GATE 2015 CS – Question 29
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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Correct answer: 4
Explanation
The number of pages is $\frac{2^{32}}{2^{12}} = 2^{20}$. At 4 bytes per entry the page table is $2^{20} \times 4 = 2^{22}$ bytes, which is 4 MB.