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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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?
What is the minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?
- a)2
- b)4
- c)8
- d)16
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A computer uses 46-bit virtual address, 32-bit physical address, and a...
1 MB 16-way set associative virtually indexed physically tagged cache(VIPT).
The cache block size is 64 bytes.
The cache block size is 64 bytes.
No of blocks is 2^20/2^6 = 2^14.
No of sets is 2^14/2^4 = 2^10.
VA(46)
+-------------------------------+
tag(30) , Set(10) , block offset(6)
+-------------------------------+
+-------------------------------+
tag(30) , Set(10) , block offset(6)
+-------------------------------+
In VIPT if the no. of bits of page offset = (Set+block offset) then only one page color is sufficient. but we need 8 colors because the number bits where the cache set index and physical page number over lap is 3 so 2^3 page colors is required.(option
c is ans).
c is ans).
Most Upvoted Answer
A computer uses 46-bit virtual address, 32-bit physical address, and a...
To understand the minimum number of page colours required to guarantee that no two synonyms map to different sets in the processor cache, let's break down the problem step by step:
1. Cache Size and Block Size:
The cache size is given as 1 MB (1 megabyte), and the block size is given as 64 bytes.
2. Number of Cache Sets:
Since the cache is 16-way set associative, we can calculate the number of cache sets using the formula:
Number of Cache Sets = (Cache Size) / (Associativity * Block Size)
Substituting the given values, we get:
Number of Cache Sets = (1 MB) / (16 * 64 bytes)
= (2^20 bytes) / (16 * 2^6 bytes)
= 2^(20-6-4) sets = 2^10 sets = 1024 sets
3. Number of Bits for Cache Index:
The number of bits required to represent the cache index is given by:
Number of Bits for Cache Index = log2(Number of Cache Sets)
= log2(1024) = 10 bits
4. Virtual Address Space:
The computer uses a 46-bit virtual address. This means the virtual address space is 2^46.
5. Number of Bits for Virtual Page Number:
To determine the number of bits used for the virtual page number, we need to subtract the number of bits used for the page offset from the total number of bits in the virtual address. The page offset is determined by the block size, which is 64 bytes or 2^6.
Number of Bits for Virtual Page Number = Total Bits - Bits for Page Offset
= 46 bits - 6 bits
= 40 bits
6. Number of Levels in Page Table:
The page table organization is three-level paged, which means we have three levels of page tables: T1, T2, and T3.
7. Number of Bits for Page Table Entry:
The size of each page table entry (PTE) is given as 32 bits.
8. Number of Entries in Each Level:
Since the base address of T1 occupies exactly one page, and each entry of T1 stores the base address of a page of T2, and each entry of T2 stores the base address of a page of T3, we can calculate the number of entries in each level using the formula:
Number of Entries = (Size of Page) / (Size of PTE)
= (2^12 bytes) / (2^5 bytes)
= 2^7 entries = 128 entries
9. Number of Bits for Each Level:
To determine the number of bits required to represent each level of the page table, we need to calculate the number of bits required to represent the number of entries in each level.
Number of Bits for Each Level = log2(Number of Entries)
= log2(128) = 7 bits
10. Number of Bits for Page Table Offset:
To determine the number of bits required for the page table offset, we need to subtract the number of bits used for the page table entry from the total number of bits in the virtual address. The page table entry is 32 bits or 2^5.
Number of Bits for Page Table Offset = Total Bits - Bits for PTE
= 46 bits - 5 bits
= 41
1. Cache Size and Block Size:
The cache size is given as 1 MB (1 megabyte), and the block size is given as 64 bytes.
2. Number of Cache Sets:
Since the cache is 16-way set associative, we can calculate the number of cache sets using the formula:
Number of Cache Sets = (Cache Size) / (Associativity * Block Size)
Substituting the given values, we get:
Number of Cache Sets = (1 MB) / (16 * 64 bytes)
= (2^20 bytes) / (16 * 2^6 bytes)
= 2^(20-6-4) sets = 2^10 sets = 1024 sets
3. Number of Bits for Cache Index:
The number of bits required to represent the cache index is given by:
Number of Bits for Cache Index = log2(Number of Cache Sets)
= log2(1024) = 10 bits
4. Virtual Address Space:
The computer uses a 46-bit virtual address. This means the virtual address space is 2^46.
5. Number of Bits for Virtual Page Number:
To determine the number of bits used for the virtual page number, we need to subtract the number of bits used for the page offset from the total number of bits in the virtual address. The page offset is determined by the block size, which is 64 bytes or 2^6.
Number of Bits for Virtual Page Number = Total Bits - Bits for Page Offset
= 46 bits - 6 bits
= 40 bits
6. Number of Levels in Page Table:
The page table organization is three-level paged, which means we have three levels of page tables: T1, T2, and T3.
7. Number of Bits for Page Table Entry:
The size of each page table entry (PTE) is given as 32 bits.
8. Number of Entries in Each Level:
Since the base address of T1 occupies exactly one page, and each entry of T1 stores the base address of a page of T2, and each entry of T2 stores the base address of a page of T3, we can calculate the number of entries in each level using the formula:
Number of Entries = (Size of Page) / (Size of PTE)
= (2^12 bytes) / (2^5 bytes)
= 2^7 entries = 128 entries
9. Number of Bits for Each Level:
To determine the number of bits required to represent each level of the page table, we need to calculate the number of bits required to represent the number of entries in each level.
Number of Bits for Each Level = log2(Number of Entries)
= log2(128) = 7 bits
10. Number of Bits for Page Table Offset:
To determine the number of bits required for the page table offset, we need to subtract the number of bits used for the page table entry from the total number of bits in the virtual address. The page table entry is 32 bits or 2^5.
Number of Bits for Page Table Offset = Total Bits - Bits for PTE
= 46 bits - 5 bits
= 41
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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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer?
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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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?a)2b)4c)8d)16Correct answer is option 'C'. 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 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?a)2b)4c)8d)16Correct answer is option 'C'. 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 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?a)2b)4c)8d)16Correct answer is option 'C'. 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 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?a)2b)4c)8d)16Correct answer is option 'C'. 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 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of 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 Computer Science Engineering (CSE).
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ample number of questions to practice 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 minimum number of page colours needed to guarantee that no two synonyms map to different sets in the processor cache of this computer?a)2b)4c)8d)16Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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