GATE | GATE-CS-2001 | Question 46

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?

(A)

16 MB

(B)

8 MB

(C)

2 MB

(D)

24 MB

Answer: (C)
Explanation:

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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