Question

A computer has 32-bit virtual addresses and the page size is 2^3 KB. Suppose the text, data, and stack segments of a process need 5 page(s) each. If the computer uses two-level page tables, with 3 bits for the second-level index, how many page table entries (PTEs) are in each second-level page table?

Answer 1

Here the page size = 2^3KB => Page offset = log 2^3KB = 13 bits.

Therefore page number bits = 32 - 13 = 19 bits.

Here we are using 2 level page table.

These 19 bit page number will be divided into 2 level page table.

But in the question, total # of page occupied is 15.

Therefore page number bits = ceil ( log (15)) = 4

Here it's mentioned that computer is using 3 bits for second level index.

Therefore # of bits for 1st level page table = 4 - 3 = 1

Thus the bits distribution is :

Virtual Address

First Level Page Table (1)

Second Level Page Table (3 )

Page Offset (13 bits)

Now # of bits for second level page table is 3.

This means # of page table entries in each second level page table is 2^3 = 8.

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