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On a system using paging and segmentation, the virtual address space consists of up to 8 segments where each segment can be up to bytes long. The hardware pages each segment into 256 bytes pages. Bits in the virtual address are- Page number
- a)14
- b)17
- c)21
- d)None
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
On a system using paging and segmentation, the virtual address space c...
with 256 = 28 byte pages, a 229 byte segment can have 229/28=221 pages. Thus, 21 bits are needed to specify the page number.
Most Upvoted Answer
On a system using paging and segmentation, the virtual address space c...
Explanation:
In a system using paging and segmentation, the virtual address space consists of multiple segments. Each segment can be up to a certain length, and within each segment, the system pages the data into smaller pages.
Given Information:
- The virtual address space consists of up to 8 segments.
- Each segment can be up to a certain length.
- The hardware pages each segment into 256 bytes pages.
- The bits in the virtual address represent the page number.
Calculating the Number of Bits Required for Page Number:
To determine the number of bits required for the page number, we need to consider the following factors:
- The number of segments: up to 8 segments.
- The length of each segment: up to a certain length.
- The page size: 256 bytes.
Since each segment can be up to a certain length, we need to calculate the maximum number of pages that can be present in each segment.
Let's assume the maximum length of each segment is 'L' bytes.
The number of pages in each segment would be L/256.
Since we have up to 8 segments, the total number of pages in the virtual address space would be 8 * (L/256).
To represent the total number of pages in the virtual address space, we need to calculate the number of bits required to represent this number.
Let's assume the number of bits required for the page number is 'B'.
Given that the maximum number of pages in the virtual address space is 8 * (L/256), we can calculate B using the following equation:
2^B = 8 * (L/256)
B = log2(8 * (L/256))
B = log2(L/32)
Conclusion:
The number of bits required for the page number in the virtual address is determined by the length of each segment. The correct answer is option 'C', which states that the number of bits required for the page number is 21.
In a system using paging and segmentation, the virtual address space consists of multiple segments. Each segment can be up to a certain length, and within each segment, the system pages the data into smaller pages.
Given Information:
- The virtual address space consists of up to 8 segments.
- Each segment can be up to a certain length.
- The hardware pages each segment into 256 bytes pages.
- The bits in the virtual address represent the page number.
Calculating the Number of Bits Required for Page Number:
To determine the number of bits required for the page number, we need to consider the following factors:
- The number of segments: up to 8 segments.
- The length of each segment: up to a certain length.
- The page size: 256 bytes.
Since each segment can be up to a certain length, we need to calculate the maximum number of pages that can be present in each segment.
Let's assume the maximum length of each segment is 'L' bytes.
The number of pages in each segment would be L/256.
Since we have up to 8 segments, the total number of pages in the virtual address space would be 8 * (L/256).
To represent the total number of pages in the virtual address space, we need to calculate the number of bits required to represent this number.
Let's assume the number of bits required for the page number is 'B'.
Given that the maximum number of pages in the virtual address space is 8 * (L/256), we can calculate B using the following equation:
2^B = 8 * (L/256)
B = log2(8 * (L/256))
B = log2(L/32)
Conclusion:
The number of bits required for the page number in the virtual address is determined by the length of each segment. The correct answer is option 'C', which states that the number of bits required for the page number is 21.
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