GATE CS 2025 Set 2 Q48 — Memory Management and Virtual Memory

To find the value of $b$, we can break down the logical address into its constituent parts based on the given specifications:

1. Logical Address Size:

   The logical address space is $2^{32}$ bytes. Therefore, the logical address size is $32$ bits.

2. Page Offset:

   The page size is $4096$ bytes ($2^{12}$ bytes). Thus, the offset within the desired page requires $\log_2(4096) = 12$ bits.

3. Inner Page Table Index (Offset within the page of the inner page table):

   All pages in the system, including those holding the inner page tables, have a size of $4096$ bytes.

   Each entry in the inner page table is $8$ bytes ($2^3$ bytes).

   The number of entries that can fit into a single page of the inner page table is:

   $$\text{Number of entries} = \frac{\text{Page Size}}{\text{Page Table Entry Size}} = \frac{4096\text{ bytes}}{8\text{ bytes}} = 512 = 2^9$$

   Thus, to index an entry within a page of the inner page table, we need $\log_2(512) = 9$ bits.

The logical address is divided as follows:

Substituting the values:

Thus, the value of $b$ is 11.