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A system that uses a two-level page table has 212 - byte pages and 36-bit virtual addresses. The first 8-bits of the address serve as the index into the first level page table.
The number of bits specify the second level index is ________?
Correct answer is '16'. Can you explain this answer?
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Verified Answer
A system that uses a two-level page table has 212- byte pages and 36-b...
12 bits is the answer
as it is given that page size is 212 Bytes, so 12 bits are needed to specify that offset. The number of bits specify the second level index is:
36 - (12 + 8)
= 36 - 20
= 16
Most Upvoted Answer
A system that uses a two-level page table has 212- byte pages and 36-b...
Given:
- Page size = 212 bytes
- Virtual address size = 36 bits
- First level page table index size = 8 bits
To find:
- Second level page table index size
Solution:
1. Determine the number of entries in the first level page table:
- Each entry in the first level page table points to a second level page table.
- The size of a page table entry is equal to the size of a physical address, which is typically the same as the virtual address size.
- Since the page size is 212 bytes, each page can hold 212/212 = 1 entry.
- Therefore, the first level page table can hold 28 = 256 entries.
2. Determine the size of a second level page table entry:
- Each entry in the second level page table points to a physical page frame.
- The size of a page frame is also 212 bytes.
- Since the virtual address size is 36 bits, the second level page table index size is 36 - 8 - 12 = 16 bits, where:
- 8 bits are used for the first level page table index.
- 12 bits are used for the page offset (212 bytes/page).
Therefore, the second level page table index size is 16 bits.
- Page size = 212 bytes
- Virtual address size = 36 bits
- First level page table index size = 8 bits
To find:
- Second level page table index size
Solution:
1. Determine the number of entries in the first level page table:
- Each entry in the first level page table points to a second level page table.
- The size of a page table entry is equal to the size of a physical address, which is typically the same as the virtual address size.
- Since the page size is 212 bytes, each page can hold 212/212 = 1 entry.
- Therefore, the first level page table can hold 28 = 256 entries.
2. Determine the size of a second level page table entry:
- Each entry in the second level page table points to a physical page frame.
- The size of a page frame is also 212 bytes.
- Since the virtual address size is 36 bits, the second level page table index size is 36 - 8 - 12 = 16 bits, where:
- 8 bits are used for the first level page table index.
- 12 bits are used for the page offset (212 bytes/page).
Therefore, the second level page table index size is 16 bits.
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A system that uses a two-level page table has 212- byte pages and 36-bit virtual addresses. The first 8-bits of the address serve as the index into the first level page table.The number of bits specify the second level index is ________?Correct answer is '16'. Can you explain this answer?
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