GATE | GATE CS 2021 | Set 2 | Question 58
Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below:
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The page size is 4 KB = (1KB = 210 bytes) and page table entry size at every level is 8 bytes. A process P is currently using 2 GB (1 GB = 230 bytes) virtual memory which OS mapped to 2 GB of physical memory. The minimum amount of memory required for the page table of P across all levels is _________ KB
Explanation: Given that Process P is using 2GB of physical memory and page size is 2^12 Bytes,
Number of pages = 2^31 / 2^12 = 2^19
Each page need a entry in 3rd level and there are 2^9 entries per page table in 3rd level.
So we need 2^19/2^9 = 2^10 page tables in 3rd level.
This implies there are 2^10 entries in 2nd level and there are 2^9 entries per page table in 2nd level.
So we need 2^10/2^9 = 2 page tables in 2nd level.
Now we have 2 entries in 1st level and hence, we need only 1 page table in 1st level.
So on total we need,
2^10 +2 +1 = 1027 page tables each with 2^9 entries of size 8 bytes.
Total size of page tables = 1027*2^9*8 Bytes
= 1027*2^12 Bytes
= 1027*4 KB
= 4108 KB
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