As shown in Figure 4-33, a Unix i-node has 10 direct address entries, and one si
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Question
As shown in Figure 4-33, a Unix i-node has 10 direct address entries, and one single, double, and triple indirect address entries. If a disk block has 4K bytes and a disk address uses 4 bytes, what would be the largest file size based those assumptions? Please show the intermediate results as asked below:
2.1) Disk space pointed by the 10 direct addresses: __________________________________________;
2.2) Disk space pointed by the single indirect block: _________________________________________;
2.3) Disk space pointed by the double indirect block: _________________________________________;
2.4) Disk space pointed by the triple indirect block: __________________________________________;
2.5) The total space pointed by the i-node: __________________________________________________;
I-node Attributes Single Double indirect block Triple indirect block Figure 4-33. A UNIX -node. Addresses of data blocksExplanation / Answer
.1) Disk space pointed by the 10 direct addresses:
Each Address is 4 bytes, so 10 * 4 bytes = 40 bytes of disk space pointed by the 10 direct addresses.
2.2) Disk space pointed by the single indirect block:
Each Address is 4 bytes.
Disk block size is 4K bytes.
So, 4K bytes / 4 bytes = 1K blocks
Therefore Disk space pointed by single indirect block is: 1K blocks * 4KB for each block size = 4 MB
2.3) Disk space pointed by the double indirect block:
Each Address is 4 bytes.
Disk block size is 4K bytes.
So, 4K bytes / 4 bytes = 1K blocks
Therefore Disk space pointed by Double indirect block is: 1K Blocks * 1K Blocks * 4KB for each block size = 4 GB
2.4) Disk space pointed by the triple indirect block:
Each Address is 4 bytes.
Disk block size is 4K bytes.
So, 4K bytes / 4 bytes = 1K blocks
Therefore Disk space pointed by Triple indirect block is: 1K blocks * 1K blocks * 1K blocks * 4KB for each block size = 4 TB
2.5) The total space pointed by the i-node:
Summing from 2.1 to 2.2 answers gives:
40 bytes + 4 MB + 4 GB + 4 TB, which is approximately 4 TB.
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