CS 620 (Spring 2008)

Answers for Assignment 2

1. (5 points)

One way to use file system's allocation of a disk and not suffer from holes is to compact the disk (i.e., move the files together to remove the holes) every time a file is deleted. Since all files are contiguous, copying a file requires a seek and rotational delay to read the file, followed by the transfer at full speed. Writing the file back requires the same work. Assuming a seek time of 6ms, a rotational delay of 4ms, a transfer rate of 5MB/sec, and an average file size of 8KB, how long does it take to read a file into main memory and then write it back to the disk at a new location? Using these numbers, how long would it take to compact half of a 16GB disk?

[Answer]

Read and write time for a file = 2 * (seek time + rotational delay + transfer time) 

 

Seek time = 6 ms

Rotational delay = 4 ms

Transfer rate = 5MB/sec = 5 * 1024 KB/sec           ( 5000 KB/sec is OK too)

Average file size = 8KB

Transfer time = 8/(5*1024) s                    (8/5000 s is OK too)

Transfer time = 1.5625 ms                         (1.6 ms is OK too)

Overall time = 6 + 4 + 1.5625 = 11.5625 ms             (11.6 ms is OK too)

Read and write time = 2 * 11.5625 = 23.125 ms         (23.2 ms is OK too)

 

A 16 GB disk = (approx) 1,000,000 * 16 KB


The number of files that Half of 16 GB disk can have:

(1024 * 1024 * 16 KB * 0.5)/8KB = 1,048,576 KB  (1,000,000 is OK)
 

Assuming we have to move all the files during the compaction, 
the maximum time taken = 23.125 * 1,048,576 = 24,248,320 ms = 24,248 s (23,125 s is OK)

2. (3 points)

In light of the answer to the previous question, does compacting the disk ever make sense? Explain your answer.

[Answer]

Compacting a disk after every file removal is a very time consuming job. In the example above every file removal if followed by a disk compaction might take hours to finish. But if a 16GB disk is occupied using only half the disk and the rest of the disk is being wasted as holes. In this situation, a disk compaction may make sense although it takes a long time. The advantage would be that the free space could be useful.

3. (3 points)

Two computer science students, John and Mary, are having a discussion about i-nodes. John maintains that memories have gotten so large and so cheap that when a file is opened, it is simpler and faster just to fetch a new copy of the i-node into the i-node table, rather than search the entire table to see is it is already there. Mary disagrees. Who is right? Justify your answer.

[ANSWER]

Mary is right because it takes a lot of time to fetch the i-node from the disk every time it is needed. Also when a process is using a file the i-node of the file would be in memory and any changes made to the i-node might not have been written back to the disk. This causes consistency problems.

4. (4 points)

List four ways a system could use to determine which blocks on a disk are free.
Give advantages of each way.

[ANSWER]

(a). Bit vector: the free space list is implemented as a bit map or bit vector. Each block is represented by a bit with 1 for free and 0 for allocated.
Advantage: it is simple to implement and is efficient tin finding the first 1 or n free blocks.
(b). Linked list: all the free disk blocks are linked together and keep a pointer to the first block in a special location on the disk and cache it in the memory.
Advantage: takes no useful place.
(c). Grouping: store the addresses of n free blocks in the first free block. The first n-1 blocks are free and the last block contains the addresses of another n free blocks, and so on.
Advantage: it can find the address of any free block quickly, unlike the linked list.
(d). Counting: keep the address of the first free block and the number n of free contiguous blocks that follow the first block. Each entry in the free list consists of a disk address and a count.
Advantage: This is a fact for use in contiguous storage search.

5. (4 points)

In some systems, a subdirectory can be read and written by an authorized user (not just the owner), just as ordinary files can be.
(a) Describe a protection problem that could arise.
(b) Suggest a scheme for dealing with the protection problem you named in part (a).

[ANSWER]

(a) One piece of information kept in a directory entry is file location. If a user could modify this location, then he could access other files defeating the access-protection scheme.
(b) Do NOT allow the user to directly write onto the subdirectory. Rather, provide system operations to do so.

6. (4 points)

(a). What problems might arise if a file being currently opened by several users is deleted?
(b). How can we solve this problem?

[ANSWER]

(a) Copier of file might delete the original shared file, depriving rest of users. They have a pointer to a deleted directory entry pointing to the original file or one overwritten by other users of the system, or a new entry pointing to a new file created by the original user.
(b) Keep a count of the number of links to a file in original directory. As each person deletes a file, the count decreases by 1.

7. (3 points)

What are the advantages and disadvantages of recording the name of the creating program with the file's attributes (as is done in the Macintosh operating system)?

[ANSWER]
Advantage: If the name of the creating program is recorded with the file's attributes, then the operating system knows exactly which program should be invoked automatically in order to load the file.
Disadvantage: The operating system needs to implement extra functionality (to set the correct attribute during the create () call) so that the attribute's use is enforced and supported by the system. Also there is lack of flexibility.

8. (8 points)

Suppose that an inode is 128 bytes, pointers are 4 bytes long, and the status information takes up 68 bytes. Assume a block size of 8K bytes and block pointers of 32 bits each.

(a) How much room is there for direct pointers in the inode?

(b) How big a file can be represented with direct pointers?

(c) How big a file can be represented with single Indirect pointers?

(d) How big a file can be represented with Double Indirect pointers?

(e) How big a file can be represented with Triple Indirect pointers?

[ANSWER]

(8KB = 8192 bytes)
The single, double, and triple indirect pointers take 4 bytes each, so
128(total) - 68(status) - 12(indirect) = 48

Therefore 48 bytes are available for direct pointers.
48/4 = 12 direct pointers can be implemented.
 
Thus a file size of 12 x 8192 = 98,304 bytes can be represented solely with direct pointers.

If the block size is 8K bytes, the single indirect pointer addresses an 8K block that can hold 
8192 / 4 = 2048 pointers to data blocks. Thus, the single indirect pointer provides the capability of addressing an additional
2048 x 8192 = 16,777,216 bytes (or 16 MB) of information. 

Double indirect addressing provide 2048 x 2048 pointers with the capability of addressing 
2048 x 2048 x 8192 = 32 GB (additional to the 98KB + 16MB)

Triple indirect addressing provides 2048 x 2048 x 2048 pointers with the capability of addressing an additional 
2048 x 2048 x 2048 x 8192 = 64 terabytes.

9. (2 points)

Consider a system that supports 5000 users. Suppose you want to allow 4990 of these users to be able to access one file. How would you specify this protection scheme in UNIX?

[ANSWER]
A new group of the 4990 users must be created and associated with the file to be accessed. The owner of the file can then set the access rights for the group (whether the group can read, write and execute the file).

10. (6 points)

Consider the following backup scheme:

Day 1. Copy to a backup medium all files from the disk.
Day 2. Copy to another medium all files changed since day 1.
Day 3. Copy to another medium all files changed since day 1.
This differs from the schedule given in Section 11.7.2 in the textbook by having all subsequent backups copy all files modified since the first full backup.
(a). What are the benefits of this system over the one in Section 11.7.2? (2 points)
(b). What are the drawbacks? (2 points)
(c). Are restore operations made easier or more difficult? Explain your answer (2 points).

[ANSWER]

(a). Benefit: Restores are easier and faster and no intermediate tapes need be read.
(b). Drawbacks: More tape is used as more files change.
(c). Just as mentioned in (a), restores are easier since you can go to the last backup tape, rather than the full tape.

11. (6 points)

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 and each disk address contains 1 bytes. For each of the three allocation strategies (Contiguous, linked, and indexed), answer these questions:

(a) How is the logical-to-physical address mapping accomplished in this system? (For the indexed allocation, assume that a file is always less than 512 blocks long.)
(b) If we 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?

[ANSWER]
The logical block number of a file always starts at 1. Let Z be the starting file address for physical block number, then the last physical block would be:

Contiguous: Divide the logical address by 512 with X and Y the resulting quotient and remainder respectively.

(a). Add X to Z to obtain the physical block number. Y is the displacement into that block.
(b). 1 block must be read from the disk.

Linked: Divide the logical physical address by 511 with X and Y the resulting quotient and remainder respectively.
(a). Chase down the linked list (getting X+1 blocks). Y+1 is the displacement into the last physical block.
(b). 4 blocks need to be read from the disk.
Indexed: Divide the logical address by 512 with X and Y the resulting quotient and remainder respectively.
(a). Get the index block into memory. Physical block address is contained in the index block at location X. Y is the displacement into the desired physical block.
(b). 2 blocks need to be read from the disk.

12. (2 points)

How many disk operations are needed to fetch the i-node for the file /home/teacher/courses/os/handout.html? Explain your answer. Assume that the root directory is in memory, but nothing else along the path is in memory. Assume also that each directory file fits in one block.

[ANSWER]
The root directory is in memory. Therefore there is no need for a disk access. But, none of the other elements in the path are in memory. For each of the directories home, teacher, courses and os, two disk operations will be needed. One for loading the inode and the second to load the directory file (data for the directory file). Finally, one disk operation will be required to load the inode for handout.html file.
Thus in all, 9 disk operations will be required to fetch the file /home/teacher/courses/os/handout.html.

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