GATE | GATE-CS-2015 (Set 1) | Question 65
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 ___________
(A) 2
(B) 4
(C) 8
(D) 16
Answer: (B)
Explanation:
Number of entries in page table = 232 / 4Kbyte
= 232 / 212
= 220
Size of page table = (No. page table entries)*(Size of an entry)
= 220 * 4 bytes
= 222
= 4 Megabytes
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Last Updated :
28 Jun, 2021
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