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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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Data

1 word (W) = 1 byte (B)

Logical address = 232 B

page size = 4 KB = 212 B

page table entry (PTE) = y B

Formula:

Page table size (PTS) = number of pages × y

Calculation:

number of pages = \(\frac{{{2^{32}}}}{{{2^{12}}}} = {2^{20}}\)

Page table size (PTS) = number of pages × y

\(PTS = {2^{20}} \times 4B\)

∴ PTS = 4 MB

Important points

1 MB = 220 B

where B stands for byte

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