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Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)?
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Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 byt...
Chained Allocation Strategy and Internal Fragmentation
To determine the number of physical blocks needed to store the four files using a chained allocation strategy, we need to consider the file sizes and the fixed physical block size of 512 bytes. Let's analyze each file and calculate the required number of blocks:
1. File F1: Size = 7178 bytes
- Number of blocks needed = ceil(7178 / 512) = ceil(14.01) = 15 blocks
2. File F2: Size = 572 bytes
- Number of blocks needed = ceil(572 / 512) = ceil(1.12) = 2 blocks
3. File F3: Size = 499 bytes
- Number of blocks needed = ceil(499 / 512) = ceil(0.97) = 1 block
4. File F4: Size = 1195 bytes
- Number of blocks needed = ceil(1195 / 512) = ceil(2.33) = 3 blocks
Total number of physical blocks:
The total number of physical blocks required to store these four files using the chained allocation strategy is the sum of the blocks needed for each file:
15 (F1) + 2 (F2) + 1 (F3) + 3 (F4) = 21 blocks
Internal Fragmentation:
Internal fragmentation refers to the wasted space within a block that is not utilized by the file stored in it. In the chained allocation strategy, each block contains some overhead information (5 bytes) to determine the next block in the link. This overhead reduces the usable space within each block.
The file that results in the maximum internal fragmentation is the one with the smallest size relative to the physical block size. In this case, the file F3 with a size of 499 bytes and requiring only 1 block results in the maximum internal fragmentation.
To calculate the internal fragmentation percentage for file F3, we need to consider the space wasted due to the overhead information within the block:
- Usable space within a block = 512 bytes - 5 bytes (overhead) = 507 bytes
- Internal fragmentation = Usable space - File size = 507 - 499 = 8 bytes
The internal fragmentation percentage for file F3 can be calculated as:
(Internal fragmentation / File size) * 100 = (8 / 499) * 100 ≈ 1.60%
Therefore, file F3 has approximately 1.60% internal fragmentation, which is the maximum among the given files.
In summary, using the chained allocation strategy, a total of 21 physical blocks would be needed to store the four files. File F3 results in the maximum internal fragmentation, with approximately 1.60% of wasted space within its allocated block.
To determine the number of physical blocks needed to store the four files using a chained allocation strategy, we need to consider the file sizes and the fixed physical block size of 512 bytes. Let's analyze each file and calculate the required number of blocks:
1. File F1: Size = 7178 bytes
- Number of blocks needed = ceil(7178 / 512) = ceil(14.01) = 15 blocks
2. File F2: Size = 572 bytes
- Number of blocks needed = ceil(572 / 512) = ceil(1.12) = 2 blocks
3. File F3: Size = 499 bytes
- Number of blocks needed = ceil(499 / 512) = ceil(0.97) = 1 block
4. File F4: Size = 1195 bytes
- Number of blocks needed = ceil(1195 / 512) = ceil(2.33) = 3 blocks
Total number of physical blocks:
The total number of physical blocks required to store these four files using the chained allocation strategy is the sum of the blocks needed for each file:
15 (F1) + 2 (F2) + 1 (F3) + 3 (F4) = 21 blocks
Internal Fragmentation:
Internal fragmentation refers to the wasted space within a block that is not utilized by the file stored in it. In the chained allocation strategy, each block contains some overhead information (5 bytes) to determine the next block in the link. This overhead reduces the usable space within each block.
The file that results in the maximum internal fragmentation is the one with the smallest size relative to the physical block size. In this case, the file F3 with a size of 499 bytes and requiring only 1 block results in the maximum internal fragmentation.
To calculate the internal fragmentation percentage for file F3, we need to consider the space wasted due to the overhead information within the block:
- Usable space within a block = 512 bytes - 5 bytes (overhead) = 507 bytes
- Internal fragmentation = Usable space - File size = 507 - 499 = 8 bytes
The internal fragmentation percentage for file F3 can be calculated as:
(Internal fragmentation / File size) * 100 = (8 / 499) * 100 ≈ 1.60%
Therefore, file F3 has approximately 1.60% internal fragmentation, which is the maximum among the given files.
In summary, using the chained allocation strategy, a total of 21 physical blocks would be needed to store the four files. File F3 results in the maximum internal fragmentation, with approximately 1.60% of wasted space within its allocated block.
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Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 byt...
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Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)?
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Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)?.
Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose we have files F1 to F4 in sizes of 7178, 572, 499 and 1195 bytes. Our disks have fixed physical block size of 512 bytes for allocation. How many physical blocks would be needed to store these four files if we were to use a chained allocation strategy assuming that we need 5 bytes of information to determine the next block in the link? Which file results in the maximum internal fragmentation (measured as a percentage of the file size itself)?.
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