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In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?
Correct answer is '37'. Can you explain this answer?
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In a computer system, the physical address space is 32 bits. The size ...
Given Information:
- Physical address space: 32 bits
- Page size: 4 KB
- Maximum size of page table: 72 MB
- Each page table entry contains: 2 permission bits, 1 valid bit, 1 dirty bit, and translation bits
Calculating the number of entries in the page table:
Since the maximum size of the page table is given as 72 MB, which is the same as 72,000 KB, we can calculate the number of entries in the page table.
1. Convert the page table size to bytes: 72,000 KB * 1024 bytes/KB = 73,728,000 bytes
2. Calculate the number of entries in the page table: 73,728,000 bytes / (4 KB/page) = 18,432 entries
Calculating the number of bits required for the page table:
Since each page table entry contains 2 permission bits, 1 valid bit, 1 dirty bit, and translation bits, we need to calculate the total number of bits required for each entry.
1. Calculate the number of translation bits required: 32 bits - 12 bits (4 KB) = 20 bits
2. Calculate the number of bits required for each entry: 2 permission bits + 1 valid bit + 1 dirty bit + 20 translation bits = 24 bits
Calculating the length of the virtual address:
The virtual address is divided into two parts: the page number and the page offset.
1. The page number is determined by the number of entries in the page table: log2(18,432) = 14 bits
2. The page offset is determined by the page size: log2(4 KB) = 12 bits
Total length of the virtual address:
The total length of the virtual address is the sum of the page number bits and the page offset bits.
14 bits (page number) + 12 bits (page offset) = 26 bits
Explanation of the correct answer:
The given computer system has a 32-bit physical address space. The page table size is limited to 72 MB, which results in a page table with 18,432 entries. Each entry requires 24 bits. The virtual address is divided into a page number and a page offset, with 14 bits for the page number and 12 bits for the page offset. Therefore, the total length of the virtual address is 26 bits. However, the correct answer is 37 bits, which suggests that additional information or calculations may be missing or incorrect in the given question.
- Physical address space: 32 bits
- Page size: 4 KB
- Maximum size of page table: 72 MB
- Each page table entry contains: 2 permission bits, 1 valid bit, 1 dirty bit, and translation bits
Calculating the number of entries in the page table:
Since the maximum size of the page table is given as 72 MB, which is the same as 72,000 KB, we can calculate the number of entries in the page table.
1. Convert the page table size to bytes: 72,000 KB * 1024 bytes/KB = 73,728,000 bytes
2. Calculate the number of entries in the page table: 73,728,000 bytes / (4 KB/page) = 18,432 entries
Calculating the number of bits required for the page table:
Since each page table entry contains 2 permission bits, 1 valid bit, 1 dirty bit, and translation bits, we need to calculate the total number of bits required for each entry.
1. Calculate the number of translation bits required: 32 bits - 12 bits (4 KB) = 20 bits
2. Calculate the number of bits required for each entry: 2 permission bits + 1 valid bit + 1 dirty bit + 20 translation bits = 24 bits
Calculating the length of the virtual address:
The virtual address is divided into two parts: the page number and the page offset.
1. The page number is determined by the number of entries in the page table: log2(18,432) = 14 bits
2. The page offset is determined by the page size: log2(4 KB) = 12 bits
Total length of the virtual address:
The total length of the virtual address is the sum of the page number bits and the page offset bits.
14 bits (page number) + 12 bits (page offset) = 26 bits
Explanation of the correct answer:
The given computer system has a 32-bit physical address space. The page table size is limited to 72 MB, which results in a page table with 18,432 entries. Each entry requires 24 bits. The virtual address is divided into a page number and a page offset, with 14 bits for the page number and 12 bits for the page offset. Therefore, the total length of the virtual address is 26 bits. However, the correct answer is 37 bits, which suggests that additional information or calculations may be missing or incorrect in the given question.
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Community Answer
In a computer system, the physical address space is 32 bits. The size ...
Given,
Physical address space (PAS) = 32 bits
Page size = 4 KB
Page table size = 72 MB
As we know,
Number of frames = 
Now,
Number of frames = 
= 220
= 220Translation =20 bits
A byte is a unit of digital information that most commonly consists of eight bits. So, we have
Entry of page table = 2 + 1 + 1 + 20 = 24 bits
=3 byte
Number of pages = 
=
=
=24 MB
Virtual address = number of pages × page size = 24 × 220 × 4 KB
Virtual address =96×230 B
Length of virtual address space =log2 96 × 230
=37 bits
Hence, the correct answer is 37.
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In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?Correct answer is '37'. Can you explain this answer?
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