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A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.
What is the size of a page in KB in this computer?
- a)2
- b)4
- c)8
- d)16
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A computer uses 46-bit virtual address, 32-bit physical address, and a...
Concept:
To minimize the need for contiguous allocation of physical memory, paging is used.
Logical address space or virtual address space is the set of all logical addresses generated by the program.
Physical addresses are the actual addresses available in memory.
Physical address space is the set of all physical addresses generated by the program.
Mapping of the virtual address to physical address is done by hardware device i.e. Memory Management Unit (MMU) & is called paging technique.
Physical Address space is divided into fixed-size no. of blocks called frames.
The logical address space is divided into fixed-size no. of blocks called pages.
To minimize the need for contiguous allocation of physical memory, paging is used.
Logical address space or virtual address space is the set of all logical addresses generated by the program.
Physical addresses are the actual addresses available in memory.
Physical address space is the set of all physical addresses generated by the program.
Mapping of the virtual address to physical address is done by hardware device i.e. Memory Management Unit (MMU) & is called paging technique.
Physical Address space is divided into fixed-size no. of blocks called frames.
The logical address space is divided into fixed-size no. of blocks called pages.
Calculation:
Page table entry size = 32 bits or 4 Bytes
No. of entries in one page = 2p/4
VAS = 46 bits
Consider size of page = 2p Bytes
Virtual address space (VAS) = No. of entries x page size
No. of entries for one level Page table (PT) = 2p ×4
Therefore, for 3 levels PT
246 x 26 = 24p
252 = 24p
52 = 4p
P = 13
So, the page size is 213 or 8KB.
Hence, the correct answer is “option 3”.
Page table entry size = 32 bits or 4 Bytes
No. of entries in one page = 2p/4
VAS = 46 bits
Consider size of page = 2p Bytes
Virtual address space (VAS) = No. of entries x page size
No. of entries for one level Page table (PT) = 2p ×4
Therefore, for 3 levels PT
246 x 26 = 24p
252 = 24p
52 = 4p
P = 13
So, the page size is 213 or 8KB.
Hence, the correct answer is “option 3”.
Community Answer
A computer uses 46-bit virtual address, 32-bit physical address, and a...
Understanding the Virtual Address Space
The computer has a 46-bit virtual address space. This means it can address a total of 2^46 bytes of virtual memory.
Determining the Page Size
1. Three-Level Paging:
- The virtual address is divided into three parts: for the first-level page table (T1), the second-level (T2), and the third-level (T3).
2. Page Table Entries:
- Each PTE is 32 bits (or 4 bytes). Thus, each entry in T3 can store an address that points to a page frame in physical memory.
3. Calculating Levels:
- Since T1 occupies one page, and we know that a page table entry (PTE) is 4 bytes, we can calculate the number of entries in T1.
- A page is typically 4KB (2^12 bytes). Therefore, T1 can have 4KB / 4B = 1024 entries.
4. Page Size Calculation:
- Given that T1 has 1024 entries, and each entry points to a second-level page table (T2) and so forth, we need 10 bits to index into T1 (2^10 = 1024).
- Similar calculations apply for T2 and T3, using the remaining bits for offset.
5. Offset Bits:
- The remaining bits after accounting for T1, T2, and T3 leaves us with 12 bits (46 - 10 - 10 - 10 = 12).
- This allows for 2^12 = 4096 bytes per page, which is equivalent to 4KB.
Conclusion
Thus, each page in this computer is 4KB in size, making option 'c' (8) incorrect. The correct answer for the size of a page in KB is indeed 8, which is a misalignment with our analysis. The correct size in KB should be 4, or option 'b'.
The computer has a 46-bit virtual address space. This means it can address a total of 2^46 bytes of virtual memory.
Determining the Page Size
1. Three-Level Paging:
- The virtual address is divided into three parts: for the first-level page table (T1), the second-level (T2), and the third-level (T3).
2. Page Table Entries:
- Each PTE is 32 bits (or 4 bytes). Thus, each entry in T3 can store an address that points to a page frame in physical memory.
3. Calculating Levels:
- Since T1 occupies one page, and we know that a page table entry (PTE) is 4 bytes, we can calculate the number of entries in T1.
- A page is typically 4KB (2^12 bytes). Therefore, T1 can have 4KB / 4B = 1024 entries.
4. Page Size Calculation:
- Given that T1 has 1024 entries, and each entry points to a second-level page table (T2) and so forth, we need 10 bits to index into T1 (2^10 = 1024).
- Similar calculations apply for T2 and T3, using the remaining bits for offset.
5. Offset Bits:
- The remaining bits after accounting for T1, T2, and T3 leaves us with 12 bits (46 - 10 - 10 - 10 = 12).
- This allows for 2^12 = 4096 bytes per page, which is equivalent to 4KB.
Conclusion
Thus, each page in this computer is 4KB in size, making option 'c' (8) incorrect. The correct answer for the size of a page in KB is indeed 8, which is a misalignment with our analysis. The correct size in KB should be 4, or option 'b'.
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A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer? for Class 6 2026 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 6 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer?.
A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer? for Class 6 2026 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 6 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer?.
Solutions for A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.What is the size of a page in KB in this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 6. Download more important topics, notes, lectures and mock test series for Class 6 Exam by signing up for free.
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