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      Question

      What will be the output of the following code snippet

      implementing the Least Recently Used (LRU) page replacement algorithm? def lru(pages, capacity):     page_frame = []     page_faults = 0     for page in pages:         if page not in page_frame:             if len(page_frame) < capacity:                 page_frame.append(page)             else:                 lru_page = min(page_frame, key=lambda p: pages.index(p))                 page_frame.remove(lru_page)                 page_frame.append(page)             page_faults += 1     return page_faults pages = [7, 0, 1, 2, 0, 3, 0, 4, 2, 3] capacity = 4 faults = lru(pages, capacity) print(faults)
      A 7 Correct Answer Incorrect Answer
      B 5 Correct Answer Incorrect Answer
      C 8 Correct Answer Incorrect Answer
      D 6 Correct Answer Incorrect Answer
      E 4 Correct Answer Incorrect Answer

      Solution

      In the LRU page replacement algorithm, when a page fault occurs and the page is not in the page frame, the algorithm replaces the least recently used page. In the provided code, the sequence of page requests is [7, 0, 1, 2, 0, 3, 0, 4, 2, 3] with a capacity of 4. The pages that cause faults are 7, 0, 1, 2, 3, and 4, leading to a total of 6 page faults throughout the execution. Why Other Options Are Wrong: A) 7: This option is incorrect as it counts one additional fault that does not occur in the given page sequence. B) 5: This option is incorrect as it undercounts the number of faults that occur during the execution. C) 8: This option is incorrect as it overcounts the faults; only 6 pages lead to faults based on the implemented logic. E) 4: This option is incorrect because it suggests that only four pages caused faults, which does not reflect the sequence provided.

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