MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)

SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the
model answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may
try to assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more
Importance (Not applicable for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in
the figure. The figures drawn by candidate and model answer may vary. The examiner
may give credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed
constant values may vary and there may be some difference in the candidate’s answers
and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of
relevant answer based on candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on
equivalent concept.
8) As per the policy decision of Maharashtra State Government, teaching in English/Marathi
and Bilingual (English + Marathi) medium is introduced at first year of AICTE diploma
Programme from academic year 2021-2022. Hence if the students in first year (first and
second semesters) write answers in Marathi or bilingual language (English +Marathi), the
Examiner shall consider the same and assess the answer based on matching of concepts
with model answer.

Q. Sub Answer Marking

No Q.N. Scheme
1. Attempt any FIVE of the following: 10
a) List different types of operating system 2M
Ans.  Batch Operating System
 Multi-ProgrammingOperating System Any four
types 1/2M
 Time Shared Operating System each
 Multiprocessor System
 Distributed System
 Real Time Operating System
 Mobile Operating System
b) State any four services provided by an operating system 2M
Ans.  User Interface Any four
services
 Program Execution 1/2M each
 I/O Operation

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

 File System Manipulation

 Communication
 Error Detection
 Resource Allocation
 Accounting
 Protection and Security
c) Draw process state diagram 2M
Ans.
Correct
diagram 2M

d) State two features of non-preemptive scheduling 2M

Ans.  Once CPU is allocated to the process, the process keeps the CPU
until it releases the CPU either by terminating or by switching to Any two
the waiting state. relevant
features 1M
 A process switches from running state to waiting state or running to each
terminated state.
 Low scheduling overhead.
 Simple to implement and manage.
 Less responsive in terms of priority for urgent task.
e) Define following terms: 2M
i. Memory compaction
ii. Fragmentation Correct
Ans. i. Memory compaction: definition
1M each
The process of shuffling the memory contents so as to place all free
memory together in one large block is known as memory compaction.

ii. Fragmentation:
When processes are loaded and removed from memory, the free
memory space is broken into little pieces which is known as
fragmentation.

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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

f) Write syntax of PWD command and explain its use with the help 2M
of suitable example.
Ans. Syntax: $ pwd
PWD command is used to print (display) current working directory. Syntax 1M

Use &
Example: $ pwd example 1M
Above example displays absolute pathname.
Output: /home/poly
g) List any four file operations. 2M
Ans.  Creating a file
 Writing a file Any four
operations
 Reading a file 1/2M each
 Repositioning within a file
 Deleting a file
 Appending new information to the end of the file
 Renaming an existing file.
 Truncating a file
 Creating copy of a file, copy file to another I/O device such as
printer or display
2. Attempt any THREE of the following: 12
a) Explain Resource management of an operating system. 4M
Ans.  Resource Management in Operating System is the process to
manage all the resources efficiently like CPU, memory, Any
relevant
input/output devices, and other hardware resources among the explanation
various programs and processes running in the computer. 4M
 Computer's resources are limited and several users or programs
may need to utilize the same resources—such as memory and
CPU—at the same time.
 Operating system has to manage and ensure that all processes get
the resources they need to execute, without any problems like
deadlocks.
 Operating systems use various scheduling algorithms to ensure
that each process gets its fair share of CPU time.
 Operating systems use virtual memory techniques to manage
memory efficiently.
 Operating systems use file system management techniques to
manage disk space efficiently.
 File system management allows the operating system to create,

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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

delete, and modify files and directories in Secondary Storage

Device.
 Operating systems use various network management techniques to
manage network bandwidth efficiently.
b) Explain different components of operating system. 4M
Ans. List of System Components:
1. Process Management Any two
2. Main Memory Management components
with
3. File Management relevant
4. I/O System Management explanation
5. Secondary Storage Management 2M each

1.Process Management:
A program is a set of instructions. When CPU is allocated to a
program, it can start its execution. A program in execution is a
process. A word processing program run by a user on a PC is a
process. A process needs various system resources including CPU
time, memory, files and I/O devices to complete the job execution.
These resources can be given to the process when it is created or
allocated to it while it is running.
The operating system responsible for the following activities in
connection with process management:
 Creation and deletion of user and system processes.
 Suspension and resumption of processes.
 A mechanism for process synchronization.
 A mechanism for process communication.
 A mechanism for deadlock handling.

2. Main-Memory Management
Main memory is a large array of words or bytes, ranging in size from
hundreds of thousands to billions. Each word or byte has its own
address. Main memory is a repository of quickly accessible data
shared by the CPU and I/O devices. The central processor reads
instructions from main memory during the instruction fetch cycle and
both reads and writes data from main memory during the data fetch
cycle. The main memory is generally the only large storage device
that the CPU is able to address and access directly.
The operating system responsible for the following activities in
connection with main memory s management:

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

 Keeping track of which parts of memory are currently being used

and by whom.
 Deciding which processes (or parts thereof) and data to move into
and out of memory.
 Allocating and deallocating memory space as needed.

3. File Management
A file is a collected of related information defined by its creator.
Computer can store files on the disk (secondary storage), which
provide long term storage. Some examples of storage media are
magnetic tape, magnetic disk and optical disk. Each of these media
has its own properties like speed, capacity, and data transfer rate and
access methods. A file system normally organized into directories to
ease their use. These directories may contain files and other
directions.
The operating system responsible for the following activities in
connection with file management:
 The creation and deletion of files.
 The creation and deletion of directions.
 The support of primitives for manipulating files and directions.
 The mapping of files onto secondary storage.
 The backup of files on stable storage media.

4. I/O device Management

Input / Output device management provides an environment for the
better interaction between system and the I/O devices (such as
printers, scanners, tape drives etc.). To interact with I/O devices in an
effective manner, the operating system uses some special programs
known as device driver. The device drivers take the data that
operating system has defined as a file and then translate them into
streams of bits or a series of laser pulses (in regard with laser printer).
The I/O subsystem consists of several components:
 A memory management component that includes buffering,
caching, spooling
 A general device driver interface
 Drivers for specific hardware devices

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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

5. Secondary-Storage Management
The computer system provides secondary storage to back up main
memory. Secondary storage is required because main memory is too
small to accommodate all data and programs, and the data that it
holds is lost when power is lost. Most of the programs including
compilers, assemblers, word processors, editors, and formatters are
stored on a disk until loaded into memory. Secondary storage consists
of tapes drives, disk drives, and other media.
The operating system is responsible for the following activities in
connection with disk management:
 Free space management
 Storage allocation
 Disk scheduling.
c) Describe message passing system of interprocess communication 4M
(IPC)
Ans. Message Passing: In this model, communication takes place by Relevant
explanation
exchanging messages between cooperating processes. It allows
3M
processes to communicate and synchronize their action without
sharing the same address space. It is particularly useful in a Correct
distributed environment when communication process may reside on diagram 1M
a different computer connected by a network. Communication
requires sending and receiving messages through the kernel.
The processes that want to communicate with each other must have a
communication link between them. Between each pair of processes
exactly one communication link exists.

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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

d) What is CPU Scheduler? Explain the preemptive and non- 4M

preemptive type of scheduling
Ans. CPU Scheduler: When the CPU becomes idle, the CPU scheduler selects
CPU
one of the processes in the ready queue for execution. The CPU scheduler
scheduler
selects a process from the processes in memory that are ready to execute 2M
and allocates the CPU to that process.
Relevant
Pre-emptive scheduling: - Once the CPU is allocated to a process, the CPU explanation
is forcefully removed from that process due to occurrence of interrupt or of
completion of I/O request by the requesting process. In this scheduling, a Preemptive
process changes its status from running state to ready state or waiting state 1M
to ready state. &
non-
preemptive
Non-Preemptive scheduling: - Once the CPU is allocated to a process, the
1M
process keeps the CPU until it releases the CPU by terminating or by
switching to the waiting state. Control of CPU is with the process to which
it is allocated.In this scheduling, a process changes its status from running
to waiting or running to terminate.

3. Attempt any THREE of the following: 12

a) Define Process. Draw a Process Control Block and explain the 4M
information in PCB
Ans. Process: A process is defined as, a program under execution, which
competes for the CPU time and other resources. Define
A process is a program in execution. Process is also called as job, task process
1M
and unit of work.
Diagram of
PCB with
correct
naming /
block
1M

Any 4
blocks of
information
in PCB
2M

Fig: Process Control Block

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Following are the blocks in PCB:

1. Process Number: Each process is identified by its process
number, called Process Identification Number (PID). Every process
has a unique process-id through which it is identified. The process-id
is provided by the OS. The process id of two processes could not be
same because process-id is always unique.

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

corresponds to the relative importance of the event that it services
process priority is the preference of the one process over other
process for execution. Priority may be given by the user/system
manager or it may be given internally by OS. This field stores the
priority of a particular process.

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

process. The state may be new, ready, running, and waiting, halted,
and so on.

4. Program Counter: The counter indicates the address of the next

instruction to be executed for this process.

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

on the computer architecture. They include accumulators, index
registers, stack pointers, and general-purpose registers, plus any
condition-code information. Along with the program counter, this
state information must be saved when an interrupt occurs, to allow
the process to be continued correctly afterward.

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 the base and limit registers,
the page tables, or the segment tables, depending on the memory
system used by the operating system.

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

8. Accounting Information: This information includes the amount of

CPU and real time used, time limits, account numbers, job or process
numbers, and so on.

9. I/O Status Information: This information includes the list of I/O

devices allocated to the process, a list of open files, and so on.

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

access rights etc.

11. Pointer: Pointer points to another process control block. Pointer

is used for maintaining the scheduling list.
b) Define deadlock. State the conditions necessary for deadlock. 4M
Ans. Deadlock: Deadlock is a situation when two or more processes get
locked and cannot processed further because of inter-dependability. Define
deadlock
In real world, deadlocks can arise when two persons wait for phone 1M
calls from one another.
OR State any 3
Deadlock is defined as, "a situation where a set of processes are conditions
blocked because each process is holding a resource and waiting for necessary
for deadlock
another resource acquired by some other process".
3M
Necessary Conditions to Deadlock:
Mutual Exclusion: At least one resource is held in a non-sharable
mode, that is only one process at a time can use the resource. If
another process requests that resource, the requesting process must be
delayed until the resource has been released. Each resource is either
currently assigned to exactly one process or is available.

Hold and Wait: There must exist a process that is holding at least
one resource and is waiting to acquire additional resources that are
currently being held by another process. Process currently holding
resources granted earlier can request new resources.

No Pre-emption: Resources cannot be pre-empted; i.e. resource can

only be released voluntarily by the process holding it, after the
process has completed its task. Resources previously granted cannot
be forcibly taken away from a process. They must be explicitly
released by the process holding them.

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Circular Wait: There exist a set (P0, P1, ----- Pn) of waiting
processes such that P0 is waiting for a resource which is held by P1,
P1 is waiting for a resource which is held by P2. Pn-1 is waiting for
resources which are held by Pn and Pn is waiting for a resource which
is held by P0. Thus there must be a circular chain of two or more
processes, each of which is waiting for a resource held by the next
member of the chain.

c) Explain the following terms with respect to memory 4M

management:
i) Dynamic relocation Each term
ii) Swapping explanation
2M
Ans. i) Dynamic Relocation
When a program gets swapped out to a disk memory, then it is not
always possible that when it is swapped back into main memory then
it occupies the previous memory location, since the location may still
be occupied by another process. We may need to relocate the process
to a different area of memory. Thus there is a possibility that program
may be moved in main memory due to swapping.

ii) Swapping
Swapping is mechanism in which a process can be swapped
temporarily out of main memory (or move) to secondary storage
(disk) and make that memory available to other processes. At some
later time, the system swaps back the process from the secondary
storage to main memory.

Fig: swapping in memory management

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

d) With suitable diagram, explain how contiguous file allocation is 4M

performed?
Ans. • The contiguous allocation method requires each file to occupy a Any
set of contiguous addresses on the disk. Disk addresses define a relevant
Diagram
linear ordering on the disk. Contiguous allocation of a file is 1M
defined by the disk address of the first block and its length. If the
file is ‘n’ blocks long and starts at location ‘b’, then it occupies Explanation
blocks b, b+1, b+2, - - - - - b+n-1. The directory entry for each 3M
file indicates the address of the starting block and the length of
the area allocated for this file.
• Contiguous allocation supports both sequential and direct access.
• For direct access to block ‘i’ of a file, which starts at block ‘b’,
we can immediately access block b+i. The difficulty with
contiguous allocation is finding space for a new file.
• For direct access to block ‘i’ of a file, which starts at block ‘b’,
we can immediately access block b+i.
• The difficulty with contiguous allocation is finding space for a
new file.
• If file to be created are ‘n’ blocks long, we must search free space
list for ‘n’ free contiguous blocks.

4. Attempt any THREE of the following: 12

a) Compare between Time sharing operating system and 4M
multiprogramming operative system.

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Ans. Time sharing operating Multiprogramming Any 4

system operating system correct
points 1M
In this time sharing Operating Multiprogramming operating each
system many users/processes system allows to execute
are allocated with computer multiple processes by
resources in respective time monitoring their process states
slots. and switching in between
processes.
Processors time is shared with Processor and memory
multiple users that’s why it is underutilization problem is
called as time sharing operating resolved and multiple
system. programs runs on CPU that’s
why it is called
multiprogramming.
In this process, two or more In this, the process can be
users can use a processor in executed by a single
their terminal. processor.
Time sharing OS has fixed time Multi-programming OS has no
slice. fixed time slice.
Time sharing system minimizes Multiprogramming system
response time. maximizes processor use.
Example: Windows NT. Example: Mac OS.

b) Explain any four types of system call. 4M

Ans. 1.Process and Job Control Explanation
 A running program needs to be able to halt its execution either of any 4
system calls
normally (end) or abnormally (abort). If the program discovers an 1M each
error in its input and wants to terminate abnormally, it may also
want to define an error level.
 A process or job executing one program may want to load and
execute another program. This allows the control card interpreter
to execute program as directly by the control cards of the user job.
If we create a new job or process, we should able to control its
execution.
 We may also want to terminate a job or process that we created
(terminate process). If we find that it is incorrect or no longer
needed we need waiting time to finish execution (wait time).
Another set of system calls are helpful in debugging a program.

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Example of Process related system call:

a. End, Abort.
b. Load, Execute.
c. Create process, Terminate process.
d. Ready process, Dispatch process.
e. Suspend process, Resume process.
f. Get process attributes, Set process attributes.
g. Wait for Time.
h. Wait event, Signal event.
i. Change priority of process.

2.File Management
 We can identify several common system calls dealing with files.
We first need to be able to create and delete files such a system
call requires the name of the file and perhaps some of its
attributes.
 Once, the file is created, we need to open it and use it. We may
also read, write and reposition. Finally we need to close the file,
indicating that we are no longer using it.
Example of file related system call:
a. Create file, Delete file
b. Open a file, Close a file.
c. Create directory.
d. Read, Write, Reposition.
e. Get file attributes, Set file attributes.
f. Create a link.
g. Change working directory.

3.Device Management
 Files can be thought of as abstract or virtual devices. Thus many
of the system calls for files are also needed for devices.
 If there are multiple users of the system however we must first
request the device to ensure that we have exclusive use of it. After
we are finished with the device, we must release it. These
functions are similar to the open/close system calls for files.
 Once, the device has been requested we can read, write and
reposition the device just as with files.

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

System call related to device management:

a. Request device, Release device.
b. Read, Write, Reposition.
c. Get device attributes, set device attributes.

4. Information Maintenance
 Many system calls exist simply for the purpose of transferring
information between the user program and the operating system.
For example most systems have a system call to return the current
time and date.
 Other system calls may return information about the system such
as the number of current users, the version number of the
operating system, the amount of free memory or disk space and so
on.
 In addition the operating system keeps information about all of its
jobs and processes and there are system calls to access this
information. Generally, there are also calls to reset it. (get process
attributes and set process attributes).
Information Maintenance related system call:
a. Get Time or Date, Set Time or Date.
b. Get system Data, Set system Data.
c. Get process, File or Device attributes.
d. Set process, File or Device attributes.
5. Communication
 There are two models of inter-process communication, the
message-passing model and the shared memory model.
 Message-passing uses a common mailbox to pass messages
between processes
 Shared memory use certain system calls to create and gain access
to regions of memory owned by other processes
Communication related system calls:
a. Establish a connection.
b. Send, receive messages.
c. Terminate a connection

c) Describe how context switch is executed by operating system 4M

Ans. • When CPU switches to another process, the system must save the Diagram
state of the old process and load the saved state for the new 1M
process. This task is known as a context switch.

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Subject: Operating System Subject Code: 22516

• CPU switching from one process to another process is called a

context switch. Explanation
3M
• Context switch times are highly dependent on hardware support.
Its speed varies from machine to machine, depending on the
memory speed, the number of registers that must be copied and
the existence of special instructions.
• This enables multiple processes to share a single CPU. The
context switch is an essential feature of a multitasking operating
system.
• When the process is switched, the following information is stored:
• Program Counter
• Scheduling Information
• Base and limit register value
• Currently used register
• Changed State
• I/O State
• Accounting

Fig: Context Switching

d) Compare Short Job First (SJF) and Shortest Remaining Time 4M
(SRTN) scheduling algorithm (any four points)
Ans. Any 4
Short Job First (SJF) Shortest Remaining Time correct
(SRTN) points 1M
It is a non-preemptive It is a preemptive algorithm. each
algorithm.
It involves less overhead than It involves more overheads
SRJF. than SJF.

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MODEL ANSWER
Subject: Operating System Subject Code: 22516

It leads to comparatively lower It leads to increased

throughput. throughput as execution time
is less.
It minimizes the average It may or may not minimize
waiting time for each process. the average waiting time for
each process.
It involves lesser number of It involves higher number of
context switching. context switching.
Short processes are executed Shorter processes run fast and
first and then followed by longer processes show poor
longer processes. response time.

e) Describe variable partitioning with the help of suitable example 4M

Ans. • In variable memory partitioning the partitions can vary in number
and size. In variable memory partitioning the amount of memory Explanation
allocated is exactly the amount of memory a process requires. 2M
• The operating system keeps a table indicating which parts of Any
memory are available and which are occupied. Initially all relevant
memory is available for user programs and is considered as one Example
large block of available memory, a hole. 2M
• When a job arrives and needs memory, we search for a hole large
enough for this job. If we find one, we allocate only as much as is
needed, keeping the rest available to satisfy future requests.

For example, assume 256K memory available and a resident monitor

of 40K. This situation leaves 216K for user programs.

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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

5. Attempt any TWO of the following: 12

a) Explain the use of following OS tools: 6M
i) Device Manager
ii) Task Scheduler
Ans. i) Device Manager use:
1. Managing all the hardware or virtual devices of computer system.
Any 2 uses
2. Allow interaction with hardware devices through device driver. of Device
3. Used to install device and component-level drivers as well as Manager
associated software. 1½ M each
4. Allocate devices to the process as per process requirement and
priority.
5. Deallocate devices either temporarily or permanently depending
on condition. Any 2 uses
6. Keeping track of all device’s data and location. of Task
7. Monitoring device status like printers, storage drivers and other Scheduler
devices. 1½ M each
8. Used to enforce the predetermined policies and decides which
process receives the device when and for how long.

ii) Task scheduler use:

1. Used to automate the execution of programs, scripts, and various
tasks at specific intervals or specific events.
2. Used to select the next jobs to be admitted into the system and the
next process to run.
3. Schedules tasks on CPU. Which task to run first, which has high
priority, which has low latency is decided by schedulers.

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MODEL ANSWER
Subject: Operating System Subject Code: 22516

4. Schedulers aim to provide fair access to system resources among

competing processes or users.
5. Task schedulers strive to maximize the overall throughput of the
system by executing tasks in a manner that minimizes idle time
and maximizes the number of tasks completed within a given
timeframe.
b) Explain user level thread and Kernel level thread with its 6M
advantages and disadvantages
Ans. User Level Thread For each
Thread
 In a user thread, all of the work of thread management is done by
explanation
the application and the kernel is not aware of the existence of 1M
threads.
 The thread library contains code for creating and destroying
threads, for passing message and data between threads, for 1 advantage
of each 1M
scheduling thread execution and for saving and restoring thread
contexts.
 The application begins with a single thread and begins running in
that thread. 1 Dis-
 User level threads are generally fast to create and manage. advantage
of each
1M
Advantages of user level thread over Kernel level thread:
1. Thread switching does not require Kernel mode privileges.
2. User level thread can run on any operating system. (Diagram
3. Scheduling can be application specific. optional)
4. User level threads are fast to create and manage.

Disadvantages of user level thread:

1. Limited Parallelism: Since user-level threads are managed within
the context of a single process, they are not well-suited for
applications that need to take full advantage of multi-core
processors or for applications that require parallel execution
across different processes.
2. Blocking: If a user-level thread blocks due to I/O or other reasons,
all threads in the same process can be blocked, leading to poor
utilization of resources.
3. No True Parallelism: User-level threads cannot execute in true
parallel on multi-core systems since they are managed by a single
kernel-level thread

Page 18 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

4. It is not appropriate for a multiprocessor system.

Kernel Level Threads

 In Kernel level thread, thread management done by the Kernel.
 There is no thread management code in the application area.
 Kernel threads are supported directly by the operating system.
 Any application can be programmed to be multithreaded.
 All of the threads within an application are supported within a
single process.
 The Kernel maintains context information for the process as a
whole and for individual threads within the process.
 Scheduling by the Kernel is done on a thread basis.
 The Kernel performs thread creation, scheduling and management
in Kernel space.
 Kernel threads are generally slower to create and manage than the
user threads.

Advantages of Kernel level thread:

1. Kernel can simultaneously schedule multiple threads from the
same process on multiple process.
2. If one thread in a process is blocked, the Kernel can schedule
another thread of the same process.
3. Kernel routines themselves can multithreaded.

Disadvantages:
1. Kernel threads are generally slower to create and manage than the
user threads.
2. Transfer of control from one thread to another within same process
requires a mode switch to the Kernel.

Page 19 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

c) Consider the string: 6M

0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 7 with frame size 3 and 4,
calculate page fault in both the cases using FIFO algorithm
Ans. Note: Representation of frame can be in any order
FIFO : FIFO
implementat
i)Frame Size=3 ion with
frame size 3
3M

OR

Total Page Faults :16 FIFO

implementat
ii)Frame Size=4 ion with
frame size 4
3M

OR

Total Page Faults :08

6. Attempt any TWO of the following: 12
a) What is the average turnaround time for the following process 6M
using :
i) FCFS scheduling algorithm
ii) SJF non preemptive scheduling algorithm
iii) Round Robin Scheduling algorithm

Page 20 / 26
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Process Arrival time Burst time

P1 0 8
P2 1 4
P3 2 1
For each
Ans. Note :Any time slice can be considered by student for Round Robin algorithm:
e.g. 2, 3, 4 etc Gantt Chart
1M
i) Gantt Chart FCFS
Correct
turnaround
P1 P2 P3 time
calculation
0 8 12 13 1M each

Turnaround Time =Completion Time - Arrival Time

P1=8-0=8
P2=12-1=11
P3=13-2=11
Average turnaround time=Turnaround time of all processes /
Number of processes
= (8+11+11) /3
=10 ms / Units

ii) Gantt Chart SJF non-preemptive

P1 P3 P2

0 8 9 13

Turnaround Time =Completion Time - Arrival Time

P1=8-0=8
P2=13-1=12
P3=9-2=7
Average turnaround time=Turnaround time of all processes /
Number of processes
= (8+12+7) /3
= 9 ms/ Units

Page 21 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

iii) Gantt Chart Round Robin

Consider Time Quantum= 2 Units

P1 P2 P3 P1 P2 P1 P1

0 2 4 5 7 9 11 13

Turnaround Time =Completion Time - Arrival Time

P1=13-0=13
P2=9-1=8
P3=5-2=3
Average turnaround time=Turnaround time of all processes /
Number of processes
= (13+8+3) /3
= 8 ms / Units
(OR)

Consider Time Quantum= 3 Units

P1 P2 P3 P1 P2 P1

0 3 6 7 10 11 13

Turnaround Time =Completion Time - Arrival Time

P1=13-0=13
P2=11-1=10
P3=7-2=5
Average turnaround time=Turnaround time of all processes /
Number of processes
= (13+10+5) /3
= 9.33 ms/ Units
b) Explain bit map vector and linked list free space management 6M
techniques with its advantages and disadvantages
Ans. 1 )Bitmap Vector: Bitmap
Vector
The free-space list is implemented as a bit map or bit vector.
explanation
1M
Each block is represented by 1 bit. If the block is free, the bit is 1; if
the block is allocated, the bit is 0.

Page 22 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

For example, consider a disk where blocks 2, 3, 4, 5, 8, 9, 10, 11, 12,

13 are free and the rest of the blocks are allocated. Any 1
Advantage
The free-space bit map would be : 0011110011111100 1M,

Any 1
Disadvantag
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 e 1M

0 0 1 1 1 1 0 0 1 1 1 1 1 1 0 0

1=Free block 0= Allocated block Linked List

explanation
1M
The main advantage of this approach is its relative simplicity and its
efficiency in finding the first free block or n consecutive free blocks
on the disk. Any 1
Advantage
Advantages of Bit Map vector method 1M
1. Simple and easy to understand.
2. Consumes less memory. Any 1
3. It is efficient to find free space. Disadvantag
Disadvantages of the Bit Map vector method e 1M
1. The operating system goes through all the blocks until it finds
a free block.
(diagram is
2. It is not efficient when the disk size is large. optional)

2) Linked List
In this approach, the free disk blocks are linked together i.e. a free
block contains a pointer to the next free block. The block number of
the very first disk block is stored at a separate location on disk and is
also cache in memory. In this approach, link all the disk blocks
together, keeping a pointer to the first free block. This block contains
a pointer to the next free disk block, and so on.

Page 23 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2005 Certified)

SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Advantages of linked list:

1. Files can be extended, no fragmentation problems.
2. Sequential Access is easy: just chase links.
3. No extra space for the free list.
4. Starting and ending of the file can be easily found out by the
linked list.
Disadvantages of linked list:
1. Cannot be effectively used for direct access method.
2. Random access is virtually impossible. Lots of seeking, even
in sequential access.
3. Pointers which are required need additional memory space.
4. It is not reliable as the blocks are scattered on the disk
anywhere.
c) Explain with diagram single level directory structure and two 6M
level directory structure with its advantages and disadvantages.
Ans. Single Level Directory Structure: Single level
This is the simplest directory structure. All the files are stored in the directory
same directory which is easy to support and understand. But a unique structure
name must be assigned to each file. explanation
If the number of files in the directory increases, it may become 1M,
difficult to remember names of all files.
Advantage:
 Less time required to search the file, because all files are Any 1
stored in same directory. Advantage
 Single level directory structure is easy to implement and 1M
maintain.
 Single level directory structure, the operations like creation,
searching, deletion, updating are very easy and faster

Page 24 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Disadvantages
 Not suitable for a large number of files and more than one
Any 1
user. disadvantag
 Because of single directory files, files require unique file e 1M
names.
 Difficult to remember names of all the files as the number of Two level
files increases. MS-DOS OS allows only 11 character file directory
structure
name where as UNIX allows 255 character. explanation
1M

Any 1
Advantage
1M

Any 1
disadvantag
e 1M

Two-level directory structure: -In the two-level structures, each

user has its own user file directory (UFD). The UFD lists only files of
a single user. System contains a master file directory (MFD) which is
indexed by user name or account number. Each entry in MFD points
to the UFD for that user.

When a user refers to a particular file, only his own UFD is searched.
Different users can have files with the same name, as long as all the
file names within each UFD are unique.

When we create a file for a user, operating system searches only that
user’s UFD to find whether same name file already present in the
directory. For deleting a file again operating system checks the file
name in the user’ UFD only.

Page 25 / 26
 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2024 EXAMINATION

MODEL ANSWER
Subject: Operating System Subject Code: 22516

Advantages:
 It solves isolation. i.e. data of one user is separated from other
user.
 Searching is also efficient in this type of structure.
 Every user accesses his own directory and files.
 There is no problem of files with same names in different
directories.
Disadvantages:
 Two users can not share data or communicate with each other.
 It is difficult to search the files with same names in different
directories.

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