Consider a machine with 64 Mbyte physical memory and a 32-bit

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

1. 16 Mbyte
2. 8 Mbyte
3. 2 Mbyte
4. 24 Mbyte

Correct Option: C

Virtual Address Space = 32 Bit, Page size = 4kb = 2² x 2¹⁰ = 2¹²

PTE = V.A.S. / Page Size =	2³²	= 2²⁰ .....eq.(1)
	2¹²

Physical memory size = 64MB = 2⁶ x 2¹⁰ x 2¹⁰ = 2²⁶

Then No. of Frames in Physical Memory =	2²⁶	= 2¹⁴
	2¹²

And page table needs to store the address of all these 2¹⁴ page frames. Therefore, each page table entry will contain 14 bits address of the page frame and 1 bit for valid-invalid bit. we suppose 16 bit require it means 2 bytes (approx.).
So Size of Page table = Total PTE (from eq (1) x Size of page table entry = 2²⁰ x 2B = 2MB
Hence (c) is correct option.