Sample question paper

Que.

Questions/ Answer Marks

a) Differentiate between Batch Operating System and Time shared Operating System.(any two points)

Sr. No

Batch Operating System Time shared Operating System

1 There is an operator which takes similar jobs having same requirement and group them into batches. 

Each task is given some time to execute, so that all the tasks work smoothly

2 It is very difficult to guess or know the time required by any job to complete

Each task gets an equal opportunity

b) State any four services of Operating System.

Here is a list of common services offered by an almost all operating systems:

User Interface Program Execution File system manipulation Input / Output Operations Communication Resource Allocation Error Detection Accounting Security and protection

c) Define : Process , ProgramA process is defined as an entity which represents the basic unit of work to be implemented in the system.

A program is a piece of code which may be a single line or millions of lines. d) State two features of preemptive scheduling.

1. Preemptive Scheduling is defined as the scheduling which is done when the process changes from running state to ready state or from waiting for the state to ready state.

2. We can interrupt the process in between of execution.3. In preemptive scheduling, if a process which has high priority arrives in

the ready queue, then the process which has low priority may starve.e) Define following terms

i) Page faultii) Segmentationi) Page fault

While executing a program, if the program references a page which is not available in the main memory because it was swapped out a little ago, the processor treats this invalid memory reference as a page fault

ii) Segmentation

In Operating Systems, Segmentation is a memory management technique in which, the memory is divided into the variable size parts. Each part is known as segment which can be allocated to a process.

f) Write syntax of ps command and explain its use with the help of suitable example.

ps

PID TTY TIME CMD5763 pts/3 00:00:00 zsh8534 pts/3 00:00:00 ps

The result contains four columns of information.

PID - the number of the processTTY - the name of the console that the user is logged intoTIME- the amount of CPU in minutes and seconds that the process has been runningCMD - the name of the command that launched the process

g) List any four file attributesAttributes of the File1.NameEvery file carries a name by which the file is recognized in the file system. One directory cannot have two files with the same name.2.IdentifierAlong with the name, Each File has its own extension which identifies the type of the file. For example, a text file has the extension .txt, A video file can have the extension .mp4.3.TypeIn a File System, the Files are classified in different types such as video files, audio files, text files, executable files, etc.4.LocationIn the File System, there are several locations on which, the files can be stored. Each file carries its location as its attribute.5.SizeThe Size of the File is one of its most important attribute. By size of the file, we mean the number of bytes acquired by the file in the memory.6.ProtectionThe Admin of the computer may want the different protections for the different files. Therefore each file carries its own set of permissions to the different group of Users.7.Time and DateEvery file carries a time stamp which contains the time and date on which the file is last modified.

Q.2)a) Explain dual modes of operations of an Operating system.

There are two modes of operation in the operating system to make sure it works correctly. These are user mode and kernel mode.A diagram that illustrates the transition from user mode to kernel mode and back again is as follows:

The following are the modes:

User Mode:

The system is in user mode when the operating system is running a user application such as handling a text editor. The transition from user mode to kernel mode occurs when the application requests the help of operating system or an interrupt or a system call occurs.The mode bit is set to 1 in the user mode. It is changed from 1 to 0 when switching from user mode to kernel mode.

Kernel Mode

The system starts in kernel mode when it boots and after the operating system is loaded, it executes applications in user mode. There are some privileged instructions that can only be executed in kernel mode. These are interrupt instructions, input output management etc. If the privileged instructions are executed in user mode, it is illegal and a trap is generated.The mode bit is set to 0 in the kernel mode. It is changed from 0 to 1 when switching from kernel mode to user mode.

b) Describe essential activities done by an Operating System for protection and sharingProtectionProtection provides mechanisms for controlling which users / processes have access to which system resources.System calls allow the access mechanisms to be adjusted as needed, and for non-priveleged users to be granted elevated access permissions under carefully controlled temporary circumstances.Once only of concern on multi-user systems, protection is now important on all systems, in the age of ubiquitous network connectivity.

Sharing of physical resources is not commonly implemented, but may be done as if the virtual machines were networked together. 

c) State what is interprocess communication and explain its advantages..

Interprocess communication is the mechanism provided by the operating system that allows processes to communicate with each other. This

communication could involve a process letting another process know that some event has occurred or transferring of data from one process to another.

A diagram that illustrates interprocess communication is as follows:

The models of interprocess communication are as follows:

Shared Memory Model

Shared memory is the memory that can be simultaneously accessed by multiple processes. This is done so that the processes can communicate with each other. All POSIX systems, as well as Windows operating systems use shared memory.

Advantage of Shared Memory Model

Memory communication is faster on the shared memory model as compared to the message passing model on the same machine.

Disadvantages of Shared Memory Model

Some of the disadvantages of shared memory model are as follows:

All the processes that use the shared memory model need to make sure that they are not writing to the same memory location.

Shared memory model may create problems such as synchronization and memory protection that need to be addressed.

Message Passing Model

Multiple processes can read and write data to the message queue without being connected to each other. Messages are stored on the queue until their recipient retrieves them. Message queues are quite useful for interprocess communication and are used by most operating systems.

Advantage of Messaging Passing Model

The message passing model is much easier to implement than the shared memory model.

Disadvantage of Messaging Passing Model

The message passing model has slower communication than the shared memory model because the connection setup takes time.

A diagram that demonstrates the shared memory model and message passing model is given as follows:

d) Discribe different scheduling criteria.Different scheduling criteria are:

CPU utilization. We want to keep the CPU as busy as possible. Conceptually, CPU utilization can range from 0 to 100 percent. In a real system, it should range from 40 percent (for a lightly loaded system) to 90 percent (for a heavily used system). 

Throughput. If the CPU is busy executing processes, then work is being done. One measure of work is the number of processes that are completed per time unit, called throughput.

Turnaround time. From the point of view of a particular process, the important criterion is how long it takes to execute that process. The interval from the time of submission of a process to the time of completion is the turnaround time. Turnaround time is the sum of the periods spent waiting to get into memory, waiting in the ready queue, executing on the CPU, and doing I/0.

Waiting time. The CPU-scheduling algorithm does not affect the amount of time during which a process executes or does I/0, it affects only the amount of time that a process spends waiting in the ready queue. Waiting time is the sum of the periods spent waiting in the ready queue. 

Response time. Time from the submission of a request until the first response is produced. This measure, called response time, is the since it takes to start responding, not the time it takes to output the response. 

Q.3

12 M

a) Define PCB. List information contained in PCB and explain any two 4 M

Each process is represented as a process control block (PCB) in the operating system. It containsInformation associated with specific process.In general, a PCB may contain information regarding:1. Process Number: Each process is identified by its process number, called

process identification

number (PID).

2. Priority: Each process is assigned a certain level of priority that

corresponds to the relative importance of the event that it services.

3. Process State: This information is about the current state of the process. I.e.

whether process is in

new, ready, running, waiting or terminated state.

4. Program Counter: This contains the address of the next instruction to be

executed for this process.

5. CPU Registers: CPU registers vary in number and type, depending upon the

computer architectures. These include index registers, stack pointers and

general purpose registers etc. When an interrupt occurred, information about

the current status of the old process is saved in registers along with the

program counters. This information is necessary to allow the process to be

continued correctly after the completion of an interrupted process.

6. CPU Scheduling Information: This information includes a process priority,

pointers to scheduling queues and any other scheduling parameters.

7. Memory Management Information: This information may include such

information as the value of base and limit registers, the page table or the

segment table depending upon the memory system used by operating system.

8. Accounting: This includes actual CPU time used in executing a process in

order to charge individual user for processor time.

9. I/O Status: It includes outstanding I/O request, allocated devices

information, pending operation and so on.

10. File Management: It includes information about all open files, access

rights etc.

b) Define deadlock and state the necessary conditions for deadlock..A Deadlock is a situation where each of the computer process waits for a resource which is being assigned to some another process. In this situation, none of the process gets executed since the resource it needs, is held by some other process which is also waiting for some other resource to be released.Let us assume that there are three processes P1, P2 and P3. There are three different resources R1, R2 and R3. R1 is assigned to P1, R2 is assigned to P2 and R3 is assigned to P3.After some time, P1 demands for R1 which is being used by P2. P1 halts its execution since it can't complete without R2. P2 also demands for R3 which is being used by P3. P2 also stops its execution because it can't continue without R3. P3 also demands for R1 which is being used by P1 therefore P3 also stops its execution.

In this scenario, a cycle is being formed among the three processes. None of the process is progressing and they are all waiting. The computer becomes unresponsive since all the processes got blocked.

Four condition for deadlock.1. Mutual exclusion: Only one process at a time can use non-sharable

resource.

2. Hold and wait: A process is holding at least one resource and is waiting to

acquire additional resources held by other processes.

3. No pre-emption: A resource can be released only voluntarily by the process

holding it after that

process completes its task.

4. Circular wait: There exists a set {P0, P1, …, P0} of waiting processes such

that P0 is waiting for a

resource that is held by P1, P1 is waiting for a resource that is held by P2, …,

Pn–1 is waiting for a

resource that is held by Pn, and Pn is waiting for a resource that is held by P0.

c) Explain following terms with respect to Memory management :I. CompactionII. Swapping.

Compaction

External Fragmentation is the type of fragmentation in which the memory space satisfies the requirement of the process. The problem here is that the memory is not contiguous. So, the process is not able to use that memory.It can be reduced with the help of the compaction technique. Through this technique, we will shuffle the memory blocks in such a way so that all the free memory space get merged and converted into a large memory block which can be used by the process. This is shown in diagram clearly.So, the compaction technique reduces the wastage of memory created by the process of fragmentation.

Swapping

Swapping is a concept in which we do swapping of a process from main memory to secondary storage and vice-versa. The swapping of processes from one memory to another is explained clearly in the diagram given below. It clearly explains that how this swap in and swap out thing works. This is done so that the memory which we get free through swapping can be used by other processes.Swapping affects the performance of the system, but it is helpful or beneficial for the system when we want to run multiple processes in parallel. Due to this reason, memory compaction technique is another name given to swapping.

The total time taken by the swapping can be calculated by adding1. The time is taken in moving the whole process from main memory to secondary disk.2. The time is taken in moving back from secondary disk to main memory.3. Time is taken by the process to regain its main memoryBy adding all the above time, we can calculate the total time taken in the whole process.

d) Enlist different file allocation methods and explain any twoList different file allocation methods. Explain any one in detailFile allocation methods are:

Contiguous Allocation method Linked Allocation method Indexed Allocation method

Contiguous Allocation The contiguous allocation method requires each file to occupy a set of contiguous address on the disk. Disk addresses define a linear ordering on the disk. With this ordering, accessing block b+1 after block b normally requires no head movement. Contiguous allocation of a file is defined by the disk address and the length of the first block. If the file is n blocks long, and starts at location b, then it occupies blocks b, b+1, b+2, …, b+n-1. The directory entry for each file indicates the address of the starting block and the length of the area allocated for this file

linked Allocation: In this method, each file occupies disk blocks scattered anywhere on the disk. It is a linked list of allocated blocks. When space has to be allocated to the file, any free block can be used from the disk and system makes an entry in directory. Directory entry for allocated file contains file name, a pointer to the first allocated block and last allocated block of the file. The file pointer is initialized to nil value to indicate empty file. A write to a file, causes search of free block. After getting free block data is written to the file and that block is linked to the end of the file. To read the file, read blocks by following the pointers from block to block starting with block address specified in the directory entry. For example, a file of five blocks starting with block 9 and continue with block 16,then block 1,then block 10 an finally block 25.each allocated block contains a pointer to the next block.

Indexed Allocation:

In this method, each file has its own index block. This index block is an array of disk block addresses. When a file is created, an index block and other disk blocks according to the file size are allocated to that file. Pointer to each allocated block is stored in the index block of that file. Directory entry contains file name and address of index block. When any block is allocated to the file, its address is updated in the index block. Any free disk block can be allocated to the file. Each ith entry in the index block points to the ith block of the file. To find and read the ith block, we use the pointer in the ith index block entry.

Q.4a) Compare between Windows and LINUX Operating System.(any four points)

Description LInux Windows

Cost

Linux can be freely distributed, downloaded freely, distributed through magazines, Books etc. There are priced versions for Linux also, but they are normally cheaper than Windows.

$119 - $199.99, $69 for students, $99 to upgrade to pro

Development and Distribution

Linux is developed by Open Source development i.e. through sharing and collaboration of code and features through forums etc and it is distributed

Developed and distributed by Microsoft.

by various vendors.

Manufacturer

Linux kernel is developed by the community. Linus Torvalds oversees things.

Microsoft

User

Everyone. From home users to developers and computer enthusiasts alike.

Everyone

Usage

Linux can be installed on a wide variety of computer hardware, ranging from mobile phones, tablet computers and video game consoles, to mainframes and supercomputers.

On PC's desktops, laptops, servers and some phones.

File system support

Ext2, Ext3, Ext4, Jfs, ReiserFS, Xfs, Btrfs, FAT, FAT32, NTFS

FAT, FAT32, NTFS, exFAT

Text mode interface

BASH (Bourne Again SHell) is the Linux default shell. It can support multiple command interpreters.

Windows uses a command shell and each version of Windows has a single command interpreter with dos-like commands, recently there is the addition of the optional PowerShell that uses more Unix-like commands.

Supported platforms

All

PowerPC: versions 1.0 - NT 4.0; DEC Alpha: versions 1.0 - NT 4.0; MIPS R4000: versions 1.0 - NT 4.0; IA-32: versions 1.0 - 8; IA-64: version XP; x86-64: versions XP - 8; ARM: version RT;

b) Write any four system calls related to device management.A process may need several resources for its execution. So system calls used for asking permission from the kernel to use those resources are included in this type.

1. request device, release device2. read, write, reposition3. get device attributes, set device attributes4. logically attach or detach devices

c) Compare between short term and long term scheduler.(any four points)

BASIS FOR COMPARISON

LONG-TERM SCHEDULER

SHORT-TERM SCHEDULER

Basic It picks up the process from Job Pool/Job Queue.

It picks up the process from Ready Queue.

Frequency Long-Term Scheduler selects the process less frequently.

Short-Term Scheduler select the process more frequently.

Control It controls the Degree of Multiprogramming.

It has less control over the Degree of Multiprogramming.

Alternative Alternatively, it is called Job Scheduler.

Alternatively, it is called CPU Scheduler.

Important Long-Term Scheduler is there in Batch Systems but it may or may not be present in Time Sharing System.

Short-Term Scheduler is there in Batch System and is minimally present in the Time Sharing System also

d) Compare FCFS and SJF Scheduling algorithm with any four points.

First Come First Serve (FCFS)

Let's start with the Advantages:

FCFS algorithm  doesn't include any complex logic, it just puts the process requests in a queue and executes it one by one.

Hence, FCFS is pretty simple and easy to implement. Eventually, every process will get a chance to run, so starvation doesn't

occur.

It's time for the Disadvantages:

There is no option for pre-emption of a process. If a process is started, then CPU executes the process until it ends.

Because there is no pre-emption, if a process executes for a long time, the processes in the back of the queue will have to wait for a long time before they get a chance to be executed.

Shortest Job First (SJF)

Starting with the Advantages: of Shortest Job First scheduling algorithm.

According to the definition, short processes are executed first and then

followed by longer processes. The throughput is increased because more processes can be executed in

less amount of time.

And the Disadvantages:

The time taken by a process must be known by the CPU beforehand, which is not possible.

Longer processes will have more waiting time, eventually they'll suffer starvation.

e) Describe contiguous memory allocation done by Operating System with he help ofsuitable example.Contiguous Allocation The contiguous allocation method requires each file to occupy a set of contiguous address on the disk. Disk addresses define a linear ordering on the disk. With this ordering, accessing block b+1 after block b normally requires no head movement. Contiguous allocation of a file is defined by the disk address and the length of the first block. If the file is n blocks long, and starts at location b, then it occupies blocks b, b+1, b+2, …, b+n-1. The directory entry for each file indicates the address of the starting block and the length of the area allocated for this file

Q.5

Attempt any TWO of the following. 12 M

a) Write two uses of following Operating System tools :i. Performance Monitorii. Task scheduleriii. User ManagementPerformance MonitorPerformance Monitor is the second most common operating system performance monitoring tool for Windows. Performance Monitor acts as both a

real time and log-based performance monitoring tool for operating systems, so only the real time portion of the tool will be discussed in detail in this section, and the logging portion will be discussed later.

Like Task Manager, Performance Monitor measures performance by making system calls to retrieve system counters, but Performance Monitor makes these calls via a performance library that also supports logging of the counters. Unlike Task Manager, Performance Monitor provides an interface to monitor any selection of a huge set of system counters on a graph in real time, rather than just the limited set Task Manager uses. Counters include things like percentage of processor time, thread count, page fault rate, memory size, and elapsed time for processes. Similarly, there are counters that provide state for threads, the processor, the system, network interfaces, memory, physical disks, and many others. This level of detailed information available for monitoring from Performance Monitor is very extensive and makes Performance Monitor ideal for monitoring resource usage and performance of almost all pieces of a Windows system.

Task scheduler

Schedulers

Schedulers are special system software which handle process scheduling in various ways. Their main task is to select the jobs to be submitted into the system and to decide which process to run. Schedulers are of three types −

Long-Term Scheduler

Short-Term Scheduler

Medium-Term Scheduler

Long Term Scheduler

It is also called a job scheduler. A long-term scheduler determines which programs are admitted to the system for processing. It selects processes from the queue and loads them into memory for execution. Process loads into the memory for CPU scheduling.

The primary objective of the job scheduler is to provide a balanced mix of jobs, such as I/O bound and processor bound. It also controls the degree of multiprogramming. If the degree of multiprogramming is stable, then the average rate of process creation must be equal to the average departure rate of processes leaving the system.

On some systems, the long-term scheduler may not be available or minimal. Time-sharing operating systems have no long term scheduler. When a process changes the state from new to ready, then there is use of long-term scheduler.

Short Term Scheduler

It is also called as CPU scheduler. Its main objective is to increase system performance in accordance with the chosen set of criteria. It is the change of ready state to running state of the process. CPU scheduler selects a process

among the processes that are ready to execute and allocates CPU to one of them.

Short-term schedulers, also known as dispatchers, make the decision of which process to execute next. Short-term schedulers are faster than long-term schedulers.

Medium Term Scheduler

Medium-term scheduling is a part of swapping. It removes the processes from the memory. It reduces the degree of multiprogramming. The medium-term scheduler is in-charge of handling the swapped out-processes.

A running process may become suspended if it makes an I/O request. A suspended processes cannot make any progress towards completion. In this condition, to remove the process from memory and make space for other processes, the suspended process is moved to the secondary storage. This process is called swapping, and the process is said to be swapped out or rolled out. Swapping may be necessary to improve the process mix.

User Management

Linux is a multi-user operating system. Even systems that will be used by a single user are configured as a multi-user system. This has various advantages:

security-wise, your system is protected from malicious software execution as the software is executed as an unprivileged user rather than the system administrator

if at any time multiple users are going to work on the system, you just need to add the user to the system (no need to upgrade to a multi-user environment first)

you can easily back up all files belonging to a particular user as all user files are located inside his home directory

if you messed up your personal configuration for a particular software title, you can just remove the configuration files (or move them aside) and start the software title up again to start with a clean slate. No configuration changes made by a user are propagated system-wide

b) Write the output of the following commands.a. Kill 9042018b. Ps 07121975c. Sleep 05a. Kill 9042018

To kill, or terminate a process first find out the process identifier number or PID of the process to be killed, then pass the PID number to the kill command.

b. Ps 07121975

the ps command shows the running processes by the user running it within a

terminal window. To invoke ps simply type the following:

c. Sleep 05waiting for 5 seconds

c) Given a page reference reference string(arrival) with four page frames, calculatethe page faults with FIFO and LRU page replacement algorithms respectively :12, 3, 4, 5, 1, 2, 5, 1, 2, 3, 4, 5, 1,6,7,8,7,8,9,7,8,9,5,4,4,5,4,2.FIFO

12

3 4 5 1 2 5 1 2 3 4 5 1 6 7 8 7 8 9 7 8 9 5 4 4 5 4 2

12

12

12

5 5 5 5 5 5 3 3 3 1 1 1 8 8 8 8 8 8 8 8 4 4 4 4 4

3 3 3 1 1 1 1 1 1 4 4 4 6 6 6 6 6 9 9 9 9 9 9 9 9 9 24 4 4 2 2 2 2 2 2 5 5 5 7 7 7 7 7 7 7 7 5 5 5 5 5 5

F F F F F F F F F F F F F F F F F

No. of page fault:- 17

LRU12

3 4 5 1 2 5 1 2 3 4 5 1 6 7 8 7 8 9 7 8 8 5 4 4 5 4 2

12

12

12

5 5 5 5 5 5 3 3 3 1 1 1 8 8 8 9 9 8 8 8 4 4 4 4 2

3 3 3 1 1 1 1 1 1 4 4 4 6 6 6 6 6 6 6 6 6 6 6 6 6 6 64 4 4 2 2 2 2 2 2 5 5 5 7 7 7 7 7 7 7 7 5 5 5 5 5 5

F F F F F F F F F F F F F F F F F F

No. of page fault:- 18

Q.6

Attempt any TWO of the following 12 M

A)

Solve given problem by Using FCFS to calculate average waiting time and turnaroundtime.

Process Arrival time

Burst time

P1 0 7P2 1 4P3 2 9P4 3 6P5 4 8

P1 P2 P3 P4 P5

0 7 11 20 26 34

TAT = ENDING TIME - ARRIVAL TIME

PROCESS ENDING TIME ARRIVAL TIME TATP1 7 0 7P2 11 1 10P3 20 2 18P4 26 3 23P5 34 4 30

Avg. TAT = (7+10+18+23+30)/5 = 17.6

WT = TURN AROUND TIME - BURST TIME

PROCESS TURN AROUND TIME

BURST TIME WT

P1 7 7 0P2 10 4 6P3 18 9 9P4 23 6 17P5 30 8 22

Avg. WT = (0+6+9+17+22)/5 = 10.8

b) Compare between bitmap and linked list free space management techniques.(any sixpoints)

1. Bitmap

This technique is used to implement the free space management. When the free space is implemented as the bitmap or bit vector then each block of the disk is represented by a bit. When the block is free its bit is set to 1 and when the block is allocated the bit is set to 0. The main advantage of the bitmap is it is relatively simple and efficient in finding the first free block and also the consecutive free block in the disk. Many computers provide the bit manipulation instruction which is used by the users.

The calculation of the block number is done by the formula:

(number of bits per words) X (number of 0-value word) + Offset of first 1 bit

For Example: Apple Macintosh operating system uses the bitmap method to allocate the disk space.

Assume the following are free. Rest are allocated:

Advantages:

This technique is relatively simple. This technique is very efficient to find the free space on the disk.

Disadvantages:

This technique requires a special hardware support to find the first 1 in a word it is not 0.

This technique is not useful for the larger disks.

For example: Consider a disk where blocks 2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 17, 18, 25,26, and 27 are free and the rest of the blocks are allocated. The free-space bitmap would be: 001111001111110001100000011100000

2. Linked list

This is another technique for free space management. In this linked list of all the free block is maintained. In this, there is a head pointer which points the first free block of the list which is kept in a special location on the disk. This block contains the pointer to the next block and the next block contain the pointer of another next and this process is repeated. By using this disk it is not easy to search the free list. This technique is not sufficient to traverse the list because we have to read each disk block that requires I/O time. So traversing in the free list is not a frequent action.

Advantages:

Whenever a file is to be allocated a free block, the operating system can simply allocate the first block in free space list and move the head pointer to the next free block in the list.

Disadvantages:

Searching the free space list will be very time consuming; each block will have to be read from the disk, which is read very slowly as compared to the main memory.

Not Efficient for faster access.

In our earlier example, we see that keep block 2 is the first free block which points to another block which contains the pointer of the 3 blocks and 3 blocks contain the pointer to the 4 blocks and this contains the pointer to the 5 block then 5 block contains the pointer to the next block and this process is repeated at the last.

c) Construct and explain directory structure of a file system in terms of single level, two level and tree structureA directory is a container that is used to contain folders and file. It organizes files and folders into a hierarchical manner.

There are several logical structures of a directory, these are given below.

1. Single-level directory –Single level directory is simplest directory structure.In it all files are contained in same directory which make it easy to support and understand.A single level directory has a significant limitation, however, when the number of files increases or when the system has more than one user. Since all the files are in the same directory, they must have the unique name . if two users call their dataset test, then the unique name rule violated.

Advantages: Since it is a single directory, so its implementation is very easy. If files are smaller in size, searching will faster. The operations like file creation, searching, deletion, updating are

very easy in such a directory structure.Disadvantages:

There may chance of name collision because two files can not have the same name.

Searching will become time taking if directory will large. In this can not group the same type of files together.

2. Two-level directory –As we have seen, a single level directory often leads to confusion of files names among different users. the solution to this problem is to create a separate directory for each user.In the two-level directory structure, each user has there own user files directory (UFD). The UFDs has similar structures, but each lists only the files of a single user. system’s master file directory (MFD) is searches whenever a new user id=s logged in. The MFD is indexed by username or account number, and each entry points to the UFD for that user.

Advantages: We can give full path like /User-name/directory-name/. Different users can have same directory as well as file name. Searching of files become more easy due to path name and user-

grouping.Disadvantages:

A user is not allowed to share files with other users. Still it not very scalable, two files of the same type cannot be

grouped together in the same user.3. Tree-structured directory –

Once we have seen a two-level directory as a tree of height 2, the natural generalization is to extend the directory structure to a tree of arbitrary height.This generalization allows the user to create there own subdirectories and to organize on their files accordingly.

A tree structure is the most common directory structure. The tree has a root directory, and every file in the system have a unique path.

Advantages:

Very generalize, since full path name can be given. Very scalable, the probability of name collision is less. Searching becomes very easy, we can use both absolute path as well

as relative.Disadvantages:

Every file does not fit into the hierarchical model, files may be saved into multiple directories.

We can not share files. It is inefficient, because accessing a file may go under multiple

directories.4. Acyclic graph directory –

An acyclic graph is a graph with no cycle and allows to share subdirectories and files. The same file or subdirectories may be in two different directories. It is a natural generalization of the tree-structured directory.It is used in the situation like when two programmers are working on a joint project and they need to access files. The associated files are stored in a subdirectory, separated them from other projects and files of other programmers since they are working on a joint project so they want to the subdirectories into there own directories. The common subdirectories should be shared. So here we use Acyclic directories.

It is the point to note that shared file is not the same as copy file if any programmer makes some changes in the subdirectory it will reflect in both subdirectories.

Advantages: We can share files. Searching is easy due to different-different paths.

Disadvantages: We share the files via linking, in case of deleting it may create the

problem, If the link is soft link then after deleting the file we left with a

dangling pointer. In case of hard link, to delete a file we have to delete all the

reference associated with it.

5. General graph directory structure –In general graph directory structure, cycles are allowed within a directory structure where multiple directories can be derived from more than one parent directory.The main problem with this kind of directory structure is to calculate

total size or space that has been taken by the files and directories.

Advantages: It allows cycles. It is more flexible than other directories structure.

Disadvantages: It is more costly than others. It needs garbage collection.