GATE CS 2023 Q48 — Memory Management and Virtual Memory

To find the number of levels $L$ in the multi-level page table, we analyze the virtual address split:

1. Page Offset:

   The page size is $4\text{ KB} = 2^{12}\text{ bytes}$.
   Therefore, the page offset requires $\log_2(2^{12}) = 12$ bits.

2. Virtual Page Number (VPN):

   The total virtual address size is $57$ bits.
   The bits allocated for the Virtual Page Number (VPN) are:
   $$\text{VPN bits} = 57 - 12 = 45\text{ bits}$$

3. Page Table Entry (PTE) per Page:

   Each page table at any level must fit within a single page of size $4\text{ KB}$.
   Given that each Page Table Entry (PTE) is $8\text{ bytes}$:
   $$\text{Number of entries per page table} = \frac{4\text{ KB}}{8\text{ B}} = \frac{4096}{8} = 512 = 2^9\text{ entries}$$
   Thus, each level of the page table indexes exactly $9$ bits of the virtual address.

4. Number of Levels ($L$):

   Since each level translates $9$ bits of the VPN, the number of levels $L$ required to translate the $45$-bit VPN is:
   $$L = \frac{45}{9} = 5$$

Thus, the value of $L$ is 5.