Final answer:
The maximum size of a file that uses inodes with 10 direct blocks, 4 single indirect blocks, and 1 double indirect block is calculated by adding the capacity of all these blocks, which amounts to 528,404KB.
Step-by-step explanation:
To calculate the maximum size of a file in a file system that uses inodes, we need to consider the number of direct disk blocks, single indirect disk blocks, and double indirect disk blocks. Each disk block is 2KB in size, and each pointer to a disk block takes up 4 bytes. With these values, we can proceed with the calculation.
- Direct disk blocks: The inode has 10 direct disk blocks. Since each block is 2KB, this gives us a total of 20 KB.
- Single indirect disk blocks: There are 4 single indirect blocks. Each 2KB block can contain 512 pointers (2048 bytes / 4 bytes per pointer), and each pointer points to a 2KB block. This gives us 512 * 2KB for each single indirect block, so for 4 blocks, it's 4 * 512 * 2KB, resulting in 4,096 KB.
- Double indirect disk block: There's 1 double indirect disk block. First, calculate the number of single indirect blocks it can point to 512 pointers per 2KB block. Then, each of those can point to 512 more 2KB blocks. So, it's 512 * 512 * 2KB, which equals 524,288 KB.
Add up the values from each type of block: 20KB (direct) + 4,096KB (single indirect) + 524,288KB (double indirect). The total maximum size of the file is therefore 528,404KB.