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GATE | GATE-CS-2001 | Question 46

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Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size is 4KB, what is the approximate size of the page table?


16 MB


8 MB


2 MB


24 MB

Answer: (C)


Physical Address Space = 64MB = 226B
Virtual Address = 32-bits,  ∴ Virtual Address Space = 232B
Page Size = 4KB = 212B
Number of pages = 232/212 = 220 pages.

Number of frames = 226/212 = 214 frames.

∴ Page Table Size = 220×14-bits ≈ 220×16-bits ≈ 220×2B = 2MB.

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Last Updated : 28 Jun, 2021
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