OS-Multi-Level Paging

Question

Consider a computer system using 2-level paging with TLB.The logical address supported is 32 bits. The page table is divided into 512 pages each of size 1KB.Page table entry size at $1^{st}$ level is 2 Bytes and at second level is 4 Bytes each.

What is the memory overhead of storing the top level page table along with a page of the second level page table for a process.?

Here is my analysis.

Since, the second level page table is page aligned, with each table size=1 Page size=1KB. So, offset bits=10.

Also, 10 bits(P2) would be required to index to a second level page table.Since, at second level, we have total 512 pages, each containing page table entries, at first level, we need to keep addresses to address these 512 second level page tables.9 bits at outer level(P1) would have done the job, but since the logical address is of 32 bits, we get 12 bits for the Outer level page table indexing and thus it can have maximum of $2^{12}$ entries, of which only $2^9$ would be usable.

So, size of outer level page table=$2^{12}*2B=2^{13}B=8KB$

Size of 1 page of second level PT=1KB.

So, the total required memory overhead=8+1=9KB.

Is my analysis and answer both correct?

Comments

@srestha-$P_2$ is number of bits required to index into a second level page table.

Now, a second level page table contains $\frac{2^{10}}{2^2}=2^8$ entries and hence 8 bits to index.

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@Ayush

why dividing?

there are $512=2^{9}$ PTE and each PTE has size $2^{2}$ B

So, Page table size $2^{11}$

right?

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What is page-aligned paging? How it’s different from normal paging? @adad20