Given 4 GB (= 2^32 bytes, approximately 4.3 x 10^9 bytes) of virtual address…
2023
Given 4 GB (= 2^32 bytes, approximately 4.3 x 10^9 bytes) of virtual address space, page size of 4 KB, and each page table entry of 5 bytes, how many virtual pages are there and what is the size of the complete page table?
Answer: A. 1,048,576 pages and 5,242,880 bytes — To find the number of virtual pages, divide the virtual address space by the page size. 4 GB = 2^32 bytes. 4 KB = 2^12 bytes. Number of virtual pages = 2^32 /…
- A.
1,048,576 pages and 5,242,880 bytes
- B.
215000 and 40960 bytes
- C.
10750 and 10240 bytes
- D.
43000 and 1024 bytes
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Show answer & explanation
Correct answer: A
To find the number of virtual pages, divide the virtual address space by the page size.
4 GB = 2^32 bytes.
4 KB = 2^12 bytes.
Number of virtual pages = 2^32 / 2^12 = 2^20 = 1,048,576 pages.
Each page table entry is 5 bytes, so the complete page table size is:
1,048,576 x 5 = 5,242,880 bytes.
Therefore, option A is correct: 1,048,576 pages and 5,242,880 bytes.