Assuming page size of KB and that each page table
Duration: 3 min
Assuming a page size of 1 KB and that each page table entry (PTE) takes 4bytes, how many levels of page tables would be required to map a 34-bit virtual address if every page table fits into a single page.
Answer: B. 3 — Step 1: Compute page offset: page size 1 KB = 2^10, so the offset is 10 bits. Step 2: Compute index bits per page-table level: each page is 1024 bytes and…
- A.
2
- B.
3
- C.
4
- D.
5
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Show answer & explanation
Correct answer: B
Step 1: Compute page offset: page size 1 KB = 2^10, so the offset is 10 bits.
Step 2: Compute index bits per page-table level: each page is 1024 bytes and each PTE is 4 bytes, so entries per page = 1024 / 4 = 256 = 2^8. Therefore each level supplies 8 index bits.
Step 3: Compute remaining bits to cover with levels: remaining bits = 34 (virtual address) − 10 (offset) = 24 bits.
Step 4: Compute number of levels: levels = ceil(24 / 8) = ceil(3) = 3.
Conclusion: 3 levels of page tables are required.
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