Consider processor has virtual address 50-bits, the physical address is
Duration: 9 min
Consider a processor has virtual address 50-bits, the physical address is 32 bits and the page size is 16KB. Each individual table at any level of the page table occupies a single page of physical memory. Using multilevel paging, consider page table entry requires additional 14 bits for the special purpose?Which of the following is the correct division of virtual address?
Answer: A. <12, 12, 12, 14 > — Key calculations: determine the page offset, PTE size, entries per page, and index widths. Page size = 16 KB = 2^14 bytes, so the page offset requires 14…
- A.
<12, 12, 12, 14 >
- B.
<6, 10, 10, 10, 14 >
- C.
<10, 12, 14, 14 >
- D.
<14, 12, 10, 14>
Attempted by 71 students.
Show answer & explanation
Correct answer: A
Key calculations: determine the page offset, PTE size, entries per page, and index widths.
Page size = 16 KB = 2^14 bytes, so the page offset requires 14 bits.
Physical address = 32 bits, so the physical frame number uses 32 - 14 = 18 bits.
Each page table entry requires an extra 14 bits, so PTE size = 18 + 14 = 32 bits = 4 bytes.
Entries per page = page size / PTE size = 16,384 bytes / 4 bytes = 4096 = 2^12, so each page-table index must be 12 bits.
Virtual address bits above the offset = 50 - 14 = 36 bits. Dividing these into 12-bit indices gives three levels (3 x 12 = 36).
Final division: <12, 12, 12, 14> — three 12-bit page-table indices and a 14-bit page offset.
A video solution is available for this question — log in and enroll to watch it.