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Consider a simple paging system with 1024 MB of physical memory and the page size of 4KB with a logical address space of 128 pages. Find the number of bits used to represent physical address, frame number and page number. 
  • a)
    30 , 18, 7
  • b)
    10 , 2 , 7
  • c)
    28, 12, 7
  • d)
    28, 18, 7
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider a simple paging system with 1024 MB of physical memory and th...
Physical memory = 1024 MB = 1024 × 1024 × 1024 bytes = 210 × 210 × 210 = 230
Hence, 30 bits are required to represent physical address.
Page size = 4 KB = 4 × 210 = 212
Number of frames =
Hence, 18 bits are required to represent frame number.
128 pages i.e. 27 pages being accessed. Hence, 7 bit represent page number.
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Most Upvoted Answer
Consider a simple paging system with 1024 MB of physical memory and th...
Given:
- Physical memory size: 1024 MB
- Page size: 4 KB
- Logical address space: 128 pages

To find:
- Number of bits used to represent physical address
- Number of bits used to represent frame number
- Number of bits used to represent page number

Solution:
1. Physical Address:
Since the physical memory size is given in MB and the page size is given in KB, we need to convert the physical memory size to the same unit as the page size (KB).
1024 MB = 1024 * 1024 KB = 1048576 KB

To represent the physical address, we need to find the number of bits required to represent 1048576 KB.
Number of bits = log2(1048576 * 1024)
= log2(1073741824)
= 30

Therefore, the number of bits used to represent the physical address is 30.

2. Frame Number:
The frame number represents the frame or block number within the physical memory. Since the page size is given as 4 KB, we can calculate the number of frames as the physical memory size divided by the page size.
Number of frames = Physical memory size / Page size
= 1048576 KB / 4 KB
= 262144

To represent the frame number, we need to find the number of bits required to represent 262144 frames.
Number of bits = log2(262144)
= log2(2^18)
= 18

Therefore, the number of bits used to represent the frame number is 18.

3. Page Number:
The page number represents the page or block number within the logical address space. Since the logical address space is given as 128 pages, we can directly calculate the number of bits required to represent 128 pages.
Number of bits = log2(128)
= log2(2^7)
= 7

Therefore, the number of bits used to represent the page number is 7.

Conclusion:
The number of bits used to represent the physical address is 30, the number of bits used to represent the frame number is 18, and the number of bits used to represent the page number is 7. Hence, the correct answer is option 'A'.
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Consider a simple paging system with 1024 MB of physical memory and the page size of 4KB with a logical address space of 128 pages. Find the number of bits used to represent physical address, frame number and page number.a)30 , 18, 7b)10 , 2 , 7c)28, 12, 7d)28, 18, 7Correct answer is option 'A'. Can you explain this answer?
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Consider a simple paging system with 1024 MB of physical memory and the page size of 4KB with a logical address space of 128 pages. Find the number of bits used to represent physical address, frame number and page number.a)30 , 18, 7b)10 , 2 , 7c)28, 12, 7d)28, 18, 7Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 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 Consider a simple paging system with 1024 MB of physical memory and the page size of 4KB with a logical address space of 128 pages. Find the number of bits used to represent physical address, frame number and page number.a)30 , 18, 7b)10 , 2 , 7c)28, 12, 7d)28, 18, 7Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a simple paging system with 1024 MB of physical memory and the page size of 4KB with a logical address space of 128 pages. Find the number of bits used to represent physical address, frame number and page number.a)30 , 18, 7b)10 , 2 , 7c)28, 12, 7d)28, 18, 7Correct answer is option 'A'. Can you explain this answer?.
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