15 votes

A process has been allocated $3$ page frames. Assume that none of the pages of the process are available in the memory initially. The process makes the following sequence of page references (reference string): $\mathbf{1, 2, 1, 3, 7, 4, 5, 6, 3, 1}$

If optimal page replacement policy is used, how many page faults occur for the above reference string?

- $7$
- $8$
- $9$
- $10$

16 votes

Best answer

Optimal replacement policy means a page which is "farthest" in the future to be accessed will be replaced next.

$$\require{cancel}\begin{array}{c|c}\hline

\text{Frame 0}&1\\\hline

\text{Frame 1}&\cancel{2} \quad\cancel7\quad \cancel4 \quad \cancel5 \quad 6\\\hline

\text{Frame 2}&3\\ \hline

\end{array}$$

$3$ initial page faults for pages $1,2,3$ and then for pages $7,4,5,6 \implies 7$ page faults occur.

Answer is** (A).**

5 votes

1 2 1 3 7 4 5 6 3 1

In optimal scheduling we schedule page in place of that page which is not used for scheduling in future .........

If initially memory is empty so 3 page faults

1,2 , (1) ,3 so initially 3 pages are there now - 3 faults

7 in place of (2) because 2 is not in future - 1 fault

4 in place of 7 - 1 fault

5 in place of 4 - 1 fault

6 in place of 5 - 1 fault

so total faults = 3+1+1+1+1 = 7

so option A

In optimal scheduling we schedule page in place of that page which is not used for scheduling in future .........

If initially memory is empty so 3 page faults

1,2 , (1) ,3 so initially 3 pages are there now - 3 faults

7 in place of (2) because 2 is not in future - 1 fault

4 in place of 7 - 1 fault

5 in place of 4 - 1 fault

6 in place of 5 - 1 fault

so total faults = 3+1+1+1+1 = 7

so option A

1 vote

Initially, Page fault Count =0

After first 3 pages ,i.e 1,2,1 Page Fault count=2.

1 | 2 |

After 4 pages i.e 1,2,1,3 Page Fault count =3.

1 | 2 | 3 |

After 5 pages i.e.1,2,1,3,7 ,we have to replace one page out 3 from above situation ,candidate for page replacement is 1,2,3 we choose page No.2 .reason=we use optimal page replacement ,so we see the future reference of page and came to conclusion that page No=2 have No reference in future or if they have reference in future ,which is late bit as compare to page no.1 and 3 .

Page fault count=4

1 | 7 | 3 |

After 6 pages i.e.1,2,1,3,7,4. we choose page 7 for replacement with same logic as above. Page Fault Count =5

1 | 4 | 3 |

After 7 pages i.e.1,2,1,3,7,4,5 we choose page 4 for replacement.Page Fault count=6

1 | 5 | 3 |

After 8 pages i.e.1,2,1,3,7,4,5,6 .we choose page 5 for replacement.page fault count=7

1 | 6 | 3 |

After 10 pages i.e.1,2,1,3,7,4,5,6,3,3. page No=3 and page No=3 already present so no page fault hence page fault count=7.

**Answer 7**