
2
|
2
|
3
|
3
|
3
|
3
|
||
4
|
1
|
1
|
1
|
1
|
1
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1
|
2
|
2
|
2
|
|||
Buffer
of 4 Frame
|
||||||||||||
As per above image we have size of 4 frame. When operating system trying to access data 4
from physical memory then data is not available in physical memory so data 4
will be as page fault so operating system will search data page in virtual
memory and load the frame with data 4, afterward data 3, 2, 1 also be a page
fault. At column no. 5 demanding of data 4 which is available in physical
memory (RAM) so no page fault, afterwards same as 3 is available in physical
memory. At column no. 7 demanded of data
5 which is not available in physical memory and we have to follow FIFO page
replacement algorithm so data 4 has came first so data 4 will be removed and
data 5 will be placed as a page fault because it is not available in physical
memory. Same as all other data pages and
procedure will work and at the end we can count grayed area which is demanded
but does not available in physical memory and became page fault so total grayed
area are 10. So page fault is 10.
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