GATE CS 2001 Q2.21 — Memory Management and Virtual Memory

To find the approximate size of the page table, we first determine the number of entries in the page table and the size of each Page Table Entry (PTE).

1. Number of Entries in the Page Table

The number of entries in the page table is equal to the number of pages in the virtual address space:

• Virtual Address Space: $2^{32}$ bytes (since it is a $32$-bit address space)
• Page Size: $4\text{ KB} = 2^{12}$ bytes

Thus, the page table must contain $2^{20}$ entries.

2. Size of each Page Table Entry (PTE)

Each entry in the page table must store the physical frame number corresponding to that page:

• Physical Memory Size: $64\text{ MB} = 2^{26}$ bytes
• Number of Frames: $\frac{2^{26}}{2^{12}} = 2^{14}$ frames

To address $2^{14}$ frames, we need at least $\log_2(2^{14}) = 14$ bits for the frame number.

In practice, a page table entry also contains auxiliary control bits (such as valid/invalid, dirty, and read/write protection bits). Rounding the minimum $14$ bits up to the nearest byte boundary for alignment gives:

3. Size of the Page Table

Therefore, the approximate size of the page table is 2 MB.