Question
What will be the output of the following code snippet
implementing the Least Recently Used (LRU) page replacement algorithm? def lru(pages, capacity): Â Â page_frame = [] Â Â page_faults = 0 Â Â for page in pages: Â Â Â Â if page not in page_frame: Â Â Â Â Â Â if len(page_frame) < capacity: Â Â Â Â Â Â Â Â page_frame.append(page) Â Â Â Â Â Â else: Â Â Â Â Â Â Â Â lru_page = min(page_frame, key=lambda p: pages.index(p)) Â Â Â Â Â Â Â Â page_frame.remove(lru_page) Â Â Â Â Â Â Â Â page_frame.append(page) Â Â Â Â Â Â page_faults += 1 Â Â return page_faults pages = [7, 0, 1, 2, 0, 3, 0, 4, 2, 3] capacity = 4 faults = lru(pages, capacity) print(faults)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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