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Consider a system that has 4K pages of 512 bytes in size in the logical address space. The number of bits of logical address?
- a)21
- b)20
- c)19
- d)18
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Consider a system that has 4K pages of 512 bytes in size in the logica...
Concept:
The given data,
Number of pages = 4 K
Page size PS = 512 B = 29 Bytes
Number of pages = Logical address space / Page size
Logical address space = Number of pages X Page size
Logical address space = 4 K X 29 Bytes
Logical address space = 212 X 29 Bytes
Logical address space = 221 Bytes
Logical address bits or Virtual address bits = 21
Hence the correct answer is 21.
The given data,
Number of pages = 4 K
Page size PS = 512 B = 29 Bytes
Number of pages = Logical address space / Page size
Logical address space = Number of pages X Page size
Logical address space = 4 K X 29 Bytes
Logical address space = 212 X 29 Bytes
Logical address space = 221 Bytes
Logical address bits or Virtual address bits = 21
Hence the correct answer is 21.
Community Answer
Consider a system that has 4K pages of 512 bytes in size in the logica...
The given information states that the system has 4K pages of 512 bytes in size in the logical address space. We need to determine the number of bits of the logical address.
To calculate the number of bits in the logical address, we need to consider the number of pages and the size of each page.
- Number of Pages:
Given that the system has 4K pages, where "K" represents a thousand, we can calculate the exact number of pages by multiplying 4 by 1,000. Therefore, the number of pages is 4,000.
- Size of Each Page:
The size of each page is given as 512 bytes.
To find the number of bits required to address a page, we need to calculate the logarithm base 2 of the number of pages. This will give us the number of bits needed to address the pages.
Using the formula:
Number of bits = log2(Number of pages)
Calculating the logarithm base 2 of 4,000:
Number of bits = log2(4000)
Using a calculator or mathematical software, we find that log2(4000) is approximately 11.96.
Since the number of bits must be a whole number, we round up the result to the nearest whole number, which is 12.
Therefore, the number of bits required to address the pages is 12.
Now, let's consider the size of each page. It is given as 512 bytes.
To find the number of bits required to address the bytes within a page, we need to calculate the logarithm base 2 of the page size. This will give us the number of bits needed to address the bytes within a page.
Using the formula:
Number of bits = log2(Page size)
Calculating the logarithm base 2 of 512:
Number of bits = log2(512)
Using a calculator or mathematical software, we find that log2(512) is approximately 9.
Therefore, the number of bits required to address the bytes within a page is 9.
- Total Number of Bits:
To find the total number of bits in the logical address, we sum the number of bits required to address the pages and the number of bits required to address the bytes within a page.
Total Number of Bits = Number of bits for pages + Number of bits for bytes within a page
Total Number of Bits = 12 + 9
Total Number of Bits = 21
Hence, the correct answer is option A) 21, indicating that the logical address requires 21 bits.
To calculate the number of bits in the logical address, we need to consider the number of pages and the size of each page.
- Number of Pages:
Given that the system has 4K pages, where "K" represents a thousand, we can calculate the exact number of pages by multiplying 4 by 1,000. Therefore, the number of pages is 4,000.
- Size of Each Page:
The size of each page is given as 512 bytes.
To find the number of bits required to address a page, we need to calculate the logarithm base 2 of the number of pages. This will give us the number of bits needed to address the pages.
Using the formula:
Number of bits = log2(Number of pages)
Calculating the logarithm base 2 of 4,000:
Number of bits = log2(4000)
Using a calculator or mathematical software, we find that log2(4000) is approximately 11.96.
Since the number of bits must be a whole number, we round up the result to the nearest whole number, which is 12.
Therefore, the number of bits required to address the pages is 12.
Now, let's consider the size of each page. It is given as 512 bytes.
To find the number of bits required to address the bytes within a page, we need to calculate the logarithm base 2 of the page size. This will give us the number of bits needed to address the bytes within a page.
Using the formula:
Number of bits = log2(Page size)
Calculating the logarithm base 2 of 512:
Number of bits = log2(512)
Using a calculator or mathematical software, we find that log2(512) is approximately 9.
Therefore, the number of bits required to address the bytes within a page is 9.
- Total Number of Bits:
To find the total number of bits in the logical address, we sum the number of bits required to address the pages and the number of bits required to address the bytes within a page.
Total Number of Bits = Number of bits for pages + Number of bits for bytes within a page
Total Number of Bits = 12 + 9
Total Number of Bits = 21
Hence, the correct answer is option A) 21, indicating that the logical address requires 21 bits.
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Consider a system that has 4K pages of 512 bytes in size in the logical address space. The number of bits of logical address?a)21b)20c)19d)18Correct answer is option 'A'. Can you explain this answer?
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