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A computer system implements 8 kilobyte pages and a 32- bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ______ bits.
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- 16
- 6
- 64
- 36
Correct Option: D
Given page size = 8 kb
PAS = 32 – bit
∴ Number of pages = | = | = 219 frames | ||
P.S | 213 |
So, it is given that each page table entry contains a valid bit, a dirty bit, 3 permission bits. ⇒ 5 bits are reserved.
∴ It means one entry requires 19 + 5 = 24 bits
∴ Page Table size = n × e
24 × 220 × 3 = n × (24)
⇒ n = | = 223 pages | |
24 |
Length of the virtual address = 23 + 13 = 36 bits