Chapter 11
- 1. Chapter 11 File System Implementation
- 2. Exercise 11.1 Consider a file system that uses a modified contiguous-allocation scheme with support for extents. A file is a collection of extents, with each extent corresponding to a contiguous set of blocks. A key issue in such systems is the degree of variability in the size of extents. What are the advantages and disadvantages of the following schemes?
- 3. Exercise 11.1.a All extents are of the same size, and the size is predetermined.
- 4. Answer 11.1.a If all extents are of the same size, and the size is predetermined, then it simplifies the block allocation scheme. A simple bit map for extents would suffice. Simple
- 5. Exercise 11.1.b b. Extents can be of any size and are allocated dynamically.
- 6. Answer 11.1.b If extents can be of any size and dynamically allocated, then more complex allocation schemes are required. It might be difficult to find an extent of the appropriate size external fragmentation. Very complex
- 7. Exercise 11.1.c c. Extents can be a few fixed sizes, and these are predetermined.
- 8. Answer 11.1.c If the extents can be of a few fixed sizes, and these sizes are predetermined, the we would need to maintain a separate bit map for each possible size. This scheme is of intermediate difficulty and flexibility
- 9. Exercise 11.4 Some file systems allow disk storage to be allocated at different levels of granularity. For instance, a file system could allocate 4 KB of disk space as a single 4-KB block or as eight 512-byte blocks. How could we take advantage of this flexibility to improve performance? What modifications would have to be made to the free-space management scheme in order to support this feature?
- 10. Answer 11.4 For example, if a file is 5 KB then it would be allocated a 4-KB block and two contiguous 512-byte blocks. What will this block allocation scheme help decrease?
- 11. Answer 11.4 cont. Modification are required. We should maintain: a bitmap of free blocks an extra state regarding which of the sub blocks are currently being used inside a block. Therefore, the allocator would have to examine that extra state to allocate sub blocks and coalesce the sub blocks to obtain the larger block when all of the sub blocks are free.
- 12. Exercise 11.6 Consider a file system on a disk that has both logical and physical block sizes of 512 bytes. Assume that the information about each file is already in memory. For each of the three allocation strategies (contiguous, linked, and indexed), answer these question:
- 13. Exercise 11.6.a How is the logical-to-physical address mapping accomplished in this system? For indexed allocation, assume that a file is always less than 512 blocks long.
- 14. Answer 11.6.a The CPU will generate a logical address L and we need to convert it to a physical address P P is a disk address represented by a block number and an offset Assume the following S is the starting block B is the block O is the offset
- 15. Answer 11.6.a (Contiguous) Contiguous allocation: B = L / 512 O = L modulus 512 Physical block number = S + B O is the offset into that block So: S + B + O = physical address
- 16. Answer 11.6.a (Linked) Linked allocation: B = L / 511 O = L modulus 511 Follow the linked list for (B+1) blocks (O+1) should be the offset into the last physical block
- 17. Answer 11.6.a (Indexed) Indexed allocation: B = L / 512 O = L modulus 512 At the beginning, we need to load the index block from disk The physical address is stored at in the index block B. O is the offset into the physical block pointed to by index block B.
- 18. Exercise 11.6.b b. If we are currently at logical block 10 (the last block accessed was block 10) and want to access logical block 4, how many physical blocks must be read from the disk?
- 19. Answer 11.6.b Contiguous: 1 physical block Linked: 4 physical blocks Indexed: 2 physical blocks