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:3, 7, 1, 9, 4, 8, 0, 2, 6, 5, 9, 3, 7, 1, 8, 4, 0, 6, 2, 5Determine how many page faults will occur for the Least Recently Used, assuming there are only FOUR (4) physical page frames, and all are initially invalid.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:3, 7, 1, 9, 4, 8, 0, 2, 6, 5, 9, 3, 7, 1, 8, 4, 0, 6, 2, 5Determine how many page faults will occur for the Least Recently Used, assuming there are only FOUR (4) physical page frames, and all are initially invalid.Tip: Enter numerical data ONLY. E.g: "5" and NOT "5 Page Faults".

BlackTom AI · Accepted Answer

Question restatement: You are asked to determine how many page faults occur under the Least Recently Used (LRU) page replacement policy when there are four physical page frames, all initially invalid, given the sequence of references: 3, 7, 1, 9, 4, 8, 0, 2, 6, 5, 9, 3, 7, 1, 8, 4, 0, 6, 2, 5.

Step-by-step analysis of each reference with LRU behavior:
1) Reference 3: All frames invalid, so a fault occurs and 3 is loaded. Frames: [3, -, -, -]. Faults = 1. 
2) Reference 7: Not in frames, fault, load 7. Frames: [3,7,-,-]. Faults = 2. 
3) Reference 1: Not in frames, fault, load 1. Frames: [3,7,1,-]. Faults = 3. 
4) Reference 9: Not in frames, fault, load 9. Frames: [3,7,1,9]. Faults = 4. 
5) Reference 4: Not in frames, need to evict the least recently used among {3,7,1,9}. The least recently used is 3 (last used at step 1). Evict 3, load 4. Frames: [4,7,1,9]. Faults = 5. 
6) Reference 8: Not in frames, evict LRU among {4,7,1,9}. LRU is 4 (last used at step 5). Evict 4, load 8. Frames: [8,7,1,9]. Faults = 6. 
7) Reference 0: Not in frames, evict LRU among {8,7,1,9}. LRU is 8 (last used at step 6). Evict 8, load 0. Frames: [0,7,1,9]. Faults = 7. 
8) Reference 2: Not in frames, evict LRU among {0,7,1,9}. LRU is 0 (last used at step 7). Evict 0, load 2. Frames: [2,7,1,9]. Faults = 8. 
9) Reference 6: Not in frames, evict LRU among {2,7,1,9}. LRU is 2 (last used at step 8). Evict 2, load 6. Frames: [6,7,1,9]. Faults = 9. 
10) Reference 5: Not in frames, evict LRU among {6,7,1,9}. LRU is 6 (last used at step 9). Evict 6, load 5. Frames: [5,7,1,9]. Faults = 10. 
11) Reference 9: Already in frames, a hit. No eviction. Frames remain [5,7,1,9]. Faults = 10. 
12) Reference 3: Not in frames, evict LRU among {5,7,1,9}. LRU is 7 (last used at step 2). Evict 7, load 3. Frames: [5,3,1,9]. Faults = 11. 
13) Reference 7: Not in frames (we replaced it at step 12), evict LRU among {5,3,1,9}. LRU is 1 (last used at step 3). Evict 1, load 7. Frames: [5,3,7,9]. Faults = 12. 
14) Reference 1: Not in frames, evict LRU among {5,3,7,9}. LRU is 5 (last used at step 10). Evict 5, load 1. Frames: [1,3,7,9]. Faults = 13. 
15) Reference 8: Not in frames, evict LRU among {1,3,7,9}. LRU is 9 (last used at step 11). Evict 9, load 8. Frames: [1,3,7,8]. Faults = 14. 
16) Reference 4: Not in frames, evict LRU among {1,3,7,8}. LRU is 3 (last used at step 12). Evict 3, load 4. Frames: [1,4,7,8]. Faults = 15. 
17) Reference 0: Not in frames, evict LRU among {1,4,7,8}. LRU is 7 (last used at step 13). Evict 7, load 0. Frames: [1,4,0,8]. Faults = 16. 
18) Reference 6: Not in frames, evict LRU among {1,4,0,8}. LRU is 8 (last used at step 15). Evict 8, load 6. Frames: [1,4,0,6]. Faults = 17. 
19) Reference 2: Not in frames, evict LRU among {1,4,0,6}. LRU is 4 (last used at step 16). Evict 4, load 2. Frames: [1,2,0,6]. Faults = 18. 
20) Reference 5: Not in frames, evict LRU among {1,2,0,6}. LRU is 0 (last used at step 17). Evict 0, load 5. Frames: [1,2,5,6]. Faults = 19.

Summary: After processing all 20 references with a 4-frame LRU cache, the total number of page faults accumulated is 19. The sequence includes a mix of hits (e.g., step 11) and misses (faults) at each step where the referenced page was not present in the four frames, triggering an eviction based on the least recently used policy. The computation above tracks the exact contents of the four frames and the last-used times for each page to identify the correct eviction candidate at every fault step.

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