Given that main memory is composed of only three page frames for public use and that a seven-page program (with Pages a, b, c, d, e, f, g) that requests pages in the following order: a, c, a, b, a, d, a, c, b, d, e, f. Using the FIFO page removal algorithm, indicate the movement of pages into and out of the available page frames (called a page trace analysis). Indicate each page fault with an asterisk (*). Then compute the failure and success ratios.
pages are accessed in the order
acbd*a*c*b*d*e*f*
Bad luck, that.
4. given that main memory is composed of three page frames for public use and that a seven-page program (with pages a, b, c, d, e, f, g) requests pages in the following order: a, b, a, c, d, a, e, f, g, c, b, g
Let's analyze the movement of pages into and out of the available page frames using the FIFO page replacement algorithm.
Initially, all page frames are empty.
1. Page a is requested. Current page frames: (a*).
2. Page c is requested. Current page frames: (a*, c*).
3. Page a is requested again. However, since it is already in a page frame, there is no page fault. Current page frames: (a*, c*).
4. Page b is requested. Current page frames: (a*, c*, b*).
5. Page a is requested again. It was the first page that was loaded into memory, so it is the one that is replaced. Current page frames: (c*, b*, a*).
6. Page d is requested. Current page frames: (c*, b*, a*, d*).
7. Page a is requested again. The second oldest page, page c, is replaced. Current page frames: (a*, b*, d*, c*).
8. Page c is requested again. The oldest page, page a, is replaced. Current page frames: (c*, b*, d*, a*).
9. Page b is requested again. The oldest page, page c, is replaced. Current page frames: (b*, d*, a*, c*).
10. Page d is requested again. The oldest page, page b, is replaced. Current page frames: (d*, a*, c*, b*).
11. Page e is requested. Current page frames: (d*, a*, c*, b*, e*).
12. Page f is requested. Current page frames: (d*, a*, c*, b*, e*, f*).
Page Trace Analysis:
(a*, c*, a, b*, a*, d*, a*, c*, b*, d*, e*, f*)
Failure Ratio (Page Fault Ratio):
Total Page Faults = Number of asterisks in the page trace analysis = 6
Total Requests = Total number of requests in the page trace analysis = 12
Failure Ratio = Total Page Faults / Total Requests = 6 / 12 = 0.5
Success Ratio:
Success Ratio = 1 - Failure Ratio = 1 - 0.5 = 0.5 (or 50%)
Please note that the failure and success ratios are calculated based on the given page trace analysis and the assumption that all page faults are counted as failures.
To analyze the page trace using the FIFO (First-In-First-Out) page replacement algorithm, we need to keep track of the pages that are currently in the page frames and their order of arrival.
Given:
- Three page frames (P1, P2, P3) for public use.
- A seven-page program (with Pages a, b, c, d, e, f, g) that requests pages in the following order: a, c, a, b, a, d, a, c, b, d, e, f.
Let's go step by step to observe the movement of pages into and out of the available page frames:
1. a* (P1)
- Page a is not present in any frame, so we have a page fault. Place it in an available frame (P1).
Page Frames:
- P1: a
- P2:
- P3:
2. c* (P2)
- Page c is not present in any frame, so we have a page fault. Place it in an available frame (P2).
Page Frames:
- P1: a
- P2: c
- P3:
3. a (P3)
- Page a is already present in frame P1, so no page fault occurs.
Page Frames:
- P1: a
- P2: c
- P3: a
4. b* (P1)
- Page b is not present in any frame, so we have a page fault. Replace the oldest page in the frame (P1: a) with the new page (b).
Page Frames:
- P1: b
- P2: c
- P3: a
5. a (P2)
- Page a is already present in frame P3, so no page fault occurs.
Page Frames:
- P1: b
- P2: c
- P3: a
6. d* (P1)
- Page d is not present in any frame, so we have a page fault. Replace the oldest page in the frame (P1: b) with the new page (d).
Page Frames:
- P1: d
- P2: c
- P3: a
7. a (P2)
- Page a is already present in frame P3, so no page fault occurs.
Page Frames:
- P1: d
- P2: c
- P3: a
8. c (P1)
- Page c is already present in frame P2, so no page fault occurs.
Page Frames:
- P1: d
- P2: c
- P3: a
9. b (P2)
- Page b is already present in frame P1, so no page fault occurs.
Page Frames:
- P1: d
- P2: c
- P3: a
10. d (P3)
- Page d is already present in frame P1, so no page fault occurs.
Page Frames:
- P1: d
- P2: c
- P3: a
11. e* (P1)
- Page e is not present in any frame, so we have a page fault. Replace the oldest page in the frame (P1: d) with the new page (e).
Page Frames:
- P1: e
- P2: c
- P3: a
12. f* (P2)
- Page f is not present in any frame, so we have a page fault. Replace the oldest page in the frame (P2: c) with the new page (f).
Page Frames:
- P1: e
- P2: f
- P3: a
Finally, let's compute the failure and success ratios:
The failure ratio is the total number of page faults divided by the total number of page requests:
- Number of page faults: 4 (a*, c*, b*, e*)
- Number of page requests: 12
Failure ratio = 4 / 12 = 1/3
The success ratio is the number of successful page requests divided by the total number of page requests:
- Number of successful page requests: 12 - 4 = 8
Success ratio = 8 / 12 = 2/3
So, in this case, the failure ratio is 1/3 and the success ratio is 2/3.