Consider a machine with 64 MB physical memory and a 32-bit virtual address…
2001
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
Attempted by 3 students.
Show answer & explanation
Correct answer: C
Virtual address size = 32 bits and page size = 4 KB = 2^12 bytes.
Number of virtual pages = 2^32 / 2^12 = 2^20.
Physical memory = 64 MB = 2^26 bytes. With 4 KB pages, the number of physical frames is 2^26 / 2^12 = 2^14. Hence the frame number needs 14 bits.
Ignoring small status bits and rounding the entry size to the nearest convenient byte boundary, each page-table entry needs about 2 bytes. Therefore, page table size ≈ 2^20 × 2 bytes = 2^21 bytes = 2 MB.
So, option C is correct.