GATE 2016 CS – Question 57
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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Correct answer: 384
Explanation
A 16 KB page is $2^{14}$ bytes, so the number of pages is $\frac{2^{40}}{2^{14}} = 2^{26}$. Each entry takes 48 bits, which is 6 bytes. The page table is $2^{26} \times 6$ bytes $= 402653184$ bytes, which is 384 MB.