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A machine has 64-bit virtual addresses and 48-bit physical addresses. Pages are 16K. How many entries are needed for a conventional page table?
- a)250
- b)234
- c)214
- d)248
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A machine has 64-bit virtual addresses and 48-bit physical addresses. ...
Given information:
- 64-bit virtual addresses
- 48-bit physical addresses
- Pages are 16K
To find: Number of entries needed for a conventional page table
Solution:
1. Calculate the page size:
- Page size is 16K = 2^14 bytes
- 2^14 = 16,384 bytes
2. Calculate the number of pages:
- 64-bit virtual addresses, so the address space is 2^64 bytes
- Divide the address space by the page size to get the number of pages:
- (2^64) / (2^14) = 2^50 pages
3. Calculate the size of each page table entry:
- Physical addresses are 48 bits, so each page table entry needs to store a 48-bit physical address
- Each entry also needs to store some additional information, such as a valid/invalid bit and permission bits
- For simplicity, assume each entry is 64 bits (8 bytes) in size
4. Calculate the size of the page table:
- The page table needs to have one entry for each page, so the size of the page table is:
- (2^50) * 8 bytes = 2^53 bytes
5. Calculate the number of entries needed for the page table:
- Divide the size of the page table by the size of each entry:
- (2^53 bytes) / (8 bytes) = 2^45 entries
6. Choose the closest answer option:
- Option B: 2^34 = 17,179,869,184, which is closest to 2^45 = 35,184,372,088. Therefore, the correct answer is option B.
Final answer: The number of entries needed for a conventional page table is 2^34, which is approximately 17.2 billion.
- 64-bit virtual addresses
- 48-bit physical addresses
- Pages are 16K
To find: Number of entries needed for a conventional page table
Solution:
1. Calculate the page size:
- Page size is 16K = 2^14 bytes
- 2^14 = 16,384 bytes
2. Calculate the number of pages:
- 64-bit virtual addresses, so the address space is 2^64 bytes
- Divide the address space by the page size to get the number of pages:
- (2^64) / (2^14) = 2^50 pages
3. Calculate the size of each page table entry:
- Physical addresses are 48 bits, so each page table entry needs to store a 48-bit physical address
- Each entry also needs to store some additional information, such as a valid/invalid bit and permission bits
- For simplicity, assume each entry is 64 bits (8 bytes) in size
4. Calculate the size of the page table:
- The page table needs to have one entry for each page, so the size of the page table is:
- (2^50) * 8 bytes = 2^53 bytes
5. Calculate the number of entries needed for the page table:
- Divide the size of the page table by the size of each entry:
- (2^53 bytes) / (8 bytes) = 2^45 entries
6. Choose the closest answer option:
- Option B: 2^34 = 17,179,869,184, which is closest to 2^45 = 35,184,372,088. Therefore, the correct answer is option B.
Final answer: The number of entries needed for a conventional page table is 2^34, which is approximately 17.2 billion.
Community Answer
A machine has 64-bit virtual addresses and 48-bit physical addresses. ...
Concept:
The given data,
Logical address space = 264 Bytes.
Physical address space = 248 Bytes.
Page size PS = 16 K = 24 x 210 B = 214 Bytes.
Number of pages = Logical address space / Page size
Number of pages = 264 / 214
Number of pages = 250
The given data,
Logical address space = 264 Bytes.
Physical address space = 248 Bytes.
Page size PS = 16 K = 24 x 210 B = 214 Bytes.
Number of pages = Logical address space / Page size
Number of pages = 264 / 214
Number of pages = 250
The page table entry is the number number of bits required to get any frame number.
Number of frames = Physical address space / Page size
Number of frames = 248 / 214
Number of frames = 234
Hence the correct answer is 234.
Number of frames = Physical address space / Page size
Number of frames = 248 / 214
Number of frames = 234
Hence the correct answer is 234.
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A machine has 64-bit virtual addresses and 48-bit physical addresses. Pages are 16K. How many entries are needed for a conventional page table?a)250b)234c)214d)248Correct answer is option 'B'. Can you explain this answer?
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