A bit off topic but I am unable to solve this
Consider a virtual memory system with physical memory of 8GB, a page size of 8KB and 46 bit virtual address. Assume every page table exactly fits into a single page. If page table entry size is 4B then how many levels of page tables would be required.
Hey @anindya-gupta
Page size = 8KB = 2^13 B
Virtual address space size = 2^46 B
PTE = 4B = 2^2 B
Number of pages or number of entries in page table, = (virtual address space size) / (page size) = 2^46B/2^13 B = 2^33
Size of page table,
= (number of entries in page table)*(size of PTE) = 2^33*2^2 B = 2^35 B
To create one more level,
Size of page table > page size Number of page tables in last level, = 2^35 B / 2^13 B = 2^22
Base address of these tables are stored in page table [second last level].
Size of page table [second last level] = 2^22*2^2B = 2^24B
To create one more level,
Size of page table [second last level] > page size
Number of page tables in second last level = 2^24B/2^13 B = 2^11
Base address of these tables are stored in page table [third last level]
Size of page table [third last level] = 2^11*2^2 B = 2^13 B = page size
∴ 3 levels are required.
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