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Question: Consider a file system using the linked allocation method. The file system has 1024 blocks and block numbers require 32 bits to store. The block size is 2 KB. The file system has only one root directory in which we are creating files of the size of 998 bytes. Directories store the file name and starting block. The files created will have a file name with 24 characters. How many of these files can we create in this file system? The only options are 1022, 1024, 512, 511, 1008, 504
First, let's calculate the size of each directory entry. A directory entry will consist of the file name and the starting block. The file name is 24 characters long, and since a character typically takes 1 byte, that is 24 bytes. The starting block number requires 32 bits which is equivalent to 4 bytes. So, the size of each directory entry is 28 bytes (24 + 4). Next, consider that there are 1024 blocks in the file system, and the block size is 2 KB (2048 bytes). Since each block can store 998 bytes of file data, one file will take exactly one block due to the linked allocation method. Now, let's see how many blocks will be required for the directory entries. Since the directory entry size is 28 bytes, we can calculate the number of directory entries that can be stored in each block: 2048 bytes (block size) / 28 bytes (directory entry size) = 73.14 We can round this down to 73, since a directory entry can't be partially stored. Therefore, each block can store 73 directory entries. We need to reserve one block for the root directory, so we have 1023 blocks available for storing file's data. Now, we need to find the number of blocks required to store the directory entries for these files: 1023 files / 73 entries per block = 14.01 Rounding this up, we'll need 15 blocks for storing the directory entries for all these files. So, we have: 1023 (available blocks) - 15 (blocks for directory entries) = 1008 blocks Now, remember that each file takes exactly one block in the linked allocation method. Therefore, we can create a total of 1008 files in this file system. The correct option is **1008**.
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