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What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?
- a)211
- b)215
- c)219
- d)220
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
What is the size of the physical address space in a paging system whic...
Data
page table entries = 64
page table entry size = 11 bit
page size = 512 bytes
page table entries = 64
page table entry size = 11 bit
page size = 512 bytes
Formula:
Size of physical address = pagging bits + offset bits
Size of physical address = pagging bits + offset bits
Calculation:
paging bits = 11 - 1 = 10(as 1 valid bit is also included)
offset bits = log2 (512) = 9 bits
size of physical address = 10 + 9 = 19 bits
Hence the correct answer is 219.
paging bits = 11 - 1 = 10(as 1 valid bit is also included)
offset bits = log2 (512) = 9 bits
size of physical address = 10 + 9 = 19 bits
Hence the correct answer is 219.
Community Answer
What is the size of the physical address space in a paging system whic...
The physical address space in a paging system is determined by the number of bits used to represent the physical addresses. In this case, we need to calculate the size of the physical address space given a page table with 64 entries, each consisting of 11 bits (including a valid/invalid bit), and a page size of 512 bytes.
1. Determining the number of bits for the page offset:
- Since the page size is 512 bytes, we need log2(512) bits to represent the page offset.
- log2(512) = 9 bits
2. Determining the number of bits for the page number:
- The page table has 64 entries, and each entry consists of 11 bits (including a valid/invalid bit).
- Therefore, we need 11 - 1 = 10 bits to represent the page number.
3. Calculating the size of the physical address space:
- The physical address space is determined by the sum of the bits for the page offset and the bits for the page number.
- Size = Number of bits for page offset + Number of bits for page number
- Size = 9 bits + 10 bits
- Size = 19 bits
4. Converting the size from bits to bytes:
- Since 8 bits make up 1 byte, we divide the size by 8.
- Size (in bytes) = 19 bits / 8
- Size (in bytes) = 2.375 bytes
5. Rounding up to the nearest power of 2:
- The size of the physical address space needs to be a power of 2.
- The nearest power of 2 greater than 2.375 bytes is 4 bytes.
- Since 1 byte is 2^0, 4 bytes is 2^2.
Therefore, the size of the physical address space in this paging system is 2^19 bytes, which is equivalent to 2^19 / 2^10 = 2^9 pages. This is approximately equal to 2^9 * 512 bytes = 2^12 bytes.
Hence, the correct answer is option C) 2^19 or 219.
1. Determining the number of bits for the page offset:
- Since the page size is 512 bytes, we need log2(512) bits to represent the page offset.
- log2(512) = 9 bits
2. Determining the number of bits for the page number:
- The page table has 64 entries, and each entry consists of 11 bits (including a valid/invalid bit).
- Therefore, we need 11 - 1 = 10 bits to represent the page number.
3. Calculating the size of the physical address space:
- The physical address space is determined by the sum of the bits for the page offset and the bits for the page number.
- Size = Number of bits for page offset + Number of bits for page number
- Size = 9 bits + 10 bits
- Size = 19 bits
4. Converting the size from bits to bytes:
- Since 8 bits make up 1 byte, we divide the size by 8.
- Size (in bytes) = 19 bits / 8
- Size (in bytes) = 2.375 bytes
5. Rounding up to the nearest power of 2:
- The size of the physical address space needs to be a power of 2.
- The nearest power of 2 greater than 2.375 bytes is 4 bytes.
- Since 1 byte is 2^0, 4 bytes is 2^2.
Therefore, the size of the physical address space in this paging system is 2^19 bytes, which is equivalent to 2^19 / 2^10 = 2^9 pages. This is approximately equal to 2^9 * 512 bytes = 2^12 bytes.
Hence, the correct answer is option C) 2^19 or 219.
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What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer?
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