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UGC NET Computer Science Nov 2021 (Paper II)

Consider a machine with $16\ GB$ main memory and $32$-bits virtual address space, with page size as $4KB$. Frame size and page size is same for the given machine.

What is the size of page table for handling the given virtual address space, given that each page table entry is of size $2$ bytes?

Solution

Number of virtual pages $=2^{20}$

Each page table entry size $=2$ bytes

Page table size:

$2^{20}\times2=2^{21}$ bytes

$2^{21}$ bytes $=2MB$

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