Operating Systems
Laboratory Manual
III YEAR I Semester B.TECH
(Computer Science and Engineering)
INFORMATION TECHNOLOGY
Department of Computer Science and Engineering
Department of Information Technology
1|Page
Operating Systems
Laboratory Manual
III YEAR B.TECH
Information
(Computer
ScienceTechnology
and Engineering)
DepartmentofofComputer
InformationScience
Technology
Department
& Engineering
Date of Issue
Document No:
MLRIT/IT/LAB
MANUAL/OS
MLRIT/IT/LAB
MANUAL/OS
MLRIT/CSE/LAB
MANUAL/OS
VERSION 1.1.4
Date of Revision
Compiled by
Ms. Naga Kalyani.A
Mr. Madan Reddy
Mr. Dayakar
Verified by
Mr.G.Kiran Kumar
HOD (CSE)
Authorized by
HOD(CSE)
2|Page
PREFACE
The course provides an overview of the organization of operating systems for generalpurpose computers. Students will be exposed to several new aspects of programming
including: process scheduling, process synchronization, multi-process computation, deadlock
avoidance, file system organization, and security. Students will get knowledge about the
concepts in Operating Systems and techniques used in their implementation. This Course
presents different aspects of operating systems such as; memory management, process
scheduling, management of devices,
IO management, multi-tasking, multi-thread
programming, file systems and inter-process communication techniques.
HOD (CSE)
Principal
3|Page
Steps in Program Development
Program development is a multi step process. The steps are 1) Understand the problem 2)
Develop a solution 3) Write the program and test it.
1. Understand the problem: When you are assigned to develop a program for solving a
problem, you are given the programming requirements. Study these requirements
carefully resolve all doubts and ensure that you have understood it fully as per the
user requirements.
2. Develop a solution: Once you have understood the problem, you have to develop the
solution in terms of some programming language. The tools that help in this are:
a) Algorithm / Pseudo code
b) Flowchart
c) Programming Language, like C, C++ etc.,
d) Test cases
a) Algorithm
Definition: A method of representing the step-by-step logical procedure for solving a
problem in natural language (like English, etc.) is called as an Algorithm.
Algorithm can also be defined as an ordered sequence of well-defined and effective
operations that, when executed, will always produce a result and eventually terminate in a
finite amount of time.
On the whole an algorithm is a recipe for finding a right answer to a problem by breaking it
down into simple steps.
Properties an Algorithm should possess:
a. Generality: The algorithm must be complete in itself so that it can also be used to
solve all problems of a specific type for given input data.
b. Input / Output: Each algorithm can take zero, one or more input data and must
produce one or more output values.
c. Optimization: Unnecessary steps should be eliminated so as to make the algorithm
to terminate in finite number of steps.
d. Effectiveness: Each step must be effective in the sense that it should be primitive
(easily convertible into program statement) and can be performed exactly in a finite
amount of time.
e. Definiteness: Each step of the algorithm, should be precisely and unambiguously
stated.
4|Page
b) Flowchart:
Definition: It is a diagrammatic representation of an algorithm. It is constructed using
different types of symbols.
Standard symbols used in drawing flowcharts:
Terminal
Start/stop/begin/end
Oval
Making data available for
processing(input)
or
recording of the processed
information(output)
Show data output in the
form of document
Parallelogram
Input/output
Document
Printout
Rectangle
Process
Diamond
Decision
Any processing to be
performed.
A
process
changes or moves data. An
assignment operation
Decision or switching type
of operation
Circle
Connector
Used to connect different
parts of flowchart
Arrow
Flow
Joins two symbols and also
represents flow of execution
Annotation
Descriptive comments
explanation
Bracket with
broken line
Double sided
rectangle
Predefined
process
or
Modules or subroutines
specified elsewhere
Programming and Testing:
A computer program is the sequence of instructions written in a computer language according
to the algorithm and computer follows these in carrying out its computations.
The process of writing a program is called programming.
Steps involved in Computer Programming:
 Problem Understanding
 Problem Definition
 Program writing
 Error analysis
 Validation & Verification
 Documentation & Maintenance
5|Page
OPERATING SYSTEMS LAB SYLLABUS
OBJECTIVE


To use Linux operating System for study of operating system concepts.
To write the code to implement and modify various concepts in operating system
in Linux.
1. Simulate the following CPU scheduling algorithms
a) Round Robin b) SJF c) FCFS d) Priority
2. Simulate all file allocation strategies
a) Sequential b) Indexed c) Linked
3. Simulate MVT and MFT.
4. Simulate all file Organization Techniques
a) Single level directory b) Two Level c) Hierarchical d) DAG
5. Simulate Bankers Algorithm for Dead Lock Avoidance
6. Simulate Bankers Algorithm for Dead Lock Prevention
7. Simulate all page replacement algorithms
a) FIFO b) LRU c) LFU Etc. …
8. Simulate Paging Technique of memory management.
OUTCOMES:
 The course objectives ensure the development of students applied skills in
operating systems related areas.
 Students will gain knowledge in writing software routines modules or
implementing various concepts of operating systems.
6|Page
LAB INSTRUCTIONS
1. Students should report to the concerned lab as per the time table.
2. Students who turn up late to the labs will in no case be permitted to do the program
schedule for the day.
3. After completion of the program, certification of the concerned staff in-charge in the
observation book is necessary.
4. Student should bring a notebook of 100 pages and should enter the readings /observations
into the notebook while performing the experiment.
5. The record of observations along with the detailed experimental procedure of the
experiment in the immediate last session should be submitted and certified staff member
in-charge.
6. Not more than 3-students in a group are permitted to perform the experiment on the set.
7. The group-wise division made in the beginning should be adhered to and no mix up of
students among different groups will be permitted.
8. The components required pertaining to the experiment should be collected from stores incharge after duly filling in the requisition form.
9. When the experiment is completed, should disconnect the setup made by them, and
should return all the components/instruments taken for the purpose.
10. Any damage of the equipment or burn-out components will be viewed seriously either by
putting penalty or by dismissing the total group of students from the lab for the
semester/year.
7|Page
TABLE CONTENT(As per JNTUH Syllabus)
S.No
Week No
Name of the program
Page
No
Simulate the following cpu scheduling algorithms:
a) FCFS
b) SJF
1.
c) Round Robin
d) priority
2.
Simulate the file allocation strategies:
a) Sequential
3.
b) Indexed
4.
c) Linked
5.
Simulate MVT and MFT
6.
Simulate all file Organization Techniques
a) Single level
7.
b) Two Level
8.
c) Hierarchical
9.
d) DAG
10.
Simulate Bankers Algorithm for Deadlock Avoidance
11.
Simulate Bankers algorithm for Deadlock Prevention
12.
Simulate all page replacement Algorithms
a)FIFO
13.
b) LRU
14.
c) LFU
15.
Simulate Paging Technique of memory management.
8|Page
OPERATING SYSTEMS
EXPERIMENT : 1
NAMEOF THE EXPERIMENT: Simulate the following CPU Scheduling Algorithms
a) FCFS b) SJF c) Round Robin d) Priority
AIM: Using CPU scheduling algorithms find the min & max waiting time.
HARDWARE REQUIREMENTS: Intel based Desktop Pc RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
CPU SCHEDULING
Maximum CPU utilization obtained with multiprogramming
CPU–I/O Burst Cycle – Process execution consists of a cycle of
CPU execution and I/O wait
CPU burst distribution
a) First-Come, First-Served (FCFS) Scheduling
Process
Burst Time
P1
24
P2
3
P3
3
Suppose that the processes arrive in the order: P1 , P2 , P3
The Gantt Chart for the schedule is:
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0
30
24
27
Waiting time for P1 = 0; P2 = 24; P3 = 27
Average waiting time: (0 + 24 + 27)/3 = 17
ALGORITHM
1. Start
2. Declare the array size
3. Read the number of processes to be inserted
4. Read the processname, Burst times , arrival times of processes
5. calculate the waiting time of each process
wt[i+1]=bt[i]+wt[i]
6. calculate the turnaround time of each process
tt[i+1]=tt[i]+bt[i+1]
7. Calculate the average waiting time and average turnaround time.
8. Display the values
9. Stop
TEST CASES:
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Problem Statements:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS FOR WEEK1
P1: Simulate FCFS scheduling without arrival times to display average waiting time along
with process id.
TEST CASES:
P2: Simulate FCFS scheduling without arrival times to calculate average turnaround time.
TEST CASES:
P3: Simulate FCFS scheduling with arrival times to calculate average waiting time.
TEST CASES:
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P4: Simulate FCFS scheduling with arrival times to calculate average turnaround
time.
TEST CASES:
P5: Simulate FCFS t scheduling to display the start time and finish time of each
process(without arrival time)
TEST CASES:
P6: Simulate FCFS scheduling to display the start time and finish time of each
process (with arrival time).
TEST CASES:
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P7: Simulate FCFS scheduling to display gantt chart with arrival time.
TEST CASES:
P8: Simulate FCFS scheduling to display Gantt chart for processes without arrival
times.
TEST CASES:
13 | P a g e
P9: Simulate FCFS scheduling to display gantt chart and average waiting time with
arrival time.
TEST CASES:
P10: Simulate FCFS scheduling to display gantt chart and average Turnaround time
with arrival time.
TEST CASES:
VIVA QUESTIONS
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1. What is First-Come-First-Served (FCFS) Scheduling?
2. Why CPU scheduling is required?
3. Which technique was introduced because a single job could not keep both the CPU
and the I/O devices busy?
1) Time-sharing 2) SPOOLing 3) Preemptive scheduling 4) Multiprogramming
4. CPU performance is measured through ________.
1) Throughput 2) MHz 3) Flaps 4) None of the above
5. Which of the following is a criterion to evaluate a scheduling algorithm?
1 CPU Utilization: Keep CPU utilization as high as possible.
2 Throughput: number of processes completed per unit time.
3 Waiting Time: Amount of time spent ready to run but not running.
4 All of the above
EXPERIMENT : 1b)
NAMEOF THE EXPERIMENT: Simulate the following CPU Scheduling Algorithms
b) SJF
AIM: Using CPU scheduling algorithms find the min & max waiting time.
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
Example of Non Preemptive SJF
Process
Arrival Time
Burst Time
P1
0.0
7
P2
2.0
4
P3
4.0
1
P4
3.0
4
P1
P3
P2
0
7
8
P4
12
16
Example of Preemptive SJF
Process
P1
Arrival Time
0.0
Burst Time
7
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P2
2.0
4
P3
4.0
1
P4
3.0
4
P1
P2
0
2
P3
4
P2
5
P4
7
P1
11
16
Average waiting time = (9 + 1 + 0 +2)/4 = 3
ALGORITHM
1. Start
2. Declare the array size
3. Read the number of processes to be inserted
4. Read the Burst times of processes
5. sort the Burst times in ascending order and process with shortest burst time is first
executed.
6. calculate the waiting time of each process
wt[i+1]=bt[i]+wt[i]
7. calculate the turnaround time of each process
tt[i+1]=tt[i]+bt[i+1]
8. Calculate the average waiting time and average turnaround time.
9. Display the values
10. Stop
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS FOR WEEK2
P1: Simulate Shortest job first scheduling without arrival times to calculate average
waiting time.
TEST CASES:
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P2: Simulate Shortest job first scheduling without arrival times to calculate average
turnaround time.
TEST CASES:
P3: Simulate Shortest job first scheduling with arrival times to calculate average
waiting time.
TEST CASES:
17 | P a g e
P4: Simulate Shortest job first scheduling with arrival times to calculate average
turnaround time.
TEST CASES:
P5: Simulate Shortest job first scheduling to display only the start time and finish
time of each process.(without arrival time).
TEST CASES:
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
the number of processes:3
the name, burst time for processes
the name of the process:p1
Burst time for process P1 in ms:10
the name of the process:p2
Burst time for process P2 in ms:5
the name of the process:p3
Burst time for process P3 in ms:3
Process BT
ST
FT
-------------------------------------p3
3ms
0ms
3ms
p2
5ms
3ms
8ms
p1
10ms
8ms
18ms
P6: Simulate Shortest job first scheduling to display the start time and finish time of
each process and their average waiting time.(without arrival time).
TEST CASES:
Enter
Enter
Enter
Enter
the number of processes:3
the name, burst time for processes
the name of the process:p1
Burst time for process P1 in ms:10
18 | P a g e
Enter
Enter
Enter
Enter
the name of the process:p2
Burst time for process P2 in ms:5
the name of the process:p3
Burst time for process P3 in ms:3
Process BT
ST
FT
-------------------------------------p3
3ms
0ms
3ms
p2
5ms
3ms
8ms
p1
10ms
8ms
18ms
Average waiting time=3.33ms
P7: Simulate Shortest job first scheduling to display the start time and finish time of
each process and their average turnaround time.(without arrival time).
TEST CASES:
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
the number of processes:3
the name, burst time for processes
the name of the process:p1
Burst time for process P1 in ms:10
the name of the process:p2
Burst time for process P2 in ms:5
the name of the process:p3
Burst time for process P3 in ms:3
Process BT
ST
FT
-------------------------------------p3
3ms
0ms
3ms
p2
5ms
3ms
8ms
p1
10ms
8ms
18ms
Average turnaround time=6.6ms
P8: Simulate Shortest job first scheduling to calculate average waiting time by
defining a separate sjf function .
TEST CASES:
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
the number of processes:3
the name, burst time & arrival time for processes
the name of the process:p1
Arrival time for process P1 in ms:0
Burst time for process P1 in ms:10
the name of the process:p2
Arrival time for process P2 in ms:0
Burst time for process P2 in ms:5
the name of the process:p3
Arrival time for process P3 in ms:0
Burst time for process P3 in ms:3
Process AT
BT
WT
-------------------------------------p3
0ms
3ms
0ms
p2
0ms
5ms
3ms
p1
0ms
10ms
8ms
Avg waiting time= 3.67 ms
P9:Simulate Shortest job first scheduling to calculate average turnaround time by
defining a separate sjf function .
19 | P a g e
TEST CASES:
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
Enter
the number of processes:3
the name, burst time & arrival time for processes
the name of the process:p1
Arrival time for process P1 in ms:0
Burst time for process P1 in ms:10
the name of the process:p2
Arrival time for process P2 in ms:0
Burst time for process P2 in ms:5
the name of the process:p3
Arrival time for process P3 in ms:0
Burst time for process P3 in ms:3
Process AT
BT
TAT
-------------------------------------p3
0ms
3ms
3ms
p2
0ms
5ms
8ms
p1
0ms
10ms
18ms
Avg turn around time= 9.67 ms
P 10: Simulate Shortest job first scheduling to display gantt chart.
TEST CASES:
VIVA QUESTIONS:
1) The optimum CPU scheduling algorithm is
(A)FIFO
(B)SJF with preemption.
(C)SJF without preemption.
(D)Round Robin.
2) In terms of average wait time the optimum scheduling algorithm is
(A)FCFS
(B)SJF
(C)Priority
(D)RR
3) What are the dis-advantages of SJF Scheduling Algorithm?
4) What are the advantages of SJF Scheduling Algorithm?
5) Define CPU Scheduling algorithm?
EXPERIMENT : 1c)
20 | P a g e
NAMEOF THE EXPERIMENT: Simulate the following CPU Scheduling Algorithms
c) Round Robin
AIM: Using CPU scheduling algorithms find the min & max waiting time.
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
Round Robin:
Example of RR with time quantum=3
Process
Burst time
aaa
4
Bbb
3
Ccc
2
Ddd
5
Eee
1
ALGORITHM
1. Start
2. Declare the array size
3. Read the number of processes to be inserted
4. Read the burst times of the processes
5. Read the Time Quantum
6. if the burst time of a process is greater than time Quantum then subtract time quantum
form the burst time Else Assign the burst time to time quantum.
7.calculate the average waiting time and turn around time of the processes.
8. Display the values
9. Stop
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1. Simulate Round robin scheduling without arrival time to calculate average waiting
time.
TEST CASES:
Enter the number of processes 4
Enter the Burst Time:
8
21 | P a g e
4
9
5
Enter the time Quantum: 5
Scheduling order:
1
2
3
4
1
3
Average Waiting Time: 12.75
P2. Simulate Round robin scheduling with arrival times to calculate average
turnaround time.
TEST CASES:
Enter the number of processes: 4
Enter the arrival times:
0
1
2
3
Enter the burst Time:
8
4
9
5
Enter the time quantum: 5
Scheduling Order:
1
2
3
4
1
3
Average turnaround time: =17.5
Average waiting time = 11
P3. Simulate Round robin scheduling to display the Gantt chart.
TEST CASES:
Enter the number of processes:4
Enter the Time Quantum : 1
Enter the Arrival Time and Burst Time
Enter for Process 0 : 0 8
Enter for Process 1 : 1 4
Enter for Process 2 : 2 9
Enter the Process 3: 3 5
GANTT CHART
P0 P1 P2 P3 P0 P1 P2 P3 P0 P1 P2 P3 P0 P1 P2 P3 P0 P2 P3 P0 P2 P0 P2 P0 P2
P2
0
26
1
2
3
4
5
6
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
P4. Simulate Round robin scheduling without arrival time to calculate average
Turnaround time.
TEST CASES:
22 | P a g e
P5. Simulate Round robin scheduling with arrival times to calculate average waiting
time.
TEST CASES:
Enter the number of Process : 4
Enter the time Quantum: 4
Enter the Burst Time for Process P1: 8
Enter the Burst Time for Process P2: 4
Enter the Burst Time for Process P3: 9
Enter the Burst Time for Process P4 : 5
Enter the Arrival Time for Process 0
1
2
3
Average Waiting Time: 18.5
P6. Simulate Round robin scheduling without arrival time and print start time, finish
time of each process
TEST CASES:
P7. Simulate Round robin scheduling with arrival time and print start time, finish time
of each process
TEST CASES:
Enter the number of Process : 3
Enter the time Quantum: 4
Enter the Burst Time for Process P1: 24
Enter the Burst Time for Process P2: 3
Enter the Burst Time for Process P3: 3
Enter the Arrival Time for Process 0
0
0
PId
1
2
P ST
0
4
PET
30
7
3
7
10
P8. Simulate Round robin scheduling to display the Gantt chart with a time of 1 sec.
TEST CASES:
Enter the number of processes:4
23 | P a g e
Enter the Time Quantum : 1
Enter the Arrival Time and Burst Time
Enter for Process 0 : 0 8
Enter for Process 1 : 1 4
Enter for Process 2 : 2 9
Enter the Process 3: 3 5
GANTT CHART
P0 P1 P2 P3 P0 P1 P2 P3 P0 P1 P2 P3 P0 P1 P2 P3 P0 P2 P3 P0 P2 P0 P2 P0 P2
P2
0
26
1
2
3
4
5
6
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
P9. Simulate Round robin scheduling with arrivaltime & with time quantum of 4 sec.
TEST CASES:
Enter the number of Process : 3
Enter the time Quantum: 4
Enter the Burst Time for Process P1: 24
Enter the Burst Time for Process P2: 3
Enter the Burst Time for Process P3: 3
Enter the Arrival Time for Process 0
0
0
P1
P2
P3
P1
0
4
7
10
14
P1
18
P1
22
P1
26
P1
30
P10. Simulate Round robin scheduling with arrivaltime & with time quantum of 1 sec.
TEST CASES:
Enter the number of processes:4
Enter the Time Quantum : 1
Enter the Arrival Time and Burst Time
Enter for Process 0 : 0 8
Enter for Process 1 : 1 4
Enter for Process 2 : 2 9
Enter the Process 3: 3 5
GANTT CHART
P0 P1 P2 P3 P0 P1 P2 P3 P0 P1 P2 P3 P0 P1 P2 P3 P0 P2 P3 P0 P2 P0 P2 P0 P2
P2
0
26
1
2
3
4
5
6
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
24 | P a g e
VIVA QUESTIONS:
1.Round Robin scheduling is used in
(A)Disk scheduling. (B)CPU scheduling
(C)I/O scheduling.
(D)Multitasking
2. What are the dis-advantages of RR Scheduling Algoritm?
3.What are the advantages of RR Scheduling Algoritm?
4.Super computers typically employ _______.
1 Real time Operating system 2 Multiprocessors OS
3 desktop OS
4 None of the above
5. An optimal scheduling algorithm in terms of minimizing the average waiting time of a
given set of processes is ________.
1 FCFS scheduling algorithm 2 Round robin scheduling algorithm
3 Shortest job - first scheduling algorithm 4 None of the above
25 | P a g e
EXPERIMENT : 1d)
NAMEOF THE EXPERIMENT: Simulate the following CPU Scheduling Algorithms
d) Priority
AIM: Using CPU scheduling algorithms find the min & max waiting time.
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
In Priority Scheduling, each process is given a priority, and higher priority methods are
executed first, while equal priorities are executed First Come First Served or Round Robin.
There are several ways that priorities can be assigned:

Internal priorities are assigned by technical quantities such as memory usage, and
file/IO operations.

External priorities are assigned by politics, commerce, or user preference, such as
importance and amount being paid for process access (the latter usually being for
mainframes).
ALGORITHM
1. Start
2. Declare the array size
3. Read the number of processes to be inserted
4. Read the Priorities of processes
5. sort the priorities and Burst times in ascending order
5. calculate the waiting time of each process
wt[i+1]=bt[i]+wt[i]
6. calculate the turnaround time of each process
tt[i+1]=tt[i]+bt[i+1]
6. Calculate the average waiting time and average turnaround time.
7. Display the values
8. Stop
26 | P a g e
TEST CASES:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1: Simulate Priority scheduling without arrival times to calculate average waiting
time.
TEST CASES:
Enter the number of processes 5
Enter the Burst Time for each process:
10
1
2
1
5
Enter the Priority of each Process
3
1
4
5
2
The Average Waiting Time : 8.2ms.
P2: Simulate Priority scheduling without arrival times to calculate average turnaround
time.
TEST CASES:
Enter the number of processes 3
Enter the burst time for each process 10
11
27 | P a g e
12
Enter the priority values 3
2
1
Average turnaround time : 21ms.
P3: Simulate Priority scheduling with arrival times to calculate average waiting time.
TEST CASES:
Enter the number of process 3
Enter the CPU Burst Time
10
5
2
Enter the Priority
3
2
1
Enter Arrival Time
0
1
2
Average Waiting Time is : 4 ms
P4: Simulate Priority scheduling with arrival times to calculate average turnaround
time.
TEST CASES:
Enter the number of process 3
Enter the CPU Burst Time
10
5
2
Enter the Priority
3
2
1
Enter Arrival Time
0
1
2
Average Turn Around Time is : 10.3 ms
28 | P a g e
P5: Simulate Priority scheduling to display only the start time and finish time of each
process.(without arrival time).
TEST CASES:
Enter the no of process : 5
Enter the burst time of each process: 10
1
2
1
5
Enter the priority for each process : 3
1
4
5
2
PId PST
PET
P1
10
16
P2
0
1
P3
16
18
P4
18
19
P5
1
6
P6: Simulate Priority scheduling to display the start time and finish time of each process
and their average waiting time.(without arrival time).
TEST CASES:
Enter the no of process : 5
Enter the burst time of each process: 10
1
2
1
5
Enter the priority for each process : 3
1
4
5
2
PId PST
PET
P1
10
16
P2
0
1
P3
16
18
P4
18
19
P5
1
6
The Average Waiting Time: 8.2ms
29 | P a g e
P7: Simulate Priority scheduling to display the start time and finish time of each process
And their average turnaround time.(without arrival time).
TEST CASES:
Enter the no of processes : 5
Enter the Burst Times: 10
1
2
1
5
Enter the Priority for each process: 3
1
3
4
2
PId PST
PET
P1
6
16
P2
0
1
P3
16
18
P4
18
19
P5
1
6
Total Turn Around Time :76 ms
Average Turn Around Time : 15.2 ms
P8: Simulate Priority scheduling to calculate average waiting time by defining a
separate priority function .
TEST CASES:
Enter the no of process : 5
Enter the burst time of each process: 10
1
2
1
5
Enter the priority for each process : 3
1
4
5
2
PId PST
PET
P1
10
16
2
0
1
P3
16
18
P4
18
19
P5
1
6
30 | P a g e
The Average Waiting Time: 8.2ms
P9:Simulate Priority scheduling to calculate average turnaround time by defining a
separate Priority function .
TEST CASES:
Enter the no of processes : 5
Enter the Burst Times: 10
1
2
1
5
Enter the Priority for each process: 3
1
3
4
2
PId
PST
PET
P1
6
16
P2
0
1
P3
16
18
P4
18
19
P5
1
6
Total Turn Around Time :76 ms
Average Turn Around Time : 15.2 m
P 10: Simulate Priority scheduling to display gantt chart.
TEST CASES:
Enter the no of processes : 5
Enter the Burst Times: 10
1
2
1
5
Enter the Priority for each process: 3
1
3
4
2
P2
0
P5
1
P1
6
P3
16
P4
18
19
31 | P a g e
VIVA QUESTIONS:
1. Priority CPU scheduling would most likely be used in a _____________ os.
2. Cpu allocated process to ___________________ priority.
3. calculate avg waiting time=
4. Maximum CPU utilization obtained with _________________________
5. Using _______________algorithms find the min & max waiting time.
EXPERIMENT: 2a)
NAME OF EXPERIMENT: Simulate file Allocation strategies:
a) Sequential
AIM: Simulate the file allocation strategies using file allocation methods
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
File Allocation Strategies:
The main problem is how to allocate disk space to the files so that disk space is utilized
effectively band files can be accessed quickly. We have 3 space allocation method.
1. Contiguous allocation (Sequential)
It requires each file to occupy a set of contiguous blocks on the hard disk where disk
address define a linear ordering on the disk.
Disadvantages:
i.
Difficult for finding space for a new file.
ii.
Internal and external fragmentation will be occurred.
ALGORITHM:
1.
2.
3.
4.
5.
6.
Start
Read the number of files
For each file, read the number of blocks required and the starting block of the file.
Allocate the blocks sequentially to the file from the starting block.
Display the file name, starting block , and the blocks occupied by the file.
stop
32 | P a g e
TEST CASES:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1: simulate sequential file allocation by creating sequential file and find out no. of
sequential block occupied.
Input and Output:
Test case 1 :
Test case 2 :
33 | P a g e
P2: Simulate sequential file allocation by creating single sequential file and find out no.
of block required by taking input block size for a file.
Input and Output:
Test case :
:
P3: Simulate sequential file allocation by creating two files and find out no. of block
required by taking input size of file and block size for both files.
Input and Output:
Test case :
34 | P a g e
P4: Simulate sequential file allocation by creating three files and find out no. of block
required by taking input size of file and block size for these files.
Input and Output:
Test case :
P5: Simulate sequential file allocation by creating n number of sequential file and find out
no. of block required by taking input block size for n number of file.
Input and Output:
Test case :
35 | P a g e
P6: Simulate sequential file allocation by creating single sequential file and find out no.
of block required along with block address number.
Test case :
P7: Simulate sequential file allocation by creating two files and find out no. of block
required and block address number by taking input size of file and block size for both
files.
Test case :
36 | P a g e
P8: Simulate sequential file allocation by creating three files and find out no. of block
required and block address number by taking input size of file and block size for all
files.
Test case :
P9: Simulate sequential file allocation by creating n number of sequential file and find
out no. of block required and block address numbers by taking input block size for
n number of file.
Input and output
Test case :
37 | P a g e
P10: Simulate sequential file allocation by creating n number of sequential file and find
out no. of block required and block address numbers. Repeat the procedure.
Test case :
VIVA QUESTIONS:
1.What file access pattern is particularly suited to chained file allocation on disk?
2. Define Sequential File allocation
4. Why we use file allocation strategies?
5.what are the advantages and dis-advantages of Sequential File allocation?
6. The average waiting time =________________________________.
38 | P a g e
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
EXPERIMENT: 2 b)
NAME OF EXPERIMENT: Simulate file Allocation strategies:
b) Indexed
AIM: Simulate the file allocation strategies using file allocation methods
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
Indexed allocation
In linked allocation it is difficult to maintain FAT – so instead of that method indexed
allocation method is used. Indexed allocation method solves all the problems in the linked
allocation by bringing all the pointers together into one location called index block.
ALGORITHM:
1.
2.
3.
4.
5.
6.
7.
Start
Read the number of files
Read the index block for each file.
For each file, read the number of blocks occupied and number of blocks of the file.
Link all the blocks of the file to the index block.
Display the file name, index block , and the blocks occupied by the file.
stop
TESTCASES
VIVA QUESTIONS:
1. What file allocation strategy is most appropriate for random access files?
2.Define File?
3.Define Directory?
4. Why we use file allocation strategies?
5.what are the advantages and dis-advantages Indexed Allocation?
39 | P a g e
EXPERIMENT: 2 c)
NAME OF EXPERIMENT: Simulate file Allocation strategies:
c) Linked
AIM: Simulate the file allocation strategies using file allocation methods
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
Linked Allocation
Linked allocation of disk space overcomes all the problems of contiguous allocation. In
linked allocation each file is a linked list of disk blocks where the disk blocks may be
scattered anywhere on the disk. The directory contains a pointer to the first and last blocks of
the file.
Disadvantages : Space required to maintain pointers.
ALGORITHM:
1. Start
2. Read the number of files
3. For each file, read the file name, starting block, number of blocks and block numbers
of the file.
4. Start from the starting block and link each block of the file to the next block in a
linked list fashion.
5. Display the file name, starting block, size of the file , and the blocks occupied by the
file.
6. stop
Test Case:
40 | P a g e
VIVA QUESTIONS:
1.What file access pattern is particularly suited to chained file allocation on disk?
2. What file allocation strategy is most appropriate for random access files?
3. Mention different file allocation strategies?
4. Why we use file allocation strategies?
5. What are the advantages and dis-advantages of each strategies?
6. The ______________contains a pointer to the first and last blocks of the file.
EXPERIMENT: 3a
NAME OF EXPERIMENT: Simulate MFT .
AIM: Simulate Multiple Programming with fixed Number of Tasks (MFT)
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
Multiple Programming with fixed Number of Tasks (MFT) Algorithm
Background:
IBM in their Mainframe Operating System OS/MFT implements the MFT concept.
OS/MFT uses Fixed partitioning concept to load programs into Main memory.
Fixed Partitioning:

In fixed partitioning concept, RAM is divided into set of fixed partition of
equal Size

Programs having the Size Less than the partition size are loaded into Memory

Programs Having Size more then the size of Partitions Size is rejected

The program having the size less than the partition size will lead to internal
Fragmentation.

If all partitions are allocated and a new program is to be loaded, the program
that lead to Maximum Internal Fragmentation can be replaced
ALGORITHM:
Step1: start
Step2: Declare variables.
Step3: Enter total memory size.
Step4: Read the no of partitions to be divided.
Step5: Allocate memory for os.
Step6:calculate available memory by subtracting the memory of os from total memory
41 | P a g e
Step7: calculate the size of each partition by dividing available memory with no of
partitions.
Step8: Read the number of processes and the size of each process.
Step9: If size of process<= size of partition then allocate memory to that process.
Step10: Display the wastage of memory.
Step11: Stop .
Test Case:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1. Simulate MFT which accepts only even number of memory partitions.
Test Case:
42 | P a g e
P2) Simulate MFT which accepts Odd number of memory partitions.
Test Case:
P3. Simulate MFT which accepts the memory capacity in the range from 100 to
500MB.
Test Case1:
Test Case2:
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P4. Simulate MFT which accepts the memory capacity in the range from 500 to
1000MB.
Test Case 1:
Test Case2:
44 | P a g e
P5. Simulate MFT which accepts the memory capacity in the range from 100 to 500MB
and no of partitions are even.
Test Case:
P6. Simulate MFT which accepts the memory capacity in the range from 100 to 500MB
and no of partitions are odd.
Test case:
P7. Simulate MFT which accepts the memory capacity in the range from 500 to
1000MB and no of partitions are even.
Test case:
45 | P a g e
P8. Simulate MFT which accepts the memory capacity in the range from 500 to
1000MB and no of partitions are odd.
Test case:
P9. Simulate MFT which accepts the memory capacity in the range from 1000 to
2000MB.
Test case1:
Test case2:
P10. Simulate MFT which accepts the memory capacity in the range from 2000 to
3000MB.
Test Case1:
46 | P a g e
Test case2:
VIVA QUESTIONS
1. The problem of fragmentation arises in ________.
1)Static storage allocation 2) Stack allocation storage
3 Stack allocation with dynamic binding 4 Heap allocation
2. Boundary registers ________.
1 Are available in temporary program variable storage
2 Are only necessary with fixed partitions
3 Track the beginning and ending the program
4 Track page boundaries
3.The principle of locality of reference justifies the use of ________.
1 Virtual Memory 2 Interrupts
3 Main memory 4 Cache memory
4. In memory management , a technique called as paging, physical memory is broken into
fixed-sized blocks called ___________.
1) Pages 2) Frames 3) Blocks 4) Segments
5.Demand paged memory allocation
1 allows the virtual address space to be independent of the physical memory
2 allows the virtual address space to be a multiple of the physical memory size
3 allows deadlock tobe detected in paging schemes
4 is present only in Windows NT
EXPERIMENT: 3 b)
NAME OF EXPERIMENT: multiple Programming with Varible Number of Tasks
(MVT) :
AIM: Simulate multiple Programming with Varible Number of Tasks (MVT)
47 | P a g e
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
multiple Programming with Varible Number of Tasks (MVT) Algorithm
Background:
IBM in their Mainframe Operating ‘System OS/MVT implements the MVT concept.
OSIMVT uses Dynamic Partition concept to load programs into Main memory.
Dynamic Partitioning:

Initially RAM is portioned according to the of programs to be loaded into
Memory till such time no other program can be loaded.

The Left over Memory is called a hole which is too small too fit any process.

When a new program is to be into Memory Look for the partition, Which Leads to
least External fragmentation and load the Program.

The space that is not used in a partition is called as External Fragmentation
ALGORITHM:
Step1: start
Step2: Declare variables.
Step3: Enter total memory size.
Step4: Read the no of processes
Step5: Allocate memory for os.
Step6: read the size of each process
Step7:calculate available memory by subtracting the memory of os from total
memory
Step8: If available memory >= size of process then allocate memory to that
process.
Step9: Display the wastage of memory.
Step10: Stop .
Test Cases :
48 | P a g e
VIVA QUESTIONS:
1. Explain about MFT?
2. Full form of MFT____________________
3. Full form of MVT____________________
4. Differentiate MFT and MVT?
5. The __________________ Memory is called a hole.
6. OSIMVT uses ________________concept to load programs into Main memory.
7. OS/MFT uses ____________________ concept to load programs into Main memory.
Experiment No: 4 a)
NAME OF EXPERIMENT:
Simulate all file Organization Techniques
a) Single level directory b) Two Level c) Hierarchical d) DAG
Aim: To understand how the file are organized in single level directories.
HARDWARE REQUIREMENTS: Intel based Desktop Pc RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:

The simplest directory structure is the single-level directory. All files are contained in
the same directory, which is easy to support and understand
49 | P a g e




On early personal computers, this system was common, in part because there was
only one user. The world's first supercomputer, the CDC 6600, also had only a single
directory for all files, even though it was used by many users at once.
A single-level directory has significant limitations, when the number of files increases
or when the system has more than one user.
Since all files are in the same directory, they must have unique names. If two users
call their data file test , then the unique-name rule is violated.
Even a single user on a single-level directory may find it difficult to remember the
names of all the files as the number of files increases.
Algorithm:
Step 1: Start
Step2: Read the number of directories
Repeat
Step 3: Read the names of the directory.
Step 4: Read the size/ number of files in the directory.
Step 5: Read the file names
End Repeat
Step 6: Display the content of the directory.
TEST CASES:
P1: Simulate Single level directory for directory of size 3.
TEST CASES 1
50 | P a g e
TESTCASE2:
P2: Simulate Single level directory by using separate function for size of the directory.
TESTCASES
51 | P a g e
Experiment No: 4 b)
NAME OF EXPERIMENT:
Simulate all file Organization Techniques
b) Two Level
Aim: To understand how the file are organized in two level directories.
HARDWARE REQUIREMENTS: Intel based Desktop Pc RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
It contains one master file directory. Each user has its own user file directory. Each entry in
the master file directory points to a user file directory. The issues with two level directories
are sharing, system files and grouping problems.
Algorithm:
Start
Step2: Read the number of directories
Repeat
Step 3: Read the names of the directory.
Step 4: Read the sub directory names
Step 5: Read the size/ number of files in the directory.
Step 6: Read the file names
End Repeat
Step 7: Display the content of the directory.
End
52 | P a g e
Test Case
Experiment No: 4 c)
NAME OF EXPERIMENT:
Simulate all file Organization Techniques
c) Hierarchical
Aim: To understand how the file are organized in hierarchial level directories.
HARDWARE REQUIREMENTS: Intel based Desktop Pc RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
53 | P a g e
Algorithm:
Start
Step2: Read the number of directories
Repeat
Step 3: Read the names of the directory.
Step 4: Read the sub directory names
Step 5: Read the names of sub-sub directories.
Step 6: Read the size/ number of files in the directory.
Step 7: Read the file names
End Repeat
Step 8: Display the content of the directory.
End
EXPERIMENT: 5
NAME OF EXPERIMENT: Simulate Banker’s Algorithm for Deadlock Avoidance.
AIM: Simulate Banker’s Algorithm for Deadlock Avoidance to find whether the system is in
safe state or not.
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
DEAD LOCK AVOIDANCE
To implement deadlock avoidance & Prevention by using Banker’s Algorithm.
Banker’s Algorithm:
When a new process enters a system, it must declare the maximum number of
instances of each resource type it needed. This number may exceed the total number of
resources in the system. When the user request a set of resources, the system must determine
whether the allocation of each resources will leave the system in safe state. If it will the
resources are allocation; otherwise the process must wait until some other process release the
resources.
Data structures

n-Number of process, m-number of resource types.

Available: Available[j]=k, k – instance of resource type Rj is available.
54 | P a g e

Max: If max[i, j]=k, Pi may request at most k instances resource Rj.

Allocation: If Allocation [i, j]=k, Pi allocated to k instances of resource Rj

Need: If Need[I, j]=k, Pi may need k more instances of resource type Rj,
Need[I, j]=Max[I, j]-Allocation[I, j];
Safety Algorithm
1. Work and Finish be the vector of length m and n respectively, Work=Available and
Finish[i] =False.
2. Find an i such that both

Finish[i] =False

Need<=Work
If no such I exists go to step 4.
3. work=work+Allocation, Finish[i] =True;
4. if Finish[1]=True for all I, then the system is in safe state.
Resource request algorithm
Let Request i be request vector for the process Pi, If request i=[j]=k, then process Pi
wants k instances of resource type Rj.
1. if Request<=Need I go to step 2. Otherwise raise an error condition.
2. if Request<=Available go to step 3. Otherwise Pi must since the resources are
available.
3. Have the system pretend to have allocated the requested resources to process Pi by
modifying the state as follows;
Available=Available-Request I;
Allocation I =Allocation+Request I;
Need i=Need i-Request I;
If the resulting resource allocation state is safe, the transaction is completed and process Pi is
allocated its resources. However if the state is unsafe, the Pi must wait for Request i and the
old resource-allocation state is restored.
ALGORITHM:
1. Start the program.
2. Get the values of resources and processes.
3. Get the avail value.
4. After allocation find the need value.
5. Check whether its possible to allocate.
6. If it is possible then the system is in safe state.
7. Else system is not in safety state.
55 | P a g e
8. If the new request comes then check that the system is in safety.
9. or not if we allow the request.
10. stop the program.
Test Cases:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1: Simulate Bankers algorithm for Deadlock Avoidance for the
following :
TEST CASE
56 | P a g e
P2: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Case
P3: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Case:
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P4: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Case
P5: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Case:
58 | P a g e
P6: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Cases
P7: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Cases
59 | P a g e
P8: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Cases
P9: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Cases
60 | P a g e
P10: Simulate Bankers algorithm for Deadlock Avoidance for the following :
Test Cases
VIVA QUESTIONS:
1. Differentiate deadlock avoidance and fragmentation
2.Tell me the real time example where this deadlock occurs?
3.How do we calculate the need for process?
4.What is the name of the algorithm to avoid deadlock?
5.Banker’s algorithm for resource allocation deals with (A)Deadlock prevention.
(B)Deadlock avoidance. (C)Deadlock recovery. (D)Mutual exclusion
61 | P a g e
EXPERIMENT: 6
NAME OF EXPERIMENT: Simulate Algorithm for Deadlock prevention.
AIM: Simulate Algorithm for Deadlock prevention .
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
Deadlock Definition:
A set of processes is deadlocked if each process in the set is waiting for an event that only
another process in the set can cause (including itself).Waiting for an event could be:

waiting for access to a critical section

waiting for a resource Note that it is usually a non-preemptable (resource).
Conditions for Deadlock :
•Mutual exclusion: resources cannot be shared.
•Hold and wait:processes request resources incrementally, and hold on to
What they've got.
•No preemption: resources cannot be forcibly taken from processes.
•Circular wait: circular chain of waiting, in which each process is waiting for a
resource held by the next process in the chain.
Strategies for dealing with Deadlock :
•ignore the problem altogether
•detection and recovery
•avoidance by careful resource allocation
•prevention by structurally negating one of the four necessary conditions.
Deadlock Prevention :
Difference from avoidance is that here, the system itself is built in such a way that there are
no deadlocks. Make sure atleast one of the 4 deadlock conditions is never satisfied. This may
however be even more conservative than deadlock avoidance strategy.
Algorithm:
1.Start
2.Attacking Mutex condition : never grant exclusive access. but this may not be
possible for several resources.
62 | P a g e
3..Attacking preemption: not something you want to do.
4.Attacking hold and wait condition : make a process hold at the most 1 resource
at a time.make all the requests at the beginning. All or nothing policy. If you
feel,retry. eg. 2-phase locking
5.Attacking circular wait: Order all the resources. Make sure that the requests are issued in
the correct order so that there are no cycles present in the resource graph. Resources
numbered 1 ... n. Resources can be requested only in increasing
order. ie. you cannot request a resource whose no is less than any you may be holding.
6.Stop
Test Cases:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases :
63 | P a g e
P2: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases:
P3: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases
P4: Simulate Bankers algorithm for Deadlock Prevention for the following :
TestCases
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P5: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Case :
P6: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases:
P7: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases
65 | P a g e
P8: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases:
P9: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases:
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P10: Simulate Bankers algorithm for Deadlock Prevention for the following :
Test Cases:
VIVA QUESTIONS:
1. The Banker’s algorithm is used for ___________________.
2._________ is the situation in which a process is waiting on another process,which is also waiting on another
process ... which is waiting on the first process. None of the processes involved in this circular wait are making
progress.
3.what is safe state?
4.What are the conditions that cause deadlock?
5.How do we calculate the need for process?
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EXPERIMENT: 7a
NAME OF EXPERIMENT:
7) Simulate page replacement algorithms:
a) FIFO
AIM: Simulate FIFO page replacement algorithms.
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
THEORY:
FIFO algorithm:
The simpler page replacement algorithm is a FIFO algorithm. A FIFO replacement
algorithm associates with each page the time when that page was brought into memory.
When a page must be replace, the oldest page is chosen. We can create a FIFO queue to hold
all pages in memory. We replace the page at the head of the queue when a page is brought
into memory; we insert it at the tail of the queue.
7
0
1
2
7
7
7
0
0
3
0
4
2
3
0
2
2
2
4
4
4
0
0
3
3
3
2
1
1
4
0
0
0
3
2
1
2
0
0
2
2
3
3
0
1
7
0
1
0
7
7
7
1
1
1
0
0
3
2
2
2
1
ALGORITHM:
1.
2.
3.
4.
5.
6.
7.
8.
Start
Read the number of frames
Read the number of pages
Read the page numbers
Initialize the values in frames to -1
Allocate the pages in to frames in First in first out order.
Display the number of page faults.
stop
Test Case:
68 | P a g e
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1: Simulate FIFO(First in First out) Page Replacement algorithm that reads only even
no of pages in the range from 10 to 20 and reads the page numbers from the user.
Test Case1:
Test Case2:
P2: Simulate FIFO(First in First out) Page Replacement algorithm that reads only odd
no of pages in the range from 7 to 15 and reads the page numbers from the user.
Test Case1:
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Test Case2:
P3: Simulate FIFO(First in First out) Page Replacement algorithm that accepts random
page numbers. (Note: use rand() function)
Test case:
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P4: Simulate FIFO(First in First out) Page Replacement algorithm that reads only even
no of pages in the range 6 to 14 and accepts random page numbers. (Note: use and()
function)
Test Case1:
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Test Case2:
P5: Simulate FIFO(First in First out) Page Replacement algorithm that reads only odd
no of pages 11 to 23 and accepts random page numbers. (Note: use rand() function)
Test case1:
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Test Case 2:
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P6: Simulate FIFO(First in First out) Page Replacement algorithm that accepts the
page number from the user which is greater than 0 and less than 10.
Test Case:
P7: Simulate FIFO(First in First out) Page Replacement algorithm that accepts the
page number from the user which is greater than 10 and less than 20.
Test Case:
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P8: Simulate FIFO(First in First out) Page Replacement algorithm that reads a static
array of 10 page numbers.
Test Case:
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P9: Simulate FIFO(First in First out) Page Replacement algorithm that reads a static
array of 10 page numbers and loads only five pages out of the array in to the frames. .
Test Case:
P10: Simulate FIFO(First in First out) Page Replacement algorithm that accepts even
number of random page numbers . (Note: use rand() function)
Test case1:
Test Case2:
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VIVA QUESTIONS:
1.Define FIFO?
2.Which of the following statement
a)Multiprogramming
implies
b)Multi-user
does
not
imply
c) Multitasking does not imply
d)Multithreading implies multi-user
is
not true?
multitasking
multiprocessing
multiprocessing
3.Define page?
4.Define Frame?
5.Write advantages and dis-advantages of FIFO?
EXPERIMENT: 7b)
NAME OF EXPERIMENT:
7) Simulate page replacement algorithms:
b) LRU
AIM: Simulate LRU page replacement algorithms
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
ALGORITHM :
1.
2.
3.
4.
5.
6.
Start
Read the number of frames
Read the number of pages
Read the page numbers
Initialize the values in frames to -1
Allocate the pages in to frames by selecting the page that has not been used for the
longest period of time.
7. Display the number of page faults.
8. stop
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Test Cases:
PROBLEM DEFINITIONS DIFFERENT FROM JNTU TOPICS
P1: Simulate LRU(least recently used) Page Replacement algorithm that reads only even no
of pages in the range from 10 to 20 and reads the page numbers from the user.
Test Case1:
Test Case2:
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P2: Simulate LRU(least recently used) Page Replacement algorithm that reads only odd no
of pages in the range from 7 to 15 and reads the page numbers from the user.
Test Case1:
Test Case2:
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P3: Simulate LRU(least recently used) Page Replacement algorithm that accepts
random page numbers.
Test case:
P4: Simulate LRU(least recently used) Page Replacement algorithm that reads only
even no of pages in the range 6 to 14 and accepts random page numbers.
Test Case1:
Test Case2:
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P5: Simulate LRU(least recently used) Page Replacement algorithm that reads only
odd no of pages 11 to 23 and accepts random page numbers.
Test case1:
Test Case 2:
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P6: Simulate LRU(least recently used) Page Replacement algorithm that accepts the
page number from the user which is greater than 0 and less than 10.
Test Case:
P7: Simulate LRU(least recently used) Page Replacement algorithm that accepts the
page number from the user which is greater than 10 and less than 20.
Test Case:
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P8: Simulate LRU(least recently used) Page Replacement algorithm that reads a static
array of 10 page numbers.
Test Case:
P9: Simulate LRU(least recently used) Page Replacement algorithm that reads a static
array of 10 page numbers and loads only five pages out of the array in to the frames. .
Test Case:
P10: Simulate LRU(least recently used) Page Replacement algorithm that accepts even
number of random page numbers .
Test case1:
Test
Case2:
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VIVA QUESTIONS:
1.In which of the following page replacement polici es, Bolady’s
anomaly occurs?
(A)FIFO (B)LRU (C)LFU (D)SRU
2. Explain the difference between FIFO and LRU?
3.
The operating system manages ________.
1 Memory
2 Processor
3 Disk and I/O devices 4 All of the above
4. A program at the time of executing is called ________.
1 Dynamic program 2 Static program
3 Binded Program
4 A Process
5.The principle of locality of reference justifies the use of ________.
1 Virtual Memory
2 Interrupts
3 Main memory
4 Cache memory
EXPERIMENT: 7c
NAME OF EXPERIMENT:
7) Simulate page replacement algorithms:
c)LFU
AIM: Simulate LFU page replacement algorithms .
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS:
Turbo C/ Borland C.
ALGORITHM:
1.
2.
3.
4.
5.
6.
Start
Read the number of frames
Read the number of pages
Read the page numbers
Initialize the values in frames to -1
Allocate the pages in to frames by selecting the page that will not be used for the
longest period of time.
7. Display the number of page faults.
8. stop
Test Case:
Reference NO 4->
6 5 4
Reference NO 2->
2 5 4
Reference NO 3->
2 3 4
Reference NO 1->
2 3 1
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No.of page faults...6
VIVA QUESTIONS:
1. What is the full form of LRU?
2. Explain when page replacement occurs?
3. Which is the best page replacement alg ?why?
4. FIFO scheduling is ________..
5. Explain various page replacement algorithms?
6. what do u mean by page fault?
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EXPERIMENT: 8
NAME OF EXPERIMENT: Simulate Paging Technique of memory management
AIM: Simulate Paging Technique of memory management.
HARDWARE REQUIREMENTS: Intel based Desktop Pc
RAM of 512 MB
SOFTWARE REQUIREMENTS: Turbo C/ Borland C.
THEORY:
PAGING
Logical address space of a process can be noncontiguous; process
is allocated physical memory whenever the latter is available
Divide physical memory into fixed-sized blocks called frames
(size is power of 2, between 512 bytes and 8,192 bytes)
Divide logical memory into blocks of same size called pages
Keep track of all free frames
To run a program of size n pages, need to find n free frames and
load program
ALGORITHM:
1.
2.
3.
4.
5.
6.
Start
Read the number of pages
Read the page size
Allocate the memory to the pages dynamically in non contiguous locations.
Display the pages and their addresses.
stop
VIVA QUESTIONS:
1. The mechanism that bring a page into memory only when it is needed is called
_____________
2. What are the advantages and dis-advantages of paging?
3. Define external fragmentation?
4.__________________________ into blocks of same size called pages.
5._________________________ space of a process can be noncontiguous
6. What is page table?
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