GATE CSE 2021 Set 2 Q58: Minimum page-table memory for 2 GB with a three-level table

Operating System
2 marks
NAT
No negative marking

Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below.

39-bit virtual address
Level 1 offset	Level 2 offset	Level 3 offset	Page offset
9 bits	9 bits	9 bits	12 bits

The page size is 4KB ( $1\text{KB} = 2^{10}$ bytes) and page table entry size at every level is 8 bytes. A process $P$ is currently using 2GB ( $1\text{GB} = 2^{30}$ bytes) virtual memory which is mapped to 2GB of physical memory. The minimum amount of memory required for the page table of $P$ across all levels is ________ KB.

Try it: numerical answer: type the value

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Numerical answer. No negative marking.

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Answer: 4108 (official GATE 2021 final answer key)

Concept used: Each page table at every level has $2^9$ entries of 8 bytes, i.e. exactly one 4 KB page; count the tables needed.

First step: 2 GB is $2^{19}$ pages, needing $2^{19}/2^9 = 1024$ level-3 tables; these need $1024/512 = 2$ level-2 tables and 1 level-1 table.

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Source: official GATE 2021 Set 2 CS question paper and final answer key, organised by IIT Bombay. Question paper · Answer key

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GATE CSE 2021 Set 2 Q58: Minimum page-table memory for 2 GB with a three-level table

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