Question 52

Let the page size be $2^p$ bytes, so the page offset is $p$ bits.

1. Virtual address format:

   • Total virtual address size = 46 bits.
   • Page offset = $p$ bits.
   • Virtual page number = $46 - p$ bits.

2. Page table entry size and index size:

   • Each page table entry (PTE) is 32 bits $= 4$ bytes $= 2^2$ bytes.
   • Since each table at each level ($T_1, T_2, T_3$) occupies exactly one page of size $2^p$ bytes, the number of entries per page table is:
     $$\dfrac{2^p}{2^2} = 2^{p - 2}\text{ entries}$$
   • Therefore, each indexing level requires $p - 2$ bits.

3. Relating virtual address bits:
   The 3 levels of page tables consume $(p - 2) + (p - 2) + (p - 2) = 3(p - 2)$ bits for the virtual page number:

   $$3(p - 2) + p = 46$$
   $$3p - 6 + p = 46$$
   $$4p = 52 \implies p = 13$$

4. Calculate page size:

   $$\text{Page size} = 2^{13}\text{ bytes} = 8192\text{ bytes} = 8\text{ KB}$$

Thus, the page size is 8 KB, which matches option (C).