Class 6 Exam > Class 6 Questions > What is the size of the physical address spac...
Start Learning for Free
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?
- a)211
- b)215
- c)219
- d)220
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
What is the size of the physical address space in a paging system whic...
Data
page table entries = 64
page table entry size = 11 bit
page size = 512 bytes
page table entries = 64
page table entry size = 11 bit
page size = 512 bytes
Formula:
Size of physical address = pagging bits + offset bits
Size of physical address = pagging bits + offset bits
Calculation:
paging bits = 11 - 1 = 10(as 1 valid bit is also included)
offset bits = log2 (512) = 9 bits
size of physical address = 10 + 9 = 19 bits
Hence the correct answer is 219.
paging bits = 11 - 1 = 10(as 1 valid bit is also included)
offset bits = log2 (512) = 9 bits
size of physical address = 10 + 9 = 19 bits
Hence the correct answer is 219.
Community Answer
What is the size of the physical address space in a paging system whic...
The physical address space in a paging system is determined by the number of bits used to represent the physical addresses. In this case, we need to calculate the size of the physical address space given a page table with 64 entries, each consisting of 11 bits (including a valid/invalid bit), and a page size of 512 bytes.
1. Determining the number of bits for the page offset:
- Since the page size is 512 bytes, we need log2(512) bits to represent the page offset.
- log2(512) = 9 bits
2. Determining the number of bits for the page number:
- The page table has 64 entries, and each entry consists of 11 bits (including a valid/invalid bit).
- Therefore, we need 11 - 1 = 10 bits to represent the page number.
3. Calculating the size of the physical address space:
- The physical address space is determined by the sum of the bits for the page offset and the bits for the page number.
- Size = Number of bits for page offset + Number of bits for page number
- Size = 9 bits + 10 bits
- Size = 19 bits
4. Converting the size from bits to bytes:
- Since 8 bits make up 1 byte, we divide the size by 8.
- Size (in bytes) = 19 bits / 8
- Size (in bytes) = 2.375 bytes
5. Rounding up to the nearest power of 2:
- The size of the physical address space needs to be a power of 2.
- The nearest power of 2 greater than 2.375 bytes is 4 bytes.
- Since 1 byte is 2^0, 4 bytes is 2^2.
Therefore, the size of the physical address space in this paging system is 2^19 bytes, which is equivalent to 2^19 / 2^10 = 2^9 pages. This is approximately equal to 2^9 * 512 bytes = 2^12 bytes.
Hence, the correct answer is option C) 2^19 or 219.
1. Determining the number of bits for the page offset:
- Since the page size is 512 bytes, we need log2(512) bits to represent the page offset.
- log2(512) = 9 bits
2. Determining the number of bits for the page number:
- The page table has 64 entries, and each entry consists of 11 bits (including a valid/invalid bit).
- Therefore, we need 11 - 1 = 10 bits to represent the page number.
3. Calculating the size of the physical address space:
- The physical address space is determined by the sum of the bits for the page offset and the bits for the page number.
- Size = Number of bits for page offset + Number of bits for page number
- Size = 9 bits + 10 bits
- Size = 19 bits
4. Converting the size from bits to bytes:
- Since 8 bits make up 1 byte, we divide the size by 8.
- Size (in bytes) = 19 bits / 8
- Size (in bytes) = 2.375 bytes
5. Rounding up to the nearest power of 2:
- The size of the physical address space needs to be a power of 2.
- The nearest power of 2 greater than 2.375 bytes is 4 bytes.
- Since 1 byte is 2^0, 4 bytes is 2^2.
Therefore, the size of the physical address space in this paging system is 2^19 bytes, which is equivalent to 2^19 / 2^10 = 2^9 pages. This is approximately equal to 2^9 * 512 bytes = 2^12 bytes.
Hence, the correct answer is option C) 2^19 or 219.
Attention Class 6 Students!
To make sure you are not studying endlessly, EduRev has designed Class 6 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 6.
|
Explore Courses for Class 6 exam
|
|
Top Courses for Class 6View all
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer?
Question Description
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? for Class 6 2024 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 6 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer?.
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? for Class 6 2024 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 6 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer?.
Solutions for What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct 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.
Here you can find the meaning of What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer?, a detailed solution for What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid/invalid bit) and a page size of 512 bytes?a)211b)215c)219d)220Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 6 tests.
|
|
Explore Courses for Class 6 exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup