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GATE | GATE-CS-2015 (Set 1) | Question 65

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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 ___________ (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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