Chapter 13+14+15

Question 1

Can you simulate tree directory with flat directory using long names?

Answer 1

Yes - encode path in filename using separators like underscores. Example: /home/user/file becomes home_user_file. Less efficient than true tree structure

Question 2

How to allow 4990 of 5000 users access to one file in Unix?

Answer 2

Cannot be done efficiently with standard Unix permissions (owner/group/other model). Would need access control lists (ACLs) or special group containing 4990 users

Question 3

What protection scheme works better than Unix for large user groups?

Answer 3

Access Control Lists (ACLs) that specify individual users/groups with permissions. Capability-based systems. Role-based access control (RBAC)

Question 4

What does rwxr–r– mean in Unix permissions?

Answer 4

Owner: read write execute. Group: read only. Others: read only. Decimal equivalent: 744

Question 5

What does 755 mean in Unix permissions?

Answer 5

Owner: rwx (read write execute). Group: r-x (read execute). Others: r-x (read execute). Common for executable files

Question 6

What does rwxr-x— mean in Unix permissions?

Answer 6

Owner: read write execute. Group: read execute. Others: no permissions. Decimal equivalent: 750

Question 7

Give examples of sequential file access applications

Answer 7

Text processing (reading documents). Log file analysis. Video/audio streaming. Backup and archival operations

Question 8

Give examples of random file access applications

Answer 8

Database systems accessing records. Virtual memory page files. Image/graphics editing. Spreadsheet applications

Question 9

Give examples of index file access applications

Answer 9

Database index files. File allocation tables. Directory structures. Hash table implementations

Question 10

How can OS exploit knowledge of sequential access?

Answer 10

Read-ahead buffering to prefetch next blocks. Larger buffer allocation for sequential streams. Optimize disk scheduling for sequential patterns

Question 11

How does VFS layer support multiple file systems?

Answer 11

Provides common interface abstracting file system differences. Translates generic operations to specific file system calls. Allows transparent mounting of different file system types

Question 12

Why have multiple file system types on one system?

Answer 12

Different file systems optimized for different uses. Legacy compatibility requirements. Network file systems for remote access. Special purpose file systems (proc tmpfs)

Question 13

Why mount root file system automatically at boot?

Answer 13

System needs file system to load other components. Boot process requires access to system files. Root file system contains essential OS components and drivers

Question 14

How many I/O operations for contiguous allocation when adding block at beginning?

Answer 14

Need to shift all 100 existing blocks to make room. Requires reading 100 blocks and writing 101 blocks = 201 I/O operations

Question 15

How many I/O operations for linked allocation adding block at beginning?

Answer 15

Simply update head pointer and set new block to point to old first block. Only 1 I/O operation needed

Question 16

How many I/O operations for indexed allocation adding block at beginning?

Answer 16

Update index block to include new block pointer. Only 1 I/O operation to write updated index block

Question 17

How many I/O operations for contiguous allocation adding block in middle?

Answer 17

Must shift half the blocks (50) to make room. Read 50 + write 51 = 101 I/O operations

Question 18

How many I/O operations for linked allocation adding block in middle?

Answer 18

Must traverse to middle position then update pointers. Traverse + 2 writes = ~51 I/O operations

Question 19

How many I/O operations for indexed allocation adding block in middle?

Answer 19

Simply update index block to include new block pointer. Only 1 I/O operation

Question 20

Why must file allocation bitmap be kept on mass storage?

Answer 20

File system must survive system crashes and reboots. Bitmap represents permanent disk allocation state. Memory is volatile and lost on power failure

Question 21

What criteria determine best allocation strategy for a file?

Answer 21

File size and growth patterns. Access pattern (sequential vs random). Performance requirements. Storage efficiency needs

Question 22

What are advantages of fixed-size extents?

Answer 22

Simple allocation algorithm. Predictable performance. Easy to implement and manage. Reduced metadata overhead

Question 23

What are disadvantages of fixed-size extents?

Answer 23

Internal fragmentation if file doesn’t fit exactly. May waste space for small files. Inflexible for varying file sizes

Question 24

What are advantages of variable-size extents?

Answer 24

Efficient space utilization. Can match exact file requirements. Reduces internal fragmentation

Question 25

What are disadvantages of variable-size extents?

Answer 25

Complex allocation algorithms. External fragmentation over time. Harder to find suitable free extents

Question 26

How does contiguous allocation perform for sequential access?

Answer 26

Excellent - single seek followed by fast sequential read. Minimizes disk head movement. Best performance for sequential workloads

Question 27

How does contiguous allocation perform for random access?

Answer 27

Poor - may require seek for each access if blocks are far apart. Cannot optimize for random patterns

Question 28

How does linked allocation perform for sequential access?

Answer 28

Good - can follow links sequentially. Some overhead from following pointers. May have seeks between non-contiguous blocks

Question 29

How does linked allocation perform for random access?

Answer 29

Very poor - must traverse links from beginning to reach any block. No direct access to arbitrary blocks

Question 30

How does indexed allocation perform for sequential access?

Answer 30

Good - index provides direct access to each block. Single additional seek to read index. Can optimize read-ahead

Question 31

How does indexed allocation perform for random access?

Answer 31

Excellent - direct access to any block via index. Constant time to locate any block. Ideal for random access patterns

Question 32

How do performance optimizations affect crash consistency?

Answer 32

Caching and delayed writes can lose data on crash. Write reordering may leave file system in inconsistent state. Need journaling or logging for crash recovery

Question 33

How does logical-to-physical mapping work in contiguous allocation?

Answer 33

Start block + logical block number = physical block number. Simple arithmetic calculation. Direct mapping with no indirection

Question 34

How does logical-to-physical mapping work in linked allocation?

Answer 34

Follow chain of pointers from first block. Must traverse links sequentially. No direct calculation possible

Question 35

How does logical-to-physical mapping work in indexed allocation?

Answer 35

Use logical block number as index into block pointer array. Direct lookup in index table. Single level of indirection

Question 36

How many blocks to read going from logical block 10 to 4 in contiguous allocation?

Answer 36

Zero additional blocks - can calculate physical address directly. Direct addressing allows immediate access

Question 37

How many blocks to read going from logical block 10 to 4 in linked allocation?

Answer 37

Must traverse backward or restart from beginning. Typically restart from first block requiring 4 block reads to reach block 4

Question 38

How many blocks to read going from logical block 10 to 4 in indexed allocation?

Answer 38

Zero additional blocks - index allows direct access to any block. Can calculate physical address from index immediately

Question 39

What is maximum file size with inode having 12 direct + single/double/triple indirect?

Answer 39

12 direct blocks × 8KB = 96KB. Single indirect: (8KB/4 bytes) × 8KB = 16MB. Double indirect: 2048 × 16MB = 32GB. Triple indirect: 2048 × 32GB = 64TB. Total ≈ 64TB

Question 40

How do directory entries work in file systems?

Answer 40

Map file names to inode numbers or file control blocks. Contain metadata like permissions and timestamps. Enable hierarchical namespace organization

Question 41

What is difference between hard and soft links?

Answer 41

Hard link: multiple directory entries point to same inode. Soft link: file contains path to another file. Hard links share storage soft links are separate files

Question 42

How does file system journaling work?

Answer 42

Record intended changes in log before making them. Replay log after crash to complete or undo partial operations. Ensures consistency at cost of performance

Question 43

What is difference between metadata and data journaling?

Answer 43

Metadata journaling: only logs file system structure changes. Data journaling: logs both metadata and file content changes. Data journaling slower but more comprehensive