Consider a system with byte-addressable memory, 32-bit logical addresses, 4…
2015
Consider a system with byte-addressable memory, 32-bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is ___________
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Correct answer: 4
Solution summary: The page table size is 4 MB.
Determine page offset bits: 4 kilobytes = 4 × 1024 bytes = 2^12 bytes, so the offset uses 12 bits.
Number of pages = 2^(32 − 12) = 2^20 = 1,048,576 page table entries.
Total size = number of entries × size per entry = 1,048,576 × 4 bytes = 4,194,304 bytes = 4 MB.
Final answer: 4 MB.
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