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A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirect
block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the
size of each disk block address is 8 Bytes. The maximum possible file size in this file system is
block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the
size of each disk block address is 8 Bytes. The maximum possible file size in this file system is
- a)3 KBytes
- b)35 KBytes
- c)280 KBytes
- d)dependent on the size of the disk
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A file system with 300 GByte disk uses a file descriptor with 8 direct...
Each block size = 128 Bytes
Disk block address = 8 Bytes
Each disk can contain =
128
16
8
= addresses
Size due to 8 direct block addresses: 8 x 128
Size due to 1 indirect block address: 16 x 128
Size due to 1 doubly indirect block address: 16 x 16 x 128
Size due to 1 doubly indirect block address: 16 x 16 x 128
Disk block address = 8 Bytes
Each disk can contain =
128
16
8
= addresses
Size due to 8 direct block addresses: 8 x 128
Size due to 1 indirect block address: 16 x 128
Size due to 1 doubly indirect block address: 16 x 16 x 128
Size due to 1 doubly indirect block address: 16 x 16 x 128
Most Upvoted Answer
A file system with 300 GByte disk uses a file descriptor with 8 direct...
Understanding the Problem:
We are given a file system with a 300 GByte disk. The file system uses a file descriptor with 8 direct block addresses, 1 indirect block address, and 1 doubly indirect block address. The size of each disk block is 128 Bytes, and the size of each disk block address is 8 Bytes. We need to determine the maximum possible file size in this file system.
Key Information:
- Disk size: 300 GByte
- Disk block size: 128 Bytes
- Size of each disk block address: 8 Bytes
- File descriptor:
- 8 direct block addresses
- 1 indirect block address
- 1 doubly indirect block address
Solution:
To calculate the maximum file size, we need to consider the number of disk blocks that can be addressed using the given file descriptor.
Direct Blocks:
The file descriptor has 8 direct block addresses. Each direct block address can address one disk block. Therefore, the total number of disk blocks that can be addressed directly is:
8 direct block addresses * 1 disk block per address = 8 disk blocks
Indirect Blocks:
The file descriptor has 1 indirect block address. Each indirect block address can address a block that contains disk block addresses. Since the size of each disk block address is 8 Bytes, each indirect block can address:
Disk block size / Size of each disk block address = 128 Bytes / 8 Bytes = 16 disk block addresses
Doubly Indirect Blocks:
The file descriptor has 1 doubly indirect block address. Each doubly indirect block address can address a block that contains indirect block addresses. Since each indirect block can address 16 disk block addresses, each doubly indirect block can address:
16 indirect block addresses * 16 disk block addresses = 256 disk block addresses
Calculating Maximum File Size:
To calculate the maximum file size, we need to sum the number of disk blocks that can be addressed using direct blocks, indirect blocks, and doubly indirect blocks.
Maximum file size = (Number of direct block addresses * Disk block size) + (Number of indirect block addresses * Disk block size) + (Number of doubly indirect block addresses * Disk block size)
= (8 * 128) + (1 * 128 * 16) + (1 * 128 * 16 * 16)
= 1024 + 2048 + 32768
= 35840 Bytes
Since 1 KByte = 1024 Bytes, the maximum file size is:
35840 Bytes / 1024 Bytes per KByte = 35 KBytes
Therefore, the correct answer is option 'B', 35 KBytes.
We are given a file system with a 300 GByte disk. The file system uses a file descriptor with 8 direct block addresses, 1 indirect block address, and 1 doubly indirect block address. The size of each disk block is 128 Bytes, and the size of each disk block address is 8 Bytes. We need to determine the maximum possible file size in this file system.
Key Information:
- Disk size: 300 GByte
- Disk block size: 128 Bytes
- Size of each disk block address: 8 Bytes
- File descriptor:
- 8 direct block addresses
- 1 indirect block address
- 1 doubly indirect block address
Solution:
To calculate the maximum file size, we need to consider the number of disk blocks that can be addressed using the given file descriptor.
Direct Blocks:
The file descriptor has 8 direct block addresses. Each direct block address can address one disk block. Therefore, the total number of disk blocks that can be addressed directly is:
8 direct block addresses * 1 disk block per address = 8 disk blocks
Indirect Blocks:
The file descriptor has 1 indirect block address. Each indirect block address can address a block that contains disk block addresses. Since the size of each disk block address is 8 Bytes, each indirect block can address:
Disk block size / Size of each disk block address = 128 Bytes / 8 Bytes = 16 disk block addresses
Doubly Indirect Blocks:
The file descriptor has 1 doubly indirect block address. Each doubly indirect block address can address a block that contains indirect block addresses. Since each indirect block can address 16 disk block addresses, each doubly indirect block can address:
16 indirect block addresses * 16 disk block addresses = 256 disk block addresses
Calculating Maximum File Size:
To calculate the maximum file size, we need to sum the number of disk blocks that can be addressed using direct blocks, indirect blocks, and doubly indirect blocks.
Maximum file size = (Number of direct block addresses * Disk block size) + (Number of indirect block addresses * Disk block size) + (Number of doubly indirect block addresses * Disk block size)
= (8 * 128) + (1 * 128 * 16) + (1 * 128 * 16 * 16)
= 1024 + 2048 + 32768
= 35840 Bytes
Since 1 KByte = 1024 Bytes, the maximum file size is:
35840 Bytes / 1024 Bytes per KByte = 35 KBytes
Therefore, the correct answer is option 'B', 35 KBytes.
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A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer?
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A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer?.
A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirectblock address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and thesize of each disk block address is 8 Bytes. The maximum possible file size in this file system isa)3 KBytesb)35 KBytesc)280 KBytesd)dependent on the size of the diskCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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