Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size is 4 KB, what is the approximate size of the page table? A. 16 MB B. 8 MB C. 2 MB D. 24 MB

GATE 2001 · Operating System · Virtual Memory · medium

Answer: Page table size = 2^20 entries * 2 bytes/entry = 2 MB (Option C).

1. Compute number of virtual pages: Virtual address space = 2^32 bytes (32-bit address space) Page size = 4 KB = 2^12 bytes Number of virtual pages = 2^32 / 2^12 = 2^20 = 1,048,576 pages
2. Compute PTE size and total page table size: Physical memory = 64 MB = 2^26 bytes Number of physical frames = 2^26 / 2^12 = 2^14 Bits for frame number = 14 bits With control bits, PTE ~ 2 bytes (minimum) to 4 bytes (typical) With PTE = 2 bytes: Page table = 2^20 * 2 = 2^21 bytes = 2 MB With PTE = 4 bytes: Page table = 2^20 * 4 = 2^22 bytes = 4 MB The answer choices include 2 MB, which corresponds to 2-byte PTEs.