Practical Solutions of

OSOperating Systems
For
B.Sc.(CS) Final Year — V SEMESTER

Prepared By:

MD Waseem Raza (MCA-TS SET)

Assistant Professor in Computers

NSV Degree Colleges,Jagtial

https://www.sucomputersforum.com
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SATAVAHANA UNIVERSITY
B.Sc (Computer Science) –III Year –V Semester
Operating Systems - PRACTICAL SOLUTIONS
1. A program to simulate the FCFS CPU scheduling algorithm.
PROGRAM:
#include<stdio.h>
#include<conio.h>
void main()
{
char pn[10][10];
int arr[10],bur[10],star[10],finish[10],tat[10],wt[10],i,n;
int totwt=0,tottat=0;
clrscr();
printf("Enter the number of processes:");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter the Process Name, Arrival Time & Burst Time:");
scanf("%s%d%d",&pn[i],&arr[i],&bur[i]);
}
for(i=0;i<n;i++)
{
if(i==0)
{
star[i]=arr[i];
wt[i]=star[i]-arr[i];
finish[i]=star[i]+bur[i];
tat[i]=finish[i]-arr[i];
}
else
{
star[i]=finish[i-1];
wt[i]=star[i]-arr[i];
finish[i]=star[i]+bur[i];
tat[i]=finish[i]-arr[i];
}
}
printf("\nPName Arrtime Burtime Start TAT Finish");
for(i=0;i<n;i++)
{
printf("\n%s\t%6d\t\t%6d\t%6d\t%6d\t%6d",pn[i],arr[i],bur[i],star[i],tat[i],finish[i]);
totwt+=wt[i]; tottat+=tat[i];
}
printf("\nAverage Waiting time:%f",(float)totwt/n);
printf("\nAverage Turn Around Time:%f",(float)tottat/n);
getch();
}
OUTPUT:
Input:
Enter the number of processes: 3

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Enter the Process Name, Arrival Time & Burst Time: 1 2 3
Enter the Process Name, Arrival Time & Burst Time: 2 5 6
Enter the Process Name, Arrival Time & Burst Time: 3 6 7
Output:
PName ArrTime Bur Time Start TAT Finish
1 2 3 2 3 5
2 5 6 5 6 4
3 6 7 6 7 10
Average Waiting Time: 3.333 Average Turn Around Time: 7.000

2. A program to simulate the SJF CPU scheduling algorithm.

PROGRAM:
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
int et[20],at[10],n,i,j,temp,st[10],ft[10],wt[10],ta[10];
int totwt=0,totta=0;
float awt,ata;
char pn[10][10],t[10];
clrscr();
printf("Enter the number of process:");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter process name, arrival time & execution time:");
flushall();
scanf("%s%d%d",pn[i],&at[i],&et[i]);
}
for(i=0;i<n;i++) for(j=0;j<n;j++)
{
if(et[i]<et[j])
{
temp=at[i]; at[i]=at[j]; at[j]=temp;
temp=et[i]; et[i]=et[j]; et[j]=temp; strcpy(t,pn[i]);
strcpy(pn[i],pn[j]);
strcpy(pn[j],t);
}
}
for(i=0;i<n;i++)
{
if(i==0) st[i]=at[i]; else st[i]=ft[i-1];
wt[i]=st[i]-at[i];
ft[i]=st[i]+et[i];
ta[i]=ft[i]-at[i]; totwt+=wt[i]; totta+=ta[i];
}
awt=(float)totwt/n;
ata=(float)totta/n;
printf("\nPname\tarrivaltime\texecutiontime\twaitingtime\ttatime");
for(i=0;i<n;i++)
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printf("\n%s\t%5d\t\t%5d\t\t%5d\t\t%5d",pn[i],at[i],et[i],wt[i],ta[i]);
printf("\nAverage waiting time is:%f",awt);
printf("\nAverage turn Around time is:%f", ata);
getch();
}

OUTPUT:
Input:
Enter the number of processes: 3
Enter the Process Name, Arrival Time & Burst Time: 1 4 6
Enter the Process Name, Arrival Time & Burst Time: 2 5 15
Enter the Process Name, Arrival Time & Burst Time: 3 6 11
Output:
Pname arrivaltime executiontime waitingtime tatime
1 4 6 0 6
3 6 11 4 15
2 5 15 16 31
Average Waiting Time: 6.6667 Average Turn Around Time: 17.3333

3. A program to simulate the priority CPU scheduling algorithm.

PROGRAM:
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
int et[20],at[10],n,i,j,temp,p[10],st[10],ft[10],wt[10],ta[10];
int totwt=0,totta=0;
float awt,ata;
char pn[10][10],t[10]; clrscr();
printf("Enter the number of process:");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter process name,arrivaltime,execution time & priority:");
flushall();
scanf("%s%d%d%d",pn[i],&at[i],&et[i],&p[i]);
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
if(p[i]<p[j])
{ temp=p[i];
p[i]=p[j];
p[j]=temp;
temp=at[i];
at[i]=at[j];
at[j]=temp;
temp=et[i];
et[i]=et[j];
et[j]=temp;
strcpy(t,pn[i]);
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strcpy(pn[i],pn[j]);
strcpy(pn[j],t);
}}
for(i=0;i<n;i++)
{
if(i==0)
{
st[i]=at[i];
wt[i]=st[i]-at[i];
ft[i]=st[i]+et[i];
ta[i]=ft[i]-at[i];
}
else
{
st[i]=ft[i-1];
wt[i]=st[i]-at[i];
ft[i]=st[i]+et[i];
ta[i]=ft[i]-at[i];
}
totwt+=wt[i]; totta+=ta[i];
}
awt=(float)totwt/n;
ata=(float)totta/n;
printf("\nPname\tarrivaltime\texecutiontime\tpriority\twaitingtime\ttatime"
); for(i=0;i<n;i++)
printf("\n%s\t%5d\t\t%5d\t\t%5d\t\t%5d\t\t%5d",pn[i],at[i],et[i],p[i],wt[i],ta[i]);
printf("\nAverage waiting time is:%f",awt);
printf("\nAverage Turn Around Time is:%f",ata); getch();
}

OUTPUT:
Input:
Enter the number of processes: 3
Enter the Process Name, Arrival Time, execution time & priority: 1 2 3 1
Enter the Process Name, Arrival Time, execution time & priority: 2 4 5 2
Enter the Process Name, Arrival Time, execution time & priority: 3 5 6 3
Output:
PnamearrivaltimeexecutiontimePrioritywaitingtimetatime

1 2 3 1 0 3
2 4 5 2 1 6
3 5 6 3 5 11
Average Waiting Time: 2.0000 Average Turn Around Time: 6.6667
4. A program to simulate the Round Robin CPU scheduling
algorithm.
PROGRAM:
#include<stdio.h> #include<conio.h> void main()
{
int et[30],ts,n,i,x=0,tot=0; char pn[10][10];
clrscr();
printf("Enter the no of processes:"); scanf("%d",&n);
printf("Enter the time quantum:"); scanf("%d",&ts);
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for(i=0;i<n;i++)
{
printf("enter process name & estimated time:"); scanf("%s %d",pn[i],&et[i]);
}
printf("The processes are:"); for(i=0;i<n;i++)
printf("process %d:%s\n",i+1,pn[i]); for(i=0;i<n;i++)
tot=tot+et[i]; while(x!=tot)
{
for(i=0;i<n;i++)
{
if(et[i]>ts)
{
x=x+ts;
printf("\n %s -> %d",pn[i],ts); et[i]=et[i]-ts;
}
else if((et[i]<=ts)&&et[i]!=0)
{
x=x+et[i];
printf("\n %s -> %d",pn[i],et[i]); et[i]=0;}
}

}
printf("\n Total Estimated Time:%d",x); getch();
}

OUTPUT:
Input:
Enter the no of processes: 2
Enter the time quantum: 3
Enter the process name & estimated time: p1 12
Enter the process name & estimated time: p2 15
Output:
p1 ->3
p2 ->3
p1 ->3
p2 ->3
p1 ->3
p2 ->3
p1 ->3
p2 ->3
p2 ->3
Total Estimated Time: 27

5. A program to simulate the MVT.

PROGRAM:
#include<stdio.h>
#include<conio.h>
void main()
{
int m=0,m1=0,m2=0,p,count=0,i;
clrscr();
printf("Enter the memory capacity:");
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scanf("%d",&m);
printf("Enter the no of processes:");
scanf("%d",&p);
for(i=0;i<p;i++)
{
printf("\nEnter memory req for process%d: ",i+1);
scanf("%d",&m1);
count=count+m1; if(m1<=m)
{
if(count==m)
{
printf("There is no further memory remaining:");
}
else
{
printf("The memory allocated for process%d is: %d ",i+1,m1); m2=m-m1;
printf("\nRemaining memory is: %d",m2); m=m2;
}
}
else
{
printf("Memory is not allocated for process%d",i+1);
}
printf("\nExternal fragmentation for this process is:%d",m2);
}
getch();
}

OUTPUT:
Input:
Enter the memory capacity: 80
Enter no of processes: 2
Enter memory req for process1: 23
Output:
The memory allocated for process1 is: 23
Remaining memory is: 57
External fragmentation for this process is: 57
Enter memory req for process2: 52
The memory allocated for process2 is: 52
Remaining memoryis: 5
External fragmentation for this process is: 5

6. A Program to simulate the MFT.

PROGRAM:
#include<stdio.h> #include<conio.h> void main()
{
int m,p,s,p1;
int m1[4],i,f,f1=0,f2=0,fra1,fra2,s1;
clrscr();
printf("Enter the memory size:");
scanf("%d",&m);
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printf("Enter the no of partitions:");
scanf("%d",&p);
s=m/p;
printf("Each partn size is:%d",s);
printf("\nEnter the no of processes:");
scanf("%d",&p1);
for(i=0;i<p1;i++)
{
printf("\nEnter the memory req for process%d:",i+1);
scanf("%d",&m1[i]);
if(m1[i]<=s)
{
printf("\nProcess is allocated in partition%d",i+1);
fra1=s-m1[i];
printf("\nInternal fragmentation for process is:%d",fra1);
f1=f1+fra1;
}
else
{
printf("\nProcess not allocated in partition%d",i+1);
s1=m1[i]-s;
fra2=s-s1;
f2=f2+fra2;
printf("\nExternal fragmentation for partition is:%d",fra2);
}
}
printf("\nProcess\tmemory\tallocatedmemory");
for(i=0;i<p1;i++)
printf("\n%5d\t%5d\t%5d",i+1,s,m1[i]);
f=f1+f2;
printf("\nThe tot no of fragmentation is:%d",f);
getch();
}

OUTPUT:
Input:
Enter the memory size: 80
Enter the no of partitions: 4
Each partition size: 20
Enter the number of processes: 2
Enter the memory req for process1: 18
Output:
Process1 is allocated in partn1
Internal fragmentation for process1 is: 2
Enter the memory req for process2: 22
Process is not allocated in partition 2
External fragmentation for process2 is: 18
Process memory allocated
1 20 18
2 20 22
The tot no of fragmentation is: 20

Prepared By: Waseem Raza MDMCA-TSSET 8

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7. A program to simulate the Bankers Algorithm for Deadlock
Avoidance.
PROGRAM:
#include<stdio.h>
#include<conio.h>
void main()
{
int n,r,i,j,k,p,u=0,s=0,m;
int block[10],run[10],active[10],newreq[10];
int max[10][10],resalloc[10][10],resreq[10][10];
int totalloc[10],totext[10],simalloc[10];
clrscr();
printf("Enter the no of processes:");
scanf("%d",&n);
printf("Enter the no of resource classes:");
scanf("%d",&r);
printf("Enter the total existed resource in each class:");
for(k=1;k<=r;k++)
scanf("%d",&totext[k]);
printf("Enter the allocated resources:");
for(i=1;i<=n;i++)
for(k=1;k<=r;k++) scanf("%d",&resalloc);
printf("Enter the process making the new request:");
scanf("%d",&p);
printf("Enter the requested resource:");
for(k=1;k<=r;k++)
scanf("%d",&newreq[k]);
printf("Enter the processes which are n blocked or running:");
for(i=1;i<=n;i++)
{
if(i!=p)
{
printf("process %d:\n",i);
scanf("%d%d",&block[i],&run[i]);
}
}
block[p]=0;
run[p]=0;
for(k=1;k<=r;k++)
{

j=0;

for(i=1;i<=n;i++)
{
totalloc[k]=j+resalloc[i][k];
j=totalloc[k];
}
}
for(i=1;i<=n;i++)
{
if(block[i]==1||run[i]==1) active[i]=1;
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else active[i]=0;
}
for(k=1;k<=r;k++)
{
resalloc[p][k]+=newreq[k];
totalloc[k]+=newreq[k];
}
for(k=1;k<=r;k++)
{
if(totext[k]-totalloc[k]<0)
{
u=1;break;
}
}
if(u==0)
{
for(k=1;k<=r;k++)
simalloc[k]=totalloc[k];
for(s=1;s<=n;s++)
for(i=1;i<=n;i++)
{
if(active[i]==1)
{ j=0;
for(k=1;k<=r;k++)
{
if((totext[k]-simalloc[k])<(max[i][k]-resalloc[i][k]))
{
j=1;break;
}}}
if(j==0)
{
active[i]=0;
for(k=1;k<=r;k++)
simalloc[k]=resalloc[i][k];
}
}
m=0;
for(k=1;k<=r;k++)
resreq[p][k]=newreq[k];
printf("Deadlock willn't occur");
}
else
{
for(k=1;k<=r;k++)
{
resalloc[p][k]=newreq[k];
totalloc[k]=newreq[k];
}
printf("Deadlock will occur");
}
getch();
}

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OUTPUT:
Input:
Enter the no of resources: 4
Enter the no of resource classes: 3
Enter the total existed resources in each class: 3 2 2
Enter the allocated resources: 1 0 0 5 1 1 2 1 1 0 0 2
Enter the process making the new request: 2
Enter the requested resource: 1 1 2
Enter the processes which are n blocked or running: Process 1: 1 2
Process 3: 1 0
Process 4: 1 0
Output:
Deadlock will occur

8. A program to simulate Bankers Algorithm for Deadlock

Prevention.

PROGRAM:
#include<stdio.h>
#include<conio.h>
void main()
{
int cl[10][10],al[10][10],av[10],i,j,k,m,n,ne[10][10],flag=0;
clrscr();
printf("\nEnter the matrix");
scanf("%d %d",&m,&n);
printf("\nEnter the claim matrix:");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&cl[i][j]);
}}
printf("\nEnter allocated matrix:");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&al[i][j]);
}}
printf("\nThe need matrix:\n"); for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
ne[i][j]=cl[i][j]-al[i][j];
printf("\t%d",ne[i][j]);
}
printf("\n");
}
printf("\nEnter avaliable matrix");
for(i=0;i<n;i++)
scanf("%d",&av[i]);
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printf("Claim matrix:\n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("\t%d",cl[i][j]);
}
printf("\n");
}
printf("\n Allocated matrix:\n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("\t%d",al[i][j]);
}
printf("\n");
}
printf("Available matrix:\n");
for(i=0;i<n;i++)
{
printf("%d\t",av[i]);
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
if(av[j]>=ne[i][j])
flag=1;
else
flag=0;
}}
if(flag==0)
printf("Unsafe State");
else
printf("Safe State");
getch();
}
OUTPUT:
Input:
Enter the claim matrix:3 2 2 6 1 3 3 1 4 4 2 2
Enter allocated matrix:1 0 0 5 1 1 2 1 1 0 0 2
The need matrix:
2 22
1 02
1 03
4 20
Enter available matrix1 1 2
Output:
Claim matrix: 3 2 2
6 13
3 14
4 22
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Allocated matrix: 1 0 0
5 11
2 11
0 02
Available matrix: 1 1 2 Safe State

9. A program to simulate FIFO Page Replacement Algorithm.

PROGRAM:
#include<stdio.h> #include<conio.h> void main()
{
int a[5],b[20],n,p=0,q=0,m=0,h,k,i,q1=1;
char f='F';
clrscr();
printf("Enter the Number of Pages:");
scanf("%d",&n);
printf("Enter %d Page Numbers:",n);
for(i=0;i<n;i++)
scanf("%d",&b[i]);
for(i=0;i<n;i++)
{
if(p==0)
{
if(q>=3) q=0;
a[q]=b[i]; q++;
if(q1<3)
{
q1=q;
}
}
printf("\n%d",b[i]);
printf("\t");
for(h=0;h<q1;h++)
printf("%d",a[h]);
if((p==0)&&(q<=3))
{
printf("-->%c",f);
m++;
}
p=0;
for(k=0;k<q1;k++)
{
if(b[i+1]==a[k]) p=1;
}
}
printf("\nNo of faults:%d",m); getch();
}

OUTPUT:
Input:
Enter the Number of Pages: 12
Enter 12 Page Numbers:
232152453252
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Output:
2 2-> F
3 23-> F
2 23
1 231-> F
5 531-> F
2 521-> F
4 524-> F
5 524
3 324-> F
2 324
5 354-> F
2 352-> F
No of faults: 9

10. A program to simulate LRU Page Replacement Algorithm.

PROGRAM:
#include<stdio.h>
#include<conio.h>
void main()
{
int g=0,a[5],b[20],p=0,q=0,m=0,h,k,i,q1=1,j,u,n;
char f='F';
clrscr();
printf("Enter the number of pages:");
scanf("%d",&n);
printf("Enter %d Page Numbers:",n);
for(i=0;i<n;i++)
scanf("%d",&b[i]);
for(i=0;i<n;i++)
{
if(p==0)
{
if(q>=3)
q=0;
a[q]=b[i];
q++;
if(q1<3)
{
q1=q;
} }
printf("\n%d",b[i]);
printf("\t");
for(h=0;h<q1;h++)
printf("%d",a[h]);
if((p==0)&&(q<=3))
{
printf("-->%c",f);
m++;
}
p=0;
g=0;
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if(q1==3)
{
for(k=0;k<q1;k++)
{
if(b[i+1]==a[k])
p=1;
}
for(j=0;j<q1;j++)
{
u=0;
k=i;
while(k>=(i-1)&&(k>=0))
{
if(b[k]==a[j])
u++;
k--;
}
if(u==0)
q=j;
}
}
else
{
for(k=0;k<q;k++)
{
if(b[i+1]==a[k])
p=1;
} } }
printf("\nNo of faults:%d",m); getch();
}
OUTPUT:
Input:
Enter the Number of Pages: 12
Enter 12 Page Numbers: 2 3 2 1 5 2 4 5 3 2 5 2
Output:
2 2-> F
3 23-> F
2 23
1 231-> F
5 251-> F
2 251
4 254-> F
5 254
3 354-> F
2 352-> F
5 352
2 352
No of faults: 7

Prepared By: Waseem Raza MDMCA-TSSET 15