📝 Indexed allocation Method in File System Implementation (12 MCQs)

📖 Explanation: The index block contains the list of pointers that map each logical block of a file to its physical location on disk, which is the defining feature of indexed allocation. A superblock holds file system metadata, not per‑file pointers, so it is incorrect. An inode is a UNIX structure that may contain direct pointers, but the specific term for the pointer list in indexed allocation is the index block. Data blocks store actual file contents, not pointers, making them wrong choices.

📖 Explanation: Indexed allocation keeps an index block that holds a direct pointer for every logical block of the file. When a program requests block k, the system simply reads pointer k from the index block and jumps to the corresponding data block, avoiding any need to walk through earlier blocks. Storing data sequentially does not provide direct pointers, linking blocks requires traversal, and fixed‑size segments alone do not give random‑access capability.

📖 Explanation: Indexed allocation stores an array of pointers in an index block, so the system can jump directly to any data block without following a chain of links. This eliminates the per‑block traversal cost that linked allocation incurs for large files, where many links would have to be read. Reducing seeks is a benefit but not the primary reason; free‑space lists and metadata size are unrelated to the traversal efficiency.

📖 Explanation: The index block occupies an entire disk block even when only a handful of pointers are needed, leading to internal waste for tiny files. This overhead is the main drawback for small files. External fragmentation is not directly caused by the indexing structure, pointer indirection latency is modest, and maintaining linked lists is characteristic of linked allocation, not indexed allocation.

📖 Explanation: A first‑level index block points to many second‑level index blocks, each of which holds additional data‑block pointers. By adding this extra level, the total number of addressable data blocks grows dramatically, allowing larger files than a single‑level scheme that is limited to the pointers that fit in one block. The space occupied does not reduce capacity, and the increase is not a simple doubling but depends on the number of pointers per block.

📖 Explanation: Each pointer addresses one 4 KB data block. With 1024 pointers, the index block can reference $1024 \times 4\text{KB}=4096\text{KB}=4\text{MB}$ of data. Therefore the maximum file size reachable with one index block is 4 MB. The other options assume more pointers or larger block sizes, which are not given. The calculation directly follows from multiplying the number of pointers by the block size.

📖 Explanation: A 1 KB block holds 256 pointers, so one index block can address $256 \times 1\text{KB}=256\text{KB}$. A 10 MB file needs $10\,\text{MB} / 1\,\text{KB}=10,240$ blocks, far exceeding 256. Using a second‑level index, each first‑level pointer can refer to another index block that adds another 256 pointers, giving $256 \times 256 = 65,536$ blocks, enough for 10 MB. Thus double‑level indexing is required, while single‑level is insufficient and three‑level is unnecessary.

📖 Explanation: The block number is obtained by integer division of the offset by the block size: $5000 \div 1024 = 4$ remainder 904. Zero‑based numbering means the fifth block (index 4) contains the requested byte. If blocks were counted from one, it would be block 5, but the question asks for the block number directly, making block 4 the correct answer. The other choices correspond to off‑by‑one errors.

📖 Explanation: An index block of 4 KB equals 4096 bytes. Each pointer occupies 4 bytes, so two pointers use $2 \times 4 = 8$ bytes. The unused space is $4096 - 8 = 4088$ bytes. Option B matches this calculation. Option A would be true only if the index block were completely filled, which it is not. Option C assumes the entire block is unused, and option D represents half the block, both incorrect.

📖 Explanation: A 10 KB file needs three 4 KB blocks (12 KB total) to hold its data. The file already has one data block, so two additional data blocks must be allocated. The existing index block can simply store two new pointers, so no new index block is needed. Therefore the correct action is to allocate two more data blocks and update the index block. Allocating only one block would leave the file incomplete, and creating a new index block is unnecessary.

📖 Explanation: With linked allocation, the system must follow a pointer for each of the 100 blocks, incurring $100 \times 1\text{ ms}=100\text{ ms}$ of extra time. Indexed allocation reads the index block once and then accesses the data blocks sequentially, adding only a single 1 ms lookup. Consequently, indexed allocation yields a faster sequential read. The “both equal” option ignores the per‑block overhead, and “neither” incorrectly assumes the overhead is insignificant.

📖 Explanation: When a file is truncated, indexed allocation permits the system to modify or remove pointers in the index block directly, without walking through a linked list of blocks. This makes truncation a constant‑time operation. Storing the file size does not by itself simplify truncation, fixed‑size blocks still require deallocation, and free‑space maps are used by many schemes, not only linked allocation.