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A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.
Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)
Correct answer is '4'. Can you explain this answer?
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A certain computer system has the segmented paging architecture for vi...
Given information:
- Virtual address space size: 216 bytes
- Physical address space size: 216 bytes
- Virtual address space is divided into 8 non-overlapping equal size segments
- Memory management unit (MMU) has a hardware segment table
- Each entry in the segment table contains the physical address of the page table for the segment
- Page table is stored in main memory
- Each page table entry has 2 bytes
- Each page table entry contains: 1 valid bit, 3 bits for page protection, and 1 dirty bit
- Page size: 512 bytes
To find: Number of bits available in a page table entry for storing the aging information for the page.
Solution:
Step 1: Calculate the size of each segment
Since the virtual address space is divided into 8 equal size segments, each segment will have a size of 216 / 8 = 213 bytes.
Step 2: Calculate the number of pages in each segment
Page size is given as 512 bytes. So, each segment will have 213 / 512 = 211 pages.
Step 3: Calculate the number of bits required to address each page in a segment
Since each segment has 211 pages, we need log2(211) bits to address each page in a segment. Assuming integer division, log2(211) = 11 bits.
Step 4: Calculate the number of bits used by the page table entry
Each page table entry is 2 bytes in size. Since 1 bit is used for the valid bit, 3 bits are used for page protection, and 1 bit is used for the dirty bit, the remaining bits can be used for storing the aging information.
Total bits used by a page table entry = 2 bytes * 8 bits per byte = 16 bits
Bits used for valid bit + bits used for page protection + bits used for dirty bit = 1 + 3 + 1 = 5 bits
Bits available for aging information = Total bits used by a page table entry - Bits used for valid, page protection, and dirty bits = 16 - 5 = 11 bits
Step 5: Calculate the number of bits available for storing aging information for a page
Since each segment has 211 pages and each page can store aging information in the page table entry, the number of bits available for storing aging information for a page = Bits available for aging information / Number of pages in a segment = 11 bits / 211 pages = 0.0521 bits
Rounded to the nearest whole number, the number of bits available for storing aging information for a page is 0 bits.
Answer: The correct answer is 0 bits.
- Virtual address space size: 216 bytes
- Physical address space size: 216 bytes
- Virtual address space is divided into 8 non-overlapping equal size segments
- Memory management unit (MMU) has a hardware segment table
- Each entry in the segment table contains the physical address of the page table for the segment
- Page table is stored in main memory
- Each page table entry has 2 bytes
- Each page table entry contains: 1 valid bit, 3 bits for page protection, and 1 dirty bit
- Page size: 512 bytes
To find: Number of bits available in a page table entry for storing the aging information for the page.
Solution:
Step 1: Calculate the size of each segment
Since the virtual address space is divided into 8 equal size segments, each segment will have a size of 216 / 8 = 213 bytes.
Step 2: Calculate the number of pages in each segment
Page size is given as 512 bytes. So, each segment will have 213 / 512 = 211 pages.
Step 3: Calculate the number of bits required to address each page in a segment
Since each segment has 211 pages, we need log2(211) bits to address each page in a segment. Assuming integer division, log2(211) = 11 bits.
Step 4: Calculate the number of bits used by the page table entry
Each page table entry is 2 bytes in size. Since 1 bit is used for the valid bit, 3 bits are used for page protection, and 1 bit is used for the dirty bit, the remaining bits can be used for storing the aging information.
Total bits used by a page table entry = 2 bytes * 8 bits per byte = 16 bits
Bits used for valid bit + bits used for page protection + bits used for dirty bit = 1 + 3 + 1 = 5 bits
Bits available for aging information = Total bits used by a page table entry - Bits used for valid, page protection, and dirty bits = 16 - 5 = 11 bits
Step 5: Calculate the number of bits available for storing aging information for a page
Since each segment has 211 pages and each page can store aging information in the page table entry, the number of bits available for storing aging information for a page = Bits available for aging information / Number of pages in a segment = 11 bits / 211 pages = 0.0521 bits
Rounded to the nearest whole number, the number of bits available for storing aging information for a page is 0 bits.
Answer: The correct answer is 0 bits.
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Community Answer
A certain computer system has the segmented paging architecture for vi...
Let x bits required for storing the aging information for the page.
Physical address space = 216 Bytes
Page Size = 512 bytes = 29 bytes
216/29 = 27
7 bits are required for frame.
Page table entry = 16 bits
x + 7 + 1 + 3 + 1 = 16
x = 16 - 12 = 4 bits
Physical address space = 216 Bytes
Page Size = 512 bytes = 29 bytes
216/29 = 27
7 bits are required for frame.
Page table entry = 16 bits
x + 7 + 1 + 3 + 1 = 16
x = 16 - 12 = 4 bits
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A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer?
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A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer?.
A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer?.
Solutions for A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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ample number of questions to practice A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 216 bytes each. The virtual address space is divided into 8 non overlapping equal size segments. The memory management unit(MMU) has a hardware segment table each entry of which contains the physical address of the page table for the segment. Page table is stored in the main memory and consists of 2 bytes page table entries.Assume that each page table entry contains 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in a page table entry for storing the aging information for the page? (assume page size 512 bytes)Correct answer is '4'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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