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

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BlackTom AI · Accepted Answer

Least Recently Used (LRU) evicts the page that has not been referenced for the longest time. With two physical frames and all initially empty, every first-time load is a page fault; on each miss LRU chooses the victim.

Step-by-step simulation (Frames shown with most recently used on the right; “Fault” indicates a page fault and running total):

1: reference 1 — Frames: [1, -] ; Fault #1 (total 1)  
2: reference 2 — Frames: [1, 2] ; Fault #2 (2)  
3: reference 3 — LRU is 1 → evict 1 → Frames: [3, 2] ; Fault #3 (3)  
4: reference 4 — LRU is 2 → evict 2 → Frames: [3, 4] ; Fault #4 (4)  
5: reference 2 — LRU is 3 → evict 3 → Frames: [2, 4] ; Fault #5 (5)  
6: reference 1 — LRU is 4 → evict 4 → Frames: [2, 1] ; Fault #6 (6)  
7: reference 5 — LRU is 2 → evict 2 → Frames: [5, 1] ; Fault #7 (7)  
8: reference 6 — LRU is 1 → evict 1 → Frames: [5, 6] ; Fault #8 (8)  
9: reference 2 — LRU is 5 → evict 5 → Frames: [2, 6] ; Fault #9 (9)  
10: reference 1 — LRU is 6 → evict 6 → Frames: [2, 1] ; Fault #10 (10)  
11: reference 2 — hit (already in frames) ; no fault (10)  
12: reference 3 — LRU is 1 → evict 1 → Frames: [3, 2] ; Fault #11 (11)  
13: reference 7 — LRU is 2 → evict 2 → Frames: [3, 7] ; Fault #12 (12)  
14: reference 6 — LRU is 3 → evict 3 → Frames: [6, 7] ; Fault #13 (13)  
15: reference 3 — LRU is 7 → evict 7 → Frames: [6, 3] ; Fault #14 (14)  
16: reference 2 — LRU is 6 → evict 6 → Frames: [2, 3] ; Fault #15 (15)  
17: reference 1 — LRU is 3 → evict 3 → Frames: [2, 1] ; Fault #16 (16)  
18: reference 2 — hit ; no fault (16)  
19: reference 3 — LRU is 1 → evict 1 → Frames: [3, 2] ; Fault #17 (17)  
20: reference 6 — LRU is 2 → evict 2 → Frames: [3, 6] ; Fault #18 (18)

Why this is correct: each miss caused eviction of the least-recently-used page as shown. Common mistakes would be to use FIFO or not count initial loads; those yield different totals and are therefore incorrect. The complete LRU trace above yields 18 page faults.