MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
SUMMER– 18 EXAMINATION
Subject Name: Operating System
Important Instructions to examiners:
Model Answer Subject Code: 17512
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.

Q. Sub Answers Marking

No. Q. N. Scheme

1. a) Attempt any THREE of the following: 12 Marks

(i) List and draw a neat labelled diagram of four components of a computer system. 4M

Ans: A computer system can be divided into four components: (List:

1) The hardware. 2marks,
2) The operating system Correct
3) Application programs diagram:
4) The users 2 marks)

Diagram:

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(ii) List three main levels of data storage and explain cache storage. 4M

Ans: {{**Note: - Any other relevant explanation shall be considered. **}} (List:1
mark,
Three levels of data storage: Relevant
1. Primary Storage explanatio
2. Secondary Storage n of cache
3. Tertiary Storage storage: 3
marks)
Cache Storage:

A Cache (Pronounced as “cash”) is a small and very fast temporary storage memory. It is
designed to speed up the transfer of data and instructions. It is located inside or close to the
CPU chip. It is faster than RAM and the data/instructions that are most recently or most
frequently used by CPU are stored in cache. As CPU has to fetch instruction from main
memory speed of CPU depending on fetching speed from main memory. CPU contains
register which has fastest access but they are limited in number as well as costly. Cache is
cheaper so we can access cache. Cache memory is a very high speed memory that is placed
between the CPU and main memory, to operate at the speed of the CPU.

It is used to reduce the average time to access data from the main memory. The cache is a
smaller and faster memory which stores copies of the data from frequently used main
memory locations. Most CPUs have different independent caches, including instruction and
data.

(iii) List and draw a neat labelled diagram of process state. 4M

Ans: List: (List:2

Each process may be in one of the following states. marks,
 New State: The process is being created. Diagram: 2
 Ready State: The process is waiting to be assigned the processor. marks)
 Running State: Instructions from the process are executing.
 Waiting or Blocked: The process is waiting for some event to occur.
 Terminated State: The process has finished execution.

Process state diagram:

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(iv) List merits of I/O scheduling (Four points) and Demerits of I/O scheduling. 4M

Ans: {{**Any other relevant Merits and Demerits shall be considered*}} (Merits: 2
marks,
Merits of I/O scheduling: Demerits: 2
marks)
 It improves overall performance of the system.
 It can share device access fairly among processes.
 It helps in reducing the average waiting time for I/O to complete.
 It increases throughput of the system.
 It helps to prioritize process‟s I/O requests.

Demerits of I/O scheduling:

 There is a risk of starvation for longer processes.

 may lead to poor overlap of I/O and CPU since CPU-bound processes will force I/O
bound processes to wait for the CPU, leaving the I/O devices idle
 Difficult to know the length of the next CPU request
 An I/O bound process on a heavily loaded system will run slower
 May involve a large context switch overhead
b) Attempt any ONE of the following: 6 Marks

(i) Explain the working of Inter-process communication considering. 6M

1) Shared memory
2) Message passing
Ans: 1)Shared memory: In this model, a region of the memory residing in an address space of a (Diagram:
process creating a shared memory segment can be accessed by all processes who want to 1mark
communicate with other processes. All the processes using the shared memory segment each,
should attach to the address space of the shared memory. All the processes can exchange Explanatio
information by reading and/or writing data in shared memory segment. The form of data n :2 marks
and location are determined by these processes who want to communicate with each other. each)
These processes are not under the control of the operating system. The processes are also
responsible for ensuring that they are not writing to the same location simultaneously. After
establishing shared memory segment, all accesses to the shared memory segment are treated
as routine memory access and without assistance of kernel.

(a) Shared Memory

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2)Message Passing: In this model, communication takes place by exchanging messages
between cooperating processes. It allows processes to communicate and synchronize their
action without sharing the same address space. It is particularly useful in a distributed
environment when communication process may reside on 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 exist.

(b)Message Passing

(ii) List four Deadlock prevention condition and explain the following terms. 6M
1) Removal of “No preemption” condition.
2) Elimination of “Circular wait” related to deadlock prevention condition.
Ans: Deadlock prevention conditions:- (List: 2
1. Preventing Mutual exclusion condition marks,
2. Preventing Hold and wait condition Explanatio
3. Preventing No preemption condition n of each
4. Preventing Circular wait condition term:2
marks)
1) Removal of “No Preemption” Condition
This necessary condition specifies that there is no pre-emption of resources that have
already been allocated. To ensure that this condition does not hold, we can use the
following protocol. If a process is holding some resources and requests another resource
that cannot be immediately allocated to it (that is, the process must wait), then all resources
the process is currently holding are preempted. In other words, these resources are
implicitly released. The pre-empted resources are added to the list of resources for which
the process is waiting. The process will only be restarted when it can regain its old
resources, as well as the new ones that it is requesting.
For example: If a process requests some resources, we first check if they are available. If so
we allocate them. If they are not available, we check whether they are allocated to some
other process that is waiting for additional resources. If so, we pre-empt the desired
resources from the waiting or held by a waiting process, the requesting process must wait.
While it is waiting, some of its resources may be pre-empted, but only if another process
requests them. A process can only be restarted when it is allocated the new resources it is
requesting and recovers any resources that we pre-empted while it was waiting.

2) Elimination of “Circular wait” related to deadlock prevention condition

If a circular wait condition is prevented, the problem of the deadlock can be prevented too.

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Consider all resources are numbered as shown in figure:

Number Resource Name
0 Tape Drive
1 Printer
2 Plotter
3 Card Reader
4 Card Punch

Any process has to request for all the required resources in a numerically ascending order
during its execution. This would prevent a deadlock. Let us assume that two processes P1
and P2 are holding a tape drive and a plotter respectively. A deadlock can take place only if
P1 holds the tape drive and wants the plotter, whereas P2 holds the plotter and requests for
the tape drive, i.e. if the order in which the resources are requested by the two processes is
exactly apposite. And this contradicts our assumption. Because 0<2, a tape drive has to be
requested for before a plotter, by each process, whether it is P1 or P2.

Each process can request resources only in an increasing order of enumeration. That is, a
process can initially request any number of instances of a resource type -say, R;. After that,
the process can request instances of resource type Rj if and only if F(Rj) > F(R;). We can
demonstrate this fact by assuming that a circular wait exists. Let the set of processes
involved in the circular wait be { P0 , P1, ... , P11}, where Pi is waiting for a resource R;,
which is held by process Pi+l· (Modulo arithmetic is used on the indexes, so that P11 is
waiting for a resource R11 held by P0 .) Then, since process Pi+l is holding resource Ri
while requesting resource Ri+l' we must have F(Ri) < F(R;H) for all i. But this condition
means that F(Ro) < F(R1) < ... < F(R11) < F (Ro). By transitivity, F(Ro) < F(Ro), which is
impossible. Therefore, there can be no circular wait.

2. Attempt any FOUR of the following: 16Marks

a) Define clustered systems? List four characteristics of clustered systems. 4M

Ans: A clustered system is a collection of connected computers working together as one unit. In (Defination
this system, any member of the cluster is capable of supporting the processing functions of :2 marks,
any other member. Each computer in a cluster is referred to as a node. any four
relevant
Characteristics:- characteris
 Clustering allows two or more system to share storage closely linked via a local area tics:2
network. marks)
 A cluster has a redundant n+k configuration, where n processing nodes are actively
processing the application and k processing nodes are in a standby state, serving as
peers. In the event of a failure of an active node, the application that was running on
the failed node is moved to one of the standby nodes. The simplest redundant
configuration is active/standby, in which one node is actively processing the
application and the other node is in a standby state.
 Other common cluster configurations include simplex (one active node, no spare),
n+1 active nodes (n active nodes, 1 spare), and n active nodes. In a configuration
with n active nodes, the applications from the failed node are redistributed among
the other active nodes using a pre-specified algorithm.
 It provides excellent fault tolerance.
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b) Explain following two services of operating systems. 4M

(i) File system manipulation
(ii) Resource Allocation
Ans:  File system manipulation: While working on the computer, generally a user performs (Explanati
various types of operations on files like creating a file, opening a file, saving a file, on of each
deleting a file, search for a file with file name from the storage disk, etc. Programs need :2 marks)
to read a file or write a files and directories. The operating system gives the permission
to program for performing operations on file. Maintain details of files or directories with
their respective details. Some programs include permissions management that allow or
deny access to files or directories based on file ownership.

 Resource allocation: When multiple users or multiple jobs are running at the same
time, it is the responsibility of an operating system to allocate the required resources to
each process. Operating system manages many different types of resources such as
CPU, main memory, tape drive or secondary storage etc. For this purpose, various types
of algorithms are implemented such as process scheduling, CPU scheduling, disk
scheduling etc. There are some routines that allocate printers, modems, USB storage
drives and other peripheral devices.

c) Define synchronization 4M
Explain
(i) Blocking
(ii) Non Blocking in message passing
Ans: Process Synchronization means sharing system resources by processes in such a way that, (Any
concurrent access to shared data is handled thereby minimizing the chance of inconsistent relevant
data. Maintaining data consistency demands mechanisms to ensure synchronized execution Definition
of cooperating processes. Process Synchronization was introduced to handle problems that of
arose while multiple process executions. synchroniz
ation:
Message passing may be blocking or non-blocking, also known as synchronous and 2marks,
asynchronous. Each term
Explanatio
 Blocking send: The sending process is blocked until the message is received by the n:1mark)
receiving process or by the mailbox.
 Blocking receives: The receiver blocks until a message is available.

 Non-blocking send: The sending process sends the message and resumes operation.
 Non-blocking receive: The receiver retrieves either a valid message or a null.

d) List four process scheduling criteria and explain the term Turnaround in detail. 4M

Ans: Process scheduling criteria: (List:2

 CPU Utilization marks,
 Throughput Explanatio
 Turn-Around Time n of
 Waiting Time turnaround
 Response Time :2 marks)

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Turn-Around Time: The time interval from the time of submission of a process to the time
of completion of that process is called as turnaround time. It is the sum of time period spent
waiting to get into the memory, waiting in the ready queue, executing with the CPU, and
doing I/O operations. It indicates the time period for which a process exists in the system.

e) Explain Deadlock Avoidance with example. 4M

Ans: Most prevention algorithms have poor resource utilization, and hence result in reduced (Explanatio
throughputs. Instead, we can try to avoid deadlocks by making use prior knowledge about n of any
the usage of resources by processes including resources available, resources allocated, one
future requests and future releases by processes. Most deadlock avoidance algorithms need method: 2
every process to tell in advance the maximum number of resources of each type that it may marks,
need. Based on all this info we may decide if a process should wait for a resource or not and Example: 2
thus avoid chances for circular wait. marks)

Deadlock can be avoided by following algorithms:

Safe State: If a system is already in a safe state, we can try to stay away from an unsafe
state and avoid deadlock. Deadlocks cannot be avoided in an unsafe state. A system can be
considered to be in safe state if it is not in a state of deadlock and can allocate resources up
to the maximum available. A safe sequence of processes and allocation of resources ensures
a safe state. Deadlock avoidance algorithms try not to allocate resources to a process if it
will make the system in an unsafe state. Since resource allocation is not done right away in
some cases, deadlock avoidance algorithms also suffer from low resource utilization
problem.

Resource Allocation Graph: A resource allocation graph is generally used to avoid

deadlocks. If there are no cycles in the resource allocation graph, then there are no
deadlocks. If there are cycles, there may be a deadlock. If there is only one instance of
every resource, then a cycle implies a deadlock. Vertices of the resource allocation graph
are resources and processes. The resource allocation graph has request edges and
assignment edges. An edge from a process to resource is a request edge and an edge from a
resource to process is an allocation edge. A calm edge denotes that a request may be made
in future and is represented as a dashed line. Based on calm edges we can see if there is a
chance for a cycle and then grant requests if the system will again be in a safe state.
Example:

If R2 is allocated to p2 and if P1 request

for R2, there will be a deadlock.

Bankers Algorithm: The resource allocation graph is not much useful if there are multiple
instances for a resource. In such a case, we can use Banker‟s algorithm. In this algorithm,
every process must tell upfront the maximum resource of each type it need, subject to the
maximum available instances for each type. Allocation of resources is made only, if the
allocation ensures a safe state; else the processes need to wait. The Banker‟s algorithm can
be divided into two parts: Safety algorithm if a system is in a safe state or not. The resource
request algorithm make an assumption of allocation and see if the system will be in a safe
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state. If the new state is unsafe, the resources are not allocated and the data structures are
restored to their previous state; in this case the processes must wait for the resource.
Example:

5 processes P0 through P4;

3 resource types:
A (10 instances), B (5instances), and C (7 instances)

Snapshot at time T0:

Allocation Max Available

ABC ABC ABC

P0 010 753 332
P1 200 322
P2 302 902
P3 211 222
P4 002 433

ABC
P0 743
P1 122
P2 600
P3 011
P4 431

The system is in a safe state since the sequence < P1, P3, P4, P2, P0> satisfies safety
criteria
f) Explain “Bitmap” method in free space management technique. 4M

Ans: The free space management techniques Bitmap is also referred as Bit vector. (Relevant
Explanatio
Bit vector: n of bitmap
The free space list is implemented as a Bit Map or Bit Vector. Each block is represented by method:4
one bit. If the block is free, the bit is „1‟; if the block is allocated the bit is „0‟. marks)
Example considers a disk where blocks 2,3,4,5,8,9,10,11,12,13,17,18,25,26 and 27 are free
the remaining blocks are allocated then the free space bit map would be:
001111001111110001100000011…

The main advantage of this approach is that it is relatively simple and efficient to find the
first free blocks or n consecutive free blocks on the disk.
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3. Attempt any FOUR of the following: 16Marks

a) Explain Time sharing OS in detail. 4M

Ans: In time sharing system, the CPU executes multiple jobs by switching among them. The (Relevant
switches occur so frequently that the users can interact with each program while it is Explanatio
running. It includes an interactive computer system which provides direct communication n: 4 marks)
between the user and the system.
A time sharing system allows many users to share the computer resources
simultaneously. The time sharing system provides the direct access to a large number of
users where CPU time is divided among all the users on scheduled basis. The operating
system allocates a time slice to each user. When this time is expired, it passes control to the
next user on the system. The time allowed is extremely small and the users are given the
impression that each of them has their own CPU and they are the sole owner of the CPU. In
this time slice each user gets attention of the CPU. The objective of time sharing system is
to minimize response time of process.

Example:-
The concept of time sharing system is shown in figure:

In above figure, the user 5 is active but user 1, user 2, user 3, and user 4 are in waiting state
whereas user 6 is in ready status.

b) List types of system call and explain the system call – “Information Maintenance”. 4M

Ans: Types of System calls are as follows: (List:1

1. Process or Job control mark,
2. File Management Explanatio
3. Device Management
n of
4. Information Maintenance
5. Communications informatio
n
Information Maintenance: maintenanc
The operating system keeps information about all its processes and provides system calls to e: 3 marks)
access this information. Some system calls exist purely for transferring information between
the user program and the operating system. Transferring information between the user
program and the operating system requires system call. System information includes
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displaying current date and time, the number of current user, the version number of the
operating system, the amount of free memory or disk space and so on.

System calls Related to Information Maintenance:

 Get Time or Date, Set Time or Date
 Get System data, Set system data
 Get process, file or device attributes
 Set process, file or Device attributes.
c) Differentiate between long term scheduling and medium term scheduling. 4M

Ans: Sr. No. Long Term scheduling Medium term scheduling (Any four
points: 1
mark each)
1 In this scheduling, long term scheduler In this scheduling, medium term
selects a process from job pool and scheduler, selects a process from
loads into the memory. swapped queue and loads into the
memory.

2 It works with job pool and memory. It works with swapped process queue
and memory

3 It is a job scheduler. It is a swapped process scheduler.

4 Processes switch its state from new to Processes switch its state from
ready. A process selected from job blocked to ready. A process selected
pool enters to the memory for the first from swapped queue re-enters into the
time. memory.

5 It controls the degree of It does not the degree of

multiprogramming multiprogramming.

d) The Jobs are scheduled for execution as follows- solve the problem by using 4M
preemptive SJF (Shortest Job First). Find average waiting time using Gantt chart.

Process Arrival Time Burst Time

P1 0 10

P2 1 04

P3 2 14

P4 3 08

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Ans: ( Gantt
P1 P2 P4 P1 P3 chart:2
marks,
0 1 5 13 22 36 Average
Waiting
Waiting Time Time: 2
P1=13-1=12 marks)
P2=1-1=0
P3=22-2=20
P4=5-3=2
Average waiting time= (12+0+20+2)/4=8.5 ms
e) Explain the working of Two-level directory structure with neat labelled diagram. 4M

Ans: The two level directory 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 (Explanatio
indexed by user name or account number. Each entry in MFD points to the UFD for that n: 2 marks,
user. any correct
Diagram :
2 marks)

OR

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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 same
name file already present in the directory. For deleting a file again operating system checks
the file name in the user‟ UFD only.

4. a) Attempt any THREE of the following: 12Marks

(i) List Advantages and Disadvantages of Batch Monitoring functions. (Four points) 4M

Ans: Advantages of batch systems:- (Any

● Grouping the jobs in batch reduce the time required for loading system setup for four
execution of each job. relevant
● No need of special hardware and system support to input data in batch systems. advantages
● Best for large organizations but small organizations can also benefit from it. and
● Batch systems can work offline so it makes less stress on processor. disadvanta
● Processor consumes good time while processing that mean it knows which job to ges: 2
process next. In real time systems we don‟t have expectation time of how long the job is marks
and what is estimated time to complete it. But in batch systems the processor knows each)
how long the job is as it is queued.
● Sharing of batch system for multiple users.
● The batch systems can manage large repeated work easily.
Disadvantages of batch processing systems
● Computer operators must be trained for using batch systems.
● It is difficult to debug batch systems.
● Batch systems are sometime costly.
● If some job takes too much time i.e. if error occurs in job then other jobs will wait for
unknown time.
(ii) Explain major activities of memory management component of an operating system. 4M

Ans: 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 (Relevant
accessible data shared by the CPU and I/O devices. The central processor reads instructions Explanatio
from main memory during the instruction fetch cycle and both reads and writes data from
n: 4 marks)
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.
For a program to be executed, it must be mapped to absolute addresses and loaded into
memory. As the program executes, it accesses program instructions and data from memory
by generating these absolute addresses.
To improve both the utilization of the CPU and the speed of the computer‟s response to its
users, operating system keeps several programs in memory. To handle memory with
multiple programs memory management is necessary.
Activities of memory management.
1) Keeping track of which part of memory are currently being used and by whom.
2) Deciding which processes and data to move into and out of memory.
3) Allocating & Deallocating space as needed.

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(iii) Define the following with respect to resources. 4M
1) A preemptable resource
2) A non-preemptable resource
Ans: 1. Preemptable resource: These are the resources that can be taken away from its (Define
currently allocated process (owner) and be allocated to another process. An example is preemptabl
memory space. e and non-
preemptabl
2. Non-preemptable resource: These are the resources that cannot be taken away from its e resource:
allocated process. An example is a non sharable printer. 2 marks
each)
Memory is an example of a Preemptable resource. Consider, for example, a system with 32
MB of user memory, one printer, and two 32-MB processes that each want to print
something. Process A request and gets the printer, then start to compute the values to print.
Before it has finished with the computation, it exceeds its time quantum and is swapped out.
Process B now runs and tires, unsuccessfully, to acquire the printer. Potentially, we now
have a deadlock situation, because A has the printer and B has the memory, and neither can
proceed without the resource held by the other. Fortunately, it is possible to preempt (take
away) the memory from B by swapping it out and swapping A in. Now A can run, do its
printing, and then release the printer. No deadlock occurs.
A Non-preemptable resource, in contrast is one that cannot be taken away from its current
owner without causing the computation to fail. If a process has begun to burn a CD-ROM,
suddenly taking the CD recorder away from it and giving it to another process will result in
a garbled CD. CD recorders are not Preemptable at an arbitrary moment.

(iv) List four types of UNIX files and draw Unix file system.

Ans: Types of Unix files: (List: 2

● Ordinary files marks, Any
● Directory files relevant
● Special or Device files
Diagram:
● Fifo (Pipe) files
2 marks)

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Unix file system:

b) Attempt any ONE of the following: 6 Marks

(i) Explain working of CPU switch from process to process with diagram. 6M

Ans: A CPU switch from process to process is referred as context switch. A context switch is a
mechanism that store and restore the state or context of a CPU in Process Control block so (Explanatio
that a process execution can be resumed from the same point at a later time. When the n: 4 marks,
scheduler switches the CPU from one process to another process, the context switch saves Diagram: 2
the contents of all process registers for the process being removed from the CPU, in its marks)
process control block.

Context switch includes two operations such as state save and state restore. State save
operation stores the current information of running process into its PCB. State restore
operation restores the information of process to be executed from its PCB. Switching the
CPU from one process to another process requires performing state save operation for the
currently executing process (blocked) and a state restore operation for the process ready for
execution. This task is known as context switch.

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(ii) Explain CPU and I/O burst cycle with the help of diagram. 6M

Ans: CPU burst cycle: - It is a time period when process is busy with CPU.
(Explanatio
I/O burst cycle: - It is a time period when process is busy in working with I/O resources.
n: 4 marks,
A process execution consists of a cycle of CPU execution and I/O wait. A process starts its Diagram:2
execution when CPU is assigned to it, so process execution begins with a CPU burst cycle. This is marks)
followed by an I/O burst cycle when a process is busy doing I/O operations.

A process switch frequently from CPU burst cycle to I/O burst cycle and vice versa. The complete
execution of a process starts with CPU burst cycle, followed by I/O burst cycle, then followed by
another CPU burst cycle, then followed by another I/O burst cycle and so on. The final CPU burst
cycle ends with a system request to terminate execution.

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5. Attempt any TWO of the following: 16Marks

a) Describe concept of file, its types and operations on file attributes in detail. 8M

Ans: Concept of file: ( Relevant

description
A file is a named collection of related information that is recorded on secondary storage
of file:2
such as magnetic disks, magnetic tapes and optical disks. The operating system abstracts
marks, any
from the physical properties of its storage devices to define a logical storage unit called as
6 types:3
file. Files are mapped by the operating system onto the physical devices. Files represent
marks, any
programs and data. Data files may be numeric, alphabetic, alphanumeric or binary. A file is
relevant 6
a sequence of bits, bytes, lines or records, the meaning of which is defined by its creator and
file
user. Many different types of information may be stored in a file: source programs, object
operation:
programs, executable programs, numeric data, text payroll records, graphic images, sound
3 marks)
recording and so on. A file has a certain defined structure according to its type.

Different types of files

file type usual function

extension

executable exe,com,bin ready-to-run machine-language

program
Or none

object Obj,o compiled, machine language, not

linked

source code c,cc,java,pas Source code in various languages

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,asm,a

batch bat, sh commands to the command

interpreter

text txt ,doc textual data, documents

word wp, tex, rtf, various word-processor Formats

processor doc

library lib, a, so, dll libraries of routines for

programmers

print or view ps, pdf,jpg ASCII or binary file in a format

for printing or viewing

archive arc, zip, tar related files grouped into one file,
sometimes compressed, for
archiving or storage

multimedia mpeg,mov,r binary file containing audio or

m,mp3,avi A\V information

File Operations
Basic file operations are
1. Creating a file. Two steps are necessary to create a file.
1. Space in the file system must be found for the file.
2. An entry for the new file must be made in the directory.

2. Writing a file. To write a file, we make a system call specifying both the name of
the file and the information to be written to the file. The system must keep a write pointer
to the location in the file where the next write is to take place. The write pointer must be
updated whenever a write occurs.

3. Reading a file. To read from a file, we use a system call that specifies the name of
the file and where (in memory) the next block of the file should be put. The system needs
to keep a read pointer to the location in the file where the next read is to take place.

4. Repositioning within a file. The directory is searched for the appropriate entry, and
the current-file-position pointer is repositioned to a given value. Repositioning within a
file need not involve any actual I/O. This file operation is also known as a file seek.

5. Deleting a file. To delete a file, we search the directory for the named file. Having
found the associated directory entry, we release all file space, so that it can be reused by
other files, and erase the directory entry.

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6. Truncating a file. The user may want to erase the contents of a file but keep its
attributes. Rather than forcing the user to delete the file and then recreate it, this function
allows all attributes to remain unchanged (except for file length) but lets the file be reset
to length zero and its file space released.
b) Explain swapping in operating system with diagram and example. 8M

Ans: Swapping:A process must be in the main memory so that it can execute. Swapping is a (Explanatio
memory/process management technique used by the operating system to increase the n of
utilization of the processor. A process in execution may go into blocked state due to expiry swapping:
3 marks;
of time quantum,occurance of interrupt,etc. when a process is in blocked state and next
Diagram :
process is waiting for execution then operating system performs swapping. Swapping is a 2 marks;
process of moving blocked process from the main memory to the backing store and new Any
process from backing store to main memory. Swapping forms a queue of temporarily Example:
suspended process and the execution continues with the newly arrived process. 3 marks)

Diagram:

In the above diagram, two processes P1 and P2 are shown. A process P1 is in main memory
and in blocked state. Process P2 is in backing store waiting for its turn to execute.As P1 is
blocked, operating system swap out this process by moving it from main memory to
backing store and swap in process P2 by loading it from backing store to main memory.
This process of swap out and swap in is called as swapping of processes.

Example:Consider three processes P1,P2 and P3 are in memory. A Round-Robin CPU

scheduling algorithm is in use. A process P1 starts its execution. When a time quantum
expires memory manager swaps out the process P1 and swaps in process P2. If P1 requires
more time than the time quantum then it is added to blocked queue and waits for its turn for
execution. Once a time quantum of P2 expires ,manager swaps out it and swaps in P3. when
time quantum of P3 expires, manager swaps out it and swaps in P1. This cycle of swap out
and swap in continues till all the processes completes their execution.

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c) Comparison between Linux and UNIX. ( Four points) 8M
(i) User interface
(ii) Name of provider
(iii) Processing speed
(iv) Security
Ans: Parameter Linux Unix (Each
difference:
User interface Linux typically provides two Initially Unix was a 2 marks)
GUIs, KDE and Gnome. But command based OS, but
there are millions of later a GUI was created
alternatives such as LXDE, called Common Desktop
Xfce, Unity, Mate, twm, etc.. Environment. Most
Initially Unix was a command distributions now ship with
based OS, but later a GUI was Gnome.
created called Common
Desktop Environment. Most
distributions now ship with
Gnome.

Name of Provider Redhat, Ubantu, Fedora Osx, SolariAll LINUX

Processing speed Low: As it is GUI based High: As it is command

processing time is more as based direct interpretation
compare to UNIX of commands is done so it
takes less time as compare
to LINUX

Security Linux has had about 60- 100 A rough estimate of UNIX
viruses listed till date. None of viruses is between 85 -120
them actively is spreading viruses reported till date.
nowadays.

6. Attempt any FOUR of the following : 16Marks

a) List characteristics of operating system for smooth functioning of a computer – 4M

system. ( Eight points)

Ans: 1. It provides user interface in the form of command line interface (CLI), batch interface (Explanatio
and graphical user interface (GUI). n of any 8
2. It supports program execution by loading the contents of program file into memory. characteris
tics:
3. System provides I/O resources that may include files and I/O devices required by a
4 marks
running program. (1/2 mark
4. It supports file-system manipulation. There are many operations that are performed for each))
file creation, deletion, allocation, and naming.
5. It performs communications by message passing between systems that require messages
to be turned into packets of information, sent to the net-work controller, transmitted
across a communications medium, and reassembled by the destination system.
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6. It supports error detection: Error detection occurs at both the hardware and software
levels. At the hardware level, all data transfers must be inspected to ensure that data
have not been corrupted in transit. At the software level, media must be checked for data
consistency.
7. Computer system supports accounting of computers that keep track at which users use
how much and what kind of computer resources.
8. System allocates and deallocates resources to the process. When there are multiple users
or multiple jobs running at the same time, resources must be allocated to each of them
code.
9. It provides protection and security for owners of information stored in multiuser or
networked computer system may want to control use of the information in the form of
password or access permissions.
b) With neat labelled diagram explain the working of Booting process. 4M

Ans: The loading of the operating system is achieved by a special program called BOOT. (Explanatio
Generally this program is stored in one (or two) sectors on the disk with a pre-determined
n:2 marks,
address. This portion is normally called “BOOT Block‟ as shown in fig. The ROM
normally contains a minimum program. When one turns the computer „ON‟, the control is Diagram:
transferred to this program automatically by the hardware itself. This program in ROM
2 marks)
loads the BOOT program in pre-determined memory locations. The beauty is to keep
BOOT program as small as possible, so that the hardware can manage to load it easily and
in a very few instructions. This BOOT program in turn contains to read the rest of the
Operating System into the memory. This is depicted in figures. The mechanism gives an
impression of pulling oneself up. Therefore, the nomenclature bootstrapping or its short
form booting.

c) With neat labelled diagram explain Unix layered structure. 4M

Ans:  Kernel: The kernel is the heart of the operating system. It interacts with the hardware and (Explanatio
performs most of the tasks like memory management, task scheduling and file n:2 marks,
management. Diagram: 2
marks)
 Shell: The shell is the utility that processes your requests. When you type in a command at
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your terminal, the shell interprets the command and calls the program that you want. The
shell uses standard syntax for all commands. C Shell, Bourne Shell and Korn Shell are the
most famous shells which are available with most of the Unix variants.

 Commands and Utilities: There are various commands and utilities which you can make
use of in your day to day activities. cp, mv, cat and grep, etc. are few examples of
commands and utilities. There are over 250 standard commands plus numerous others
provided through 3rd party software. All the commands come along with various options.
Files and Directories − All the data of Unix is organized into files. All files are then
organized into directories. These directories are further organized into a tree-like structure
called the file system.

 Application programs: These are the programs that provide interface to the users through
which they can interact with the system.

d) Explain the working of semaphores. 4M

Ans: Semaphore is a synchronization tool. A semaphore S is an integer variable which is (Explanatio

initialized and accessed by only two standard operations: wait () and signal ().All the n of
modifications to the integer value of semaphore in wait () and signal () operations can be working of
done only by one process at a time. semaphores
: 4 marks)
Working of semaphore to solve synchronization problem:- Consider two concurrently
running processes P1 and P2.P1 contains statement S1 and P2 contains statement S2.When
we want to execute statement S2 only after execution of statement S1, then we can
implement it by sharing a common semaphore synch between two processes. Semaphore
synch is initialized to 0.to execute the sequence modify code for process P1 and P2.

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Process P1 contains:

S1;

signal (synch);

Process P2 contains:-

wait (synch);

S2;

As synch is initialized to 0, Process P2 will wait and process P1 will execute. Once process
P1 completes execution of statement S1, it performs signal () operation that increments
synch value. Then wait () operation checks the incremented value and starts execution of
statement S2 from Process P2.

e) Give difference between 4M

External fragmentation and
Internal fragmentation ( four points)
Ans: Internal fragmentation External fragmentation (Each
Difference:
01 mark)
1. Internal fragmentation refers to 1. External fragmentation refers
unused space from space allocated to to unused space from the
any process. memory that is not allocated to
any process.
2. Internal fragmentation occurs when 2. External fragmentation occurs
memory block allocated to a process is when free memory block is
bigger in size than the required size. available but is less than the size
required by the processes waiting
to be loaded in memory.
3. Example: 3. Example:
In fixed memory partitioning technique In variable partitioning technique
all partitions are of same size. So the each block is of variable size so
chances of occurrence of internal some tomes even though free
fragmentation are more. partition is available it cannot be
allocated to the process that
Consider all partitions are of 8K and the
requires more space. So chances
process loaded inside that partition is
of occurrence of external
only 4K then internal fragmentation
fragmentation are more.
occurs.
Consider 20K and 10K partitions
are available but space required
by the waiting process is 30K.
then external fragmentation
occurs.

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4. Consider this is fixed partitioning. 4. Consider this is variable
partitioning.

60 bytes are allocated to each process

but second process is of 57 byte so 3 Two processes are already
byte internal fragmentation is shown. loaded in 40K and 50K
partitions. Assume that one more
process is waiting to get inside
the memory with space
requirement of 25K.20K and
10K is external fragmentation i.e.
these two partitions are free but
cannot be allocated to another
process.

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