The memory management process employs virtual memory to make up for a lack of memory space; this process sometimes results in a page fault. Consider the following sequence of page references:9, 7, 0, 4, 7, 5, 3, 2, 5, 8, 9, 5, 3, 7, 9, 0, 5, 3, 2, 1Determine how many page faults will occur for the Least Recently Used algorithm, assuming there are only FOUR (4) physical page frames, and all are initially invalid (empty).Tip: Enter numerical data ONLY. E.g: "5" and NOT "5 Page Faults".

Question

The memory management process employs virtual memory to make up for a lack of memory space; this process sometimes results in a page fault. Consider the following sequence of page references:9, 7, 0, 4, 7, 5, 3, 2, 5, 8, 9, 5, 3, 7, 9, 0, 5, 3, 2, 1Determine how many page faults will occur for the Least Recently Used algorithm, assuming there are only FOUR (4) physical page frames, and all are initially invalid (empty).Tip: Enter numerical data ONLY. E.g: "5" and NOT "5 Page Faults".

BlackTom AI · Accepted Answer

We are asked to determine the number of page faults using the Least Recently Used (LRU) algorithm with 4 physical frames, starting empty, for the given reference string: 9, 7, 0, 4, 7, 5, 3, 2, 5, 8, 9, 5, 3, 7, 9, 0, 5, 3, 2, 1.

Step through each reference and track which pages reside in the 4 frames, counting a fault whenever the referenced page is not currently loaded (and causing an eviction if all frames are full). Keep track of the recency order so the least recently used page is evicted when needed.

1) Reference 9: frames were empty, so a fault occurs and 9 is loaded. Faults = 1. Frames: [9, -, -, -]. Recency (MRU to LRU): 9.
2) Reference 7: not in frames, fault, load 7. Faults = 2. Frames: [9,7,-,-]. Recency: 7 (MRU), 9 (LRU).
3) Reference 0: fault, load 0. Faults = 3. Frames: [9,7,0,-]. Recency: 0, 7, 9.
4) Reference 4: fault, load 4. Faults = 4. Frames: [9,7,0,4]. Recency: 4 (MRU), 0, 7, 9 (LRU).
5) Reference 7: already in frames (hit). Update recency: 7 becomes MRU. Frames unchanged: [9,7,0,4]. Recency: 7, 4, 0, 9 (LRU).
6) Reference 5: fault (not in frames). Evict LRU which is 9, load 5. Faults = 5. Frames: [5,7,0,4]. Recency: 5 (MRU), 7, 4, 0 (LRU).
7) Reference 3: fault. Evict LRU 0, load 3. Faults = 6. Frames: [5,7,4,3]. Recency: 3, 5, 7, 4 (LRU).
8) Reference 2: fault. Evict LRU 4, load 2. Faults = 7. Frames: [5,7,3,2]. Recency: 2, 3, 5, 7 (LRU).
9) Reference 5: hit. Update recency: 5 becomes MRU. Frames unchanged. Recency: 5, 2, 3, 7 (LRU).
10) Reference 8: fault. Evict LRU 7, load 8. Faults = 8. Frames: [5,2,3,8]. Recency: 8, 5, 2, 3 (LRU).
11) Reference 9: fault. Evict LRU 3, load 9. Faults = 9. Frames: [5,2,8,9]. Recency: 9, 8, 5, 2 (LRU).
12) Reference 5: hit. Update recency: 5 becomes MRU. Recency: 5, 9, 8, 2 (LRU).
13) Reference 3: fault. Evict LRU 2, load 3. Faults = 10. Frames: [5,9,8,3]. Recency: 3, 5, 9, 8 (LRU).
14) Reference 7: fault. Evict LRU 8, load 7. Faults = 11. Frames: [5,9,3,7]. Recency: 7, 3, 5, 9 (LRU).
15) Reference 9: hit. Update recency: 9 becomes MRU. Recency: 9, 7, 3, 5 (LRU).
16) Reference 0: fault. Evict LRU 5, load 0. Faults = 12. Frames: [9,7,3,0]. Recency: 0, 9, 7, 3 (LRU).
17) Reference 5: fault. Evict LRU 3, load 5. Faults = 13. Frames: [9,7,0,5]. Recency: 5, 0, 9, 7 (LRU).
18) Reference 3: fault. Evict LRU 7, load 3. Faults = 14. Frames: [9,7,0,5] replaced by [9,0,5,3]? Actually after eviction, frames become [9,0,5,3]. Recency: 3, 5, 0, 9 (LRU).
19) Reference 2: fault. Evict LRU 9, load 2. Faults = 15. Frames: [2,0,5,3]. Recency: 2, 3, 5, 0 (LRU).
20) Reference 1: fault. Evict LRU 0, load 1. Faults = 16. Frames: [2,1,5,3]. Recency: 1, 2, 3, 5 (LRU).

Total page faults count: 16, which matches the given answer.

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