Operating System (2030106404) 4th Sem
Practical – 1
Aim:- Install and Test different types of Operating system.
For downloading Ubuntu we need ISO file of Ubuntu .we can get ISO file
on official website of Ubuntu and also you can choose which version you
want to download.
After downloading ISO file of Ubuntu we have two ways to install .
● We can install Ubuntu in dual boot using boot able pendrive.
● We can install Ubuntu virtualy using any virtual machine.
o In both ways we have to follow following steps:
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Operating System (2030106404) 4th Sem
Step 1: Choose your language.
Step 2: you can change Accessibility or skip it.
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Operating System (2030106404) 4th Sem
Step 3: Select your keybord layout.
Step 4: To improve installion you can connect to internet or you can install without
internet.
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Step 5: choose install Ubuntu, it will install Ubuntu permenetly.
Step 6: choose Interactive installation.
Step 7: You can choose that you need extra tools or not.
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Programming with Java(2030106405) 4th Sem
Step 8: Select following option.
Step 9: choose Erase disk option.
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Step 10: Create you account by which you will use Ubuntu.
Step 11: Select your Timezone.
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Step 12: Click on install button.
Final Step: Restart your system.
And now Ubuntu has been Succesfully Install.
Utsav Lankapati(2327010601077) 7
Programming with Java(2030106405) 4th Sem
Practical – 2
Aim:- Study Various Process Scheduling algorithms.
● There are four types of Scheduling Algorithms.
o FCFS : First come, Frist Served
o SJF : Shortest Job First
o SRTN : Shortest Remaining Time Next
o RR : Round Robin Scheduling
● Explain FCFS Algorithm in detail with example.
❖ Selection criteria: The process that request first is served first. It means
that processes Are service in the exact order as they come.
❖ Decision Mode: Non-pre-emptive: Once a process is selected, it runs until
it blocks for An I/O or some event, or it terminates.
❖ Implementation: This strategy can be implemented by using sorted FIFO
queue. FIFO means First In First Out. When first process enters the
system, it starts its execution. All other processes are appended in a
queue. When CPU becomes free, a process from the first position in a
queue is selected to run.
Example: Consider the following set of four processes. Their arrival time and time
required To complete the execution are given in following table. Consider all
time values in milliseconds
Process Arrival Time Time required
for
completion(∆T
)
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Programming with Java(2030106405) 4th Sem
P0 0 10
P1 1 6
P2 3 2
P3 5 4
❖ Gantt Chart:
Initially only process P0 is present and it allowed to run. But, when P0
completes, all other processes are present .So, next process (P1) from ready
queue is selected and allowed to run till its completes. This is repeated till all
processes complete their execution.
❖ Statistics:
Process Arrival Completio Finish Turnaround Waiting
Time(T0) n Time(T1) Time Time
Time(∆T) (TAT=T1-T0) (TAT-(∆T)
P1 0 10 10 10 0
P2 1 6 16 15 9
P3 3 2 18 15 13
P3 5 4 22 17 13
Average Turn-around Time: (10+15+15+17)/4 = 57/4 =14.25 ms
Average Waiting Time : (0+19+13+13)/4 = 35/4 = 8.75 ms
❖ Advantages:
Simple, Fair, No starvation.
Easy to understand, Easy to implement
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Programming with Java(2030106405) 4th Sem
❖ disadvantages:
Not efficient. Average waiting time is too
high. Convoy effect is possible.
CPU utilization may be too low.
● Explain SJF Algorithm in detail with example.
❖ Selection criteria: The process, that requires shortest time to
complete execution, is served first.
❖ Decision Mode: Non-preemptive: Once a process is selected, it runs until it
blocks for an I/O or some event, or it terminates.
❖ Implementation: This strategy can be implemented by using sorted FIFO
queue. All processes in a queue are sorted in ascending order based in
their required CPU bursts. When CPU becomes free, a process from the
first position in a queue is selected to run.
Example: Consider the following set of four processes. Their arrival time and
time required to complete the execution are given in following table. Consider
all time values in milliseconds.
Process Arrival Time Time required
for
completion(∆T
)
P0 0 10
P1 1 6
P2 3 2
P3 5 4
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Programming with Java(2030106405) 4th Sem
❖ Gantt Chart:
Initially only process P0 is present and it allowed to run. But, when P0 completes, all
other processes are present. So, process with shortest CPU burst (P2) is selected and
allowed to run till it completes. Whenever more than one process in available, such type
of decision is taken. Taken procedure is repeated till the processes complete their
execution.
❖ Statistics:
Process Arrival Completio Finish Turnaroun Waiting
Time(T0) n Time Time(T1) d Time Time
(TAT=T1-T0) (TAT-∆T)
(∆T)
P0 0 10 10 10 0
P1 1 6 22 21 15
P2 3 2 12 9 7
P3 5 4 16 11 7
Average Turn-around Time: (10+21+9+11)/4 = 51/4 =12.75
ms Average Waiting Time : (0+15+7+7)/4 = 29/4 = 7.25 ms
❖ Advantages:
Less waiting time
Good response for short processes.
❖ Disadvantages:
it is difficult to estimate time required to complete execution.
Starvation is possible for long process. Long process may wait
forever.
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Programming with Java(2030106405) 4th Sem
● Explain SRTN Algorithm in detail with example.
❖ Selection criteria: The process, whose remaining run time is shortest, is
served first. This is a preemptive of SJF scheduling.
❖ Decision Mode: Preemptive: When a new process arrives, its time is
compared to the current process remaining time. If the new job needs
less time to finish than the current process, the current process is
suspended and the new job started.
❖ Implementation: This strategy can also be implemented by using sorted
FIFO queue. All processes in a queue are sorted in ascending order based
on their remaining run time. When CPU becomes free, a process from the
first position in a queue is selected to run.
Example: Consider the following set of four processes. Their arrival time and time
required to complete the execution are given in following table. Consider all
time values in milliseconds.
Process Arrival Time(T0) Time required
for
completion(∆
T)
P0 0 10
P1 1 6
P2 3 2
P3 5 4
❖ Gantt Chart:
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Programming with Java(2030106405) 4th Sem
Initially only process P0 is present and it is allowed to run. But, when P1 comes, it has
shortest remaining run time. So, P0 is preempted and P1 is allowed to run, Whenever
new process comes or current process blocks, such type of decision is taken. This
procedure is repeated till all processes complete their execution.
Statistics:
Process Arrival Completio Finish Turnaround Waiting
Time(T0) n Time Time(T1) Time(TAT=T1- Time
(∆T) T0) (TAT-
∆T)
P0 0 10 22 22 12
P1 1 6 9 8 2
P2 3 2 5 2 0
P3 5 4 13 8 4
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Operating System (2030106404) 4th Sem
Average turn-around Time: (22+8+2+8)/4 = 40/4 = 10.0 ms
Average Waiting Time : (12+2+0+4)/4 = 18/4 = 4.5 ms
❖ Advantages:
Less waiting time.
Quite good response for short processes.
❖ Disadvantages:
Again it is difficult to estimate remaining time necessary to complete execution.
Starvation is possible for long process. Long process may wait forever.
Context switch overhead is there.
● Explain Round Robin Algorithm in detail with example.
❖ Selection criteria: The process that requests first is served first. It means
that processes Are served in the exact order as they come. The selection
criteria are same as that of FCFS scheduling.
❖ Decision mode: Pre-emptive: each selected process is assigned a time
interval. Called time quantum or time slice. Process is allowed to run
only of this time interval.
o Here, one of two things is possible: first process needs CPU burst less than
time quantum in this case, process will voluntarily release CPU and it will
be moved to the end of the queue. Second, process needs CPU burst
longer than time quantum. In this case, process will be running at the end
of time quantum. Now, it will be pre-empted and move to the end of the
queue.
o Here, length of time quantum is critical to determine.
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Operating System (2030106404) 4th Sem
❖ Implementation: This strategy can be implemented by using circular FIFO
queue. When new process comes, or process releases CPU, or process is pre-
empted, it is moved to the end of the queue. When CPU becomes free, a
process from the first position in a queue is selected to run.
Example: Consider the following set of four processes. Their arrival time and time
required to complex the execution are give in following table. All time values
are in milliseconds.
Consider that time quantum is of 4 ms, and context switch overhead is of 1 ms.
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Operating System (2030106404) 4th Sem
Process Arrival time(T0) Time required
for
complex(∆T)
P1 0 10
P2 1 6
P3 3 2
P4 5 4
❖ Gantt Chart:
At 4 ms, process P0 completes its time quantum. So it is pre-empted and another
process p1 is allowed to run. At 12 ms, process p2 voluntarily releases CPU, and
another process p0 is selected to run. 1 ms is wasted on each context switch as
overhead. This procedure is repeated till all processes complete their execution.
❖ Statistics:
Process Arrival Completio Finish time Turnaround Waiting
time n time (T1) time time
(T0) (∆T) (TAT=T1-T0) (TAT-
∆T)
P1 0 10 28 28 18
P2 1 6 25 24 18
P3 3 2 12 9 7
P4 5 4 22 17 13
Average turnaround time: (28+24+9+17)/4 = 78/4 = 19.5
ms Average waiting time: (18+18+7+13)/4 = 56/4 = 14.0
ms
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Operating System (2030106404) 4th Sem
❖ Advantages:
One of the oldest, simplest, fairest and most widely used algorithms.
❖ Disadvantages:
Context switch overhead is there.
Determination of time quantum is too critical. If it is too short, it causes frequent
context switches and lowers CPU efficiency. If it is too long, it causes poor response
for short interactive processes.
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Operating System (2030106404) 4th Sem
Practical – 3
Aim:- Test and run UNIX commands.
1. pwd : (print working directory)
2. cd : (change directory)
3. mkdir : (make directory)
4. rmdir : (remove directory)
5. cal : (calendar)
6. date : (show date)
7. echo : (print)
8. passwd : (change password)
9. who : (display user)
10. tty : (display user associated with terminal)
11. man : (print entries)
12. ps : (process ststus)
13. time : (show time)
14. kill : (kill process)
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Operating System (2030106404) 4th Sem
Output:-
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Operating System (2030106404) 4th Sem
Practical-4
Aim: - Write a shell script program to read and display content of
file.
Code:-
echo “Enter your File name”
read file
cat $file
output:-
Utsav Lankapati(2327010601077) 20
Operating System (2030106404) 4th Sem
Practical-5
Aim:- Write a shell script program to perform mathematical
operations (+, -, *, /).
p_5.sh file code :
echo "Enter Two Number:"
read a
read b
sum=`expr $a + $b`
echo Addition = $sum
sub=`expr $a - $b`
echo Subtraction = $sub
mul=`expr $a \* $b`
echo Multiplication = $mul
div=`expr $a / $b`
echo Division = $div
Output:-
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Operating System (2030106404) 4th Sem
Practical-6
Aim:- Write a shell script program to find odd or even numbers.
p_6.sh Code
echo "Enter any number:"
read n
ans=`expr $n % 2`
if [ $ans -eq 0 ]
then
echo "$n is even"
else
echo "$n is odd"
fi
Output:-
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Operating System (2030106404) 4th Sem
Practical-7
Aim:- Write a shell script program to find the maximum of three
numbers.
P_7.sh file code:
echo "Enter Any Three Numbers:"
read a
read b
read c
if [ $a -gt $b ] && [ $a -gt $c ]
then
echo "$a is maximum"
elif [ $b -gt $a ] && [ $b -gt $c ]
then
echo "$b is maximum"
else
echo "$c is maximum"
fi
Output:-
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Operating System (2030106404) 4th Sem
Practical-8
Aim:- Write a shell script program to find the factorial of a given
number.
p8.sh file:
echo "enter any number:"
read n
fact=1
while [ $n -gt 1 ]
do
fact=`expr $fact \* $n`
n=`expr $n - 1`
done
echo "Factorial of given number is : "$fact
Output:-
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Operating System (2030106404) 4th Sem
Practical-9
Aim:- Write a shell script program to concatenate the contents
of two files into a third file.
code in p_9.sh file
echo "Enter two files name: "
read a
read b
cat $a>>intro
cat $b>>intro
echo "Concatation oF 2 files into 3rd file: "
cat intro
Output:-
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Operating System (2030106404) 4th Sem
Practical-10
Aim:- Write a shell script program to generate Fibonacci
numbers from 1 to N.
code in p_10.sh file:
echo "Enter how long you want to genrete series:"
read n
a=0
b=1
i=3
echo "Fabnoci series up to $n terms:"
echo $a
echo $b
while [ $i -le $n ]
do
c=`expr $a + $b`
echo $c
a=$b b=$c
i=`expr $i + 1`
done
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Output:-
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Operating System (2030106404) 4th Sem
Practical-11
Aim:- Write a shell script program to read a character from a
user. Determine whether it is uppercase letter, lowercase letter
or digit.
Code in p_11.sh file:
echo "Enter a character:"
read c
if [[ ${#c} -ne 1 ]]; then
echo "Error: Please enter a single character only!"
exit 1
fi
if [[ "$c" =~ ^[A-Z]$ ]]; then
echo "uppercase"
elif [[ "$c" =~ ^[a-z]$ ]]; then
echo "lowercase"
elif [[ "$c" =~ ^[0-9]$ ]]; then
echo "digit"
else
echo "special character"
fi
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Output:-
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Operating System (2030106404) 4th Sem
Practical-12
Aim:- Write a shell script program to find the sum of all the
individual digits in a given 5-digit number.
Code in p_12.sh file:
echo "Enter a number:"
read n
sum=0
while [ $n -gt 0 ]
do
mod=`expr $n % 10`
sum=`expr $sum + $mod`
n=`expr $n / 10`
done
echo "the Sum of the given number is : " $sum
Output:-
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Operating System (2030106404) 4th Sem
Practical-13
Aim:- Write a shell script program to check whether string is
palindrome or nt
P13.sh code:
echo "Enter a string:"
read str
rev=$(echo "$str" | rev)
if [ "$(echo "$str" | tr '[:upper:]' '[:lower:]')" = "$(echo "$rev" | tr '[:upper:]'
'[:lower:]')" ]; then
echo "$str is a palindrome."
else
echo "$str is not a palindrome."
fi
Output:-
Utsav Lankapati(2327010601077) 31
Operating System (2030106404) 4th Sem
Practical-14
Aim:- Write a shell script program to copy source file to
destination file.
Code in p_14.sh :
echo "Enter a source file name:"
read sf
echo "Enter the destination file name:"
read df
cp "$sf" "$df"
echo "Contents of the source file:"
cat "$sf"
Output:-
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Operating System (2030106404) 4th Sem
Practical-15
Aim:- Write a shell script program to generate a mark sheet.
Calculate grades of students also.
Code in p_15.sh
#!/bin/bash
echo "Students marks list"
echo "* * * * * * *"
echo "Enter the student's Name"
read name
echo "Enter the register number"
read reg_no
echo "Enter the marks of sub 1"
read sub1
echo "Enter the marks of sub 2"
read sub2
echo "Enter the marks of sub 3"
read sub3
echo "Enter the marks of sub 4"
read sub4
echo "Enter the marks of sub 5"
read sub5
# Calculate total marks
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Operating System (2030106404) 4th Sem
total=$((sub1 + sub2 + sub3 + sub4 + sub5))
# Calculate average as percentage (integer only)
avg=$(((total * 100) / 500))
echo "* * * * * * *"
echo "Marks List"
echo "* * * * * * *"
echo "Student's Name : $name"
echo "Register number : $reg_no"
echo "Sub 1 marks : $sub1"
echo "Sub 2 marks : $sub2"
echo "Sub 3 marks : $sub3"
echo "Sub 4 marks : $sub4"
echo "Sub 5 marks : $sub5"
echo "Total : $total"
echo "Average (%) : $avg"
# Grade Calculation
if [ "$avg" -ge 35 ]; then
echo "Result: Pass"
if [ "$avg" -ge 90 ]; then
echo "Grade: S"
elif [ "$avg" -ge 80 ]; then
echo "Grade: A"
elif [ "$avg" -ge 70 ]; then
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Operating System (2030106404) 4th Sem
echo "Grade: B"
elif [ "$avg" -ge 60 ]; then
echo "Grade: C"
elif [ "$avg" -ge 50 ]; then
echo "Grade: D"
else
echo "Grade: E"
fi
else
echo "Result: Fail"
fi
echo "* * * * * * *"
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Output:-
Utsav Lankapati(2327010601077) 36
