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A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.
Correct answer is '36'. Can you explain this answer?
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Most Upvoted Answer
A computer system implements 8-kilobyte pages and a 32 - bit physical ...
Page size = 213 Bytes = Frame size
Hence in the page table:
= 8 × 220 = 223
Hence all entries can be addressed by 23 bits.
Thus total size of virtual address = 23 + 13
= 36 bits.
Hence in the page table:
= 8 × 220 = 223
Hence all entries can be addressed by 23 bits.
Thus total size of virtual address = 23 + 13
= 36 bits.
Community Answer
A computer system implements 8-kilobyte pages and a 32 - bit physical ...
Given information:
- Page size: 8 kilobytes
- Physical address space: 32 bits
- Page table entry consists of:
- Valid bit
- Dirty bit
- 3 permission bits
- Translation
Maximum size of the page table of a process:
- The maximum size of the page table of a process is given as 24 megabytes.
Calculating the number of pages in the page table:
- Since the page table size is given in megabytes, we need to convert it to kilobytes.
- 1 megabyte = 1024 kilobytes
- Therefore, the maximum size of the page table is 24 * 1024 = 24576 kilobytes.
- Each page is 8 kilobytes in size.
- Number of pages = Maximum size of the page table / Page size
- Number of pages = 24576 / 8 = 3072 pages
Calculating the number of bits required to address the pages:
- The number of bits required to address the pages is equal to the logarithm of the number of pages to the base 2.
- Number of bits = log2(Number of pages)
- Number of bits = log2(3072)
- Number of bits = 11.934
Calculating the number of bits required for the translation:
- The physical address space is given as 32 bits.
- The page size is 8 kilobytes, which is equal to 2^13 bytes.
- The number of bits required for the translation is equal to the logarithm of the page size to the base 2.
- Number of bits = log2(Page size in bytes)
- Number of bits = log2(2^13)
- Number of bits = 13
Calculating the number of bits for the valid bit, dirty bit, and permission bits:
- Each page table entry contains a valid bit, a dirty bit, and three permission bits.
- Total number of bits for these fields = 1 (valid bit) + 1 (dirty bit) + 3 (permission bits) = 5 bits
Calculating the total number of bits required for the virtual address:
- Total number of bits = Number of bits for page addressing + Number of bits for translation + Number of bits for entry fields
- Total number of bits = 11.934 + 13 + 5
- Total number of bits = 29.934
Rounding up the total number of bits:
- Since the number of bits cannot be a fraction, we need to round up the total number of bits to the nearest integer.
- Rounded up total number of bits = ceil(29.934) = 30 bits
Conclusion:
- The length of the virtual address supported by the system is 30 bits.
- Page size: 8 kilobytes
- Physical address space: 32 bits
- Page table entry consists of:
- Valid bit
- Dirty bit
- 3 permission bits
- Translation
Maximum size of the page table of a process:
- The maximum size of the page table of a process is given as 24 megabytes.
Calculating the number of pages in the page table:
- Since the page table size is given in megabytes, we need to convert it to kilobytes.
- 1 megabyte = 1024 kilobytes
- Therefore, the maximum size of the page table is 24 * 1024 = 24576 kilobytes.
- Each page is 8 kilobytes in size.
- Number of pages = Maximum size of the page table / Page size
- Number of pages = 24576 / 8 = 3072 pages
Calculating the number of bits required to address the pages:
- The number of bits required to address the pages is equal to the logarithm of the number of pages to the base 2.
- Number of bits = log2(Number of pages)
- Number of bits = log2(3072)
- Number of bits = 11.934
Calculating the number of bits required for the translation:
- The physical address space is given as 32 bits.
- The page size is 8 kilobytes, which is equal to 2^13 bytes.
- The number of bits required for the translation is equal to the logarithm of the page size to the base 2.
- Number of bits = log2(Page size in bytes)
- Number of bits = log2(2^13)
- Number of bits = 13
Calculating the number of bits for the valid bit, dirty bit, and permission bits:
- Each page table entry contains a valid bit, a dirty bit, and three permission bits.
- Total number of bits for these fields = 1 (valid bit) + 1 (dirty bit) + 3 (permission bits) = 5 bits
Calculating the total number of bits required for the virtual address:
- Total number of bits = Number of bits for page addressing + Number of bits for translation + Number of bits for entry fields
- Total number of bits = 11.934 + 13 + 5
- Total number of bits = 29.934
Rounding up the total number of bits:
- Since the number of bits cannot be a fraction, we need to round up the total number of bits to the nearest integer.
- Rounded up total number of bits = ceil(29.934) = 30 bits
Conclusion:
- The length of the virtual address supported by the system is 30 bits.
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A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer?
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A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer? for Class 6 2024 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer? covers all topics & solutions for Class 6 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer?.
A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer? for Class 6 2024 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer? covers all topics & solutions for Class 6 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A computer system implements 8-kilobyte pages and a 32 - bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.Correct answer is '36'. Can you explain this answer?.
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