CODEWITHSUNDEEP
memoryoperating-systemvirtual-memory
jn025
jn025

Reputation: 2895

Calculating size of outer/2nd-level page table

Kind of a homework question but I am more trying to wrap my head around this concept:

If I have a 32bit virtual address in a two-level paging system where the page size is 8KB and the outer table has 1024 entries, how can I calculate how many bits are to represent (size of) the second level/outer page table?

The answer is supposedly 9 bits but I am unsure of how to calculate this or what the formula is.

Upvotes: 3

Views: 6116

Answers (1)

cadaniluk
cadaniluk

Reputation: 15229

The virtual address is divided into this:

PTI - Page Table Index

1st PTI | 2nd PTI | Page Offset

The "outer page table", the first PTI, has 1024 entries. 10 bits are needed to represent 1024 different states because 2^10 = 1024.

The page offset has to be capable of indexing 8 KiB of bytes, therefore it is 13 bits big. 8 KiB corresponds to 8192 different addresses and 2^13 = 8192.

Now, the 2nd page table is left. We can calculate its size easily by subtraction:

32 = 10 + 13 + x  = 23 + x    | - 23
9 = x
x = 9

Therefore the "inner" page table can hold up to 2^9 = 512 entries.

Upvotes: 6

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