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Consider a computer system with 40-bit virtual addressing and page size of sixteen kilobytes. If the computer system has a one-level page table per process and each page table entry requires 48 bits, then the size of the per-process page table is _____ megabytes.

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Best answer

Correct answer is 384

Given

LA = 40 bit = LAS = 240

Page size = 16KB

Page table Entry size (e) = 48 bits(or)6 bytes 

Page table size = ?

Size of the page table = n x e

∴ No. of pages (n) = LAS/PS = 240/214 = 226 = 64M

∴ Page table size = 64 x 6B = 384MB

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