NTA-UGC-NET 2021 Dec & 2022 June Paper-1
October 28, 2023GATE 2021 CS-Set-2
October 28, 2023GATE 2021 CS-Set-2
| Question 53 |
The page size is 4KB (1KB = 210bytes) and page table entry size at every level is 8 bytes. A process P is currently using 2GB (1GB = 230bytes) 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.
| 8.01 |
A system can support upto 2^39 bytes of Virtual address space. Three level paging is used with Level 1 (9 bits), Level 2 (9 bits) and Level 3 (9 bits) and page offset of 12 bits.
Page table entry is 8 bytes at every level. Therefore,
Level 3 page table size : 2^9 * 8bytes = 2^12 bytes. (Fits in one frame)
Level 2 page table size : 2^9 * 8bytes = 2^12 bytes. (Fits in one frame)
Level 1 page table size : 2^9 * 8bytes = 2^12 bytes. (Fits in one frame).
However, the process P has a virtual address space of 2^31 bytes. Thus, it is clear that the last level page table does not occupy the frame completely. Let x be the number of bits indexed in the last level.
Therefore, we can have (2^x) * (2^9) * (2^9)* (2^12) = 2^31 => x =1. Thus, the minimum size of page table for process P would be,
Page table size at level-1 + Page table size at level-2 + Page table size at level-3 =
2^1 * 8bytes + 2^9 * 8bytes + 2^9 * 8bytes = (1026)*8 bytes = 8.01 KB
A system can support upto 2^39 bytes of Virtual address space. Three level paging is used with Level 1 (9 bits), Level 2 (9 bits) and Level 3 (9 bits) and page offset of 12 bits.
Page table entry is 8 bytes at every level. Therefore,
Level 3 page table size : 2^9 * 8bytes = 2^12 bytes. (Fits in one frame)
Level 2 page table size : 2^9 * 8bytes = 2^12 bytes. (Fits in one frame)
Level 1 page table size : 2^9 * 8bytes = 2^12 bytes. (Fits in one frame).
However, the process P has a virtual address space of 2^31 bytes. Thus, it is clear that the last level page table does not occupy the frame completely. Let x be the number of bits indexed in the last level.
Therefore, we can have (2^x) * (2^9) * (2^9)* (2^12) = 2^31 => x =1. Thus, the minimum size of page table for process P would be,
Page table size at level-1 + Page table size at level-2 + Page table size at level-3 =
2^1 * 8bytes + 2^9 * 8bytes + 2^9 * 8bytes = (1026)*8 bytes = 8.01 KB