os programs

PART-B

1.a) Round Robin technique.b) Shortest Job First.c) First Come First Serve.d) Priority Scheduling.

2. Simulate all file allocation strategies.a) Sequentialb) Indexedc) Linked.

3. Simulate a) Multiprogramming with variable tasksb) Multiprogramming with fixed tasks.

4. Simulate all File organization techniques.a) Single level directoryb) Two levelc) Hierarchical

d) DAG5. Simulate banker’s algorithm for deadlock prevention.

6.a) First in first out.b) Least recently used.c) Least frequently used.

---------------------------------------- x x x -----------------------------------------

SIMULATE THE FOLLOWING CPU SCHEDULING ALGORITHMS

a) FIRST COME FIRST SERVE

#include<stdio.h>#include<conio.h>void main(){ int s[15],w[15],f[15],n,i,j,temp;

char p[15],t; clrscr( ); printf("Enter no of process:"); scanf("%d",&n); printf("Enter %d names:"); scanf("%s",p); printf("Enter %d service times:",n); for(i=0;i<n;i++) scanf("%d",&s[i]); f[0]=s[0]; for(i=1;i<n;i++) f[i]=f[i-1]+s[i]; for(i=0;i<n;i++) w[i]=f[i]-s[i]; printf("\nNAMES:\t"); for(i=0;i<n;i++)

{ printf("\t%c",p[i]);}printf("\nSERVICE TIMES:");for(i=0;i<n;i++)

{printf("\t%d",s[i]);

}printf("\nFINISH TIMES:");for(i=0;i<n;i++)

{ printf("\t%d",f[i]); }

printf("\nWAIT TIMES:");for(i=0;i<n;i++)

{printf("\t%d",w[i]);

}getch();}

INPUT:

Enter no of process: 5

Enter 5 names: ABCDE

Enter 5 service times: 2 3 5 4 9

OUTPUT:

NAMES: A B C D E

SERVICE TIMES: 2 3 5 4 9

FINISH TIMES: 2 5 10 14 23

WAIT TIMES: 0 2 5 10 14

b) SHORTEST JOB FIRST

#include<stdio.h>#include<conio.h>void main(){

int s[15],w[15],f[15],n,i,j,temp;char p[15],t;clrscr();printf("Enter no of process:");scanf("%d",&n);printf("Enter %d names:",n);scanf("%s",p);printf("Enter %d service times:",n);for(i=0;i<n;i++)scanf("%d",&s[i]);for(j=0;j<n;j++){

for(i=j+1;i<n;i++){

if(s[j]>s[i]){

//SEVICE TIMEStemp=s[j];s[j]=s[i];s[i]=temp;//NAMESt=p[j];p[j]=p[i];p[i]=t;

}}

}f[0]=s[0];for(i=1;i<n;i++)f[i]=f[i-1]+s[i];for(i=0;i<n;i++)w[i]=f[i]-s[i];printf("\nNAMES:\t");for(i=0;i<n;i++)

{printf("\t%c",p[i]);

}printf("\nSERVICE TIMES:");for(i=0;i<n;i++){

printf("\t%d",s[i]);}printf("\nFINISH TIMES:");for(i=0;i<n;i++){

printf("\t%d",f[i]);}printf("\nWAIT TIMES:");for(i=0;i<n;i++){

printf("\t%d",w[i]);}getch();

}

INPUT:

Enter no of process: 5

Enter 5 names: ABCDE


OUTPUT:

NAMES: D E A B C



WAIT TIMES: 0 1 3 8 14

c)PRIORITY SCHEDULING


int s[15],w[15],f[15],n,i,j,temp,pr[15];char p[15],t;clrscr();printf("Enter no of process:");scanf("%d",&n);printf("Enter %d names:",n);scanf("%s",p);printf("Enter %d service times:",n);for(i=0;i<n;i++)scanf("%d",&s[i]);printf("Enter %d priorities:",n);for(i=0;i<n;i++)scanf("%d",&pr[i]);for(j=0;j<n;j++){

for(i=j+1;i<n;i++){

if(pr[j]>pr[i]){

//PRIORITEStemp=pr[j];pr[j]=pr[i];pr[i]=temp;//NAMESt=p[j];p[j]=p[i];p[i]=t;//SEVICE TIMEStemp=s[j];s[j]=s[i];s[i]=temp;

}}

}f[0]=s[0];for(i=1;i<n;i++)

f[i]=f[i-1]+s[i];for(i=0;i<n;i++)w[i]=f[i]-s[i];printf("\nNAMES:\t");for(i=0;i<n;i++){

printf("\t%c",p[i]);}printf("\nSERVICE TIMES:");for(i=0;i<n;i++){

printf("\t%d",s[i]);}printf("\nPRIORITIES:");for(i=0;i<n;i++){

printf("\t%d",pr[i]);}printf("\nFINISH TIMES:");for(i=0;i<n;i++){

printf("\t%d",f[i]);}printf("\nWAIT TIMES:");for(i=0;i<n;i++){

printf("\t%d",w[i]);}getch();

}

INPUT:Enter no of process:5

Enter 5 names: ABCDEEnter 5 service times: 2 6 5 3 4Enter 5 priorities: 2 3 6 7 1

OUTPUT:NAMES: E A B C D


PRIORITIES: 1 2 3 6 7


WAIT TIMES: 0 4 6 12 1

d)ROUND ROBIN TECHNIQUE


int s[15],a[10],w[15],t[15],n,i,j,flag=1,count=0,k,ts;clrscr();printf("Enter no of process:");scanf("%d",&n);printf("Enter %d service times:",n);for(i=1;i<=n;i++)scanf("%d",&s[i]);for(i=1;i<=n;i++)a[i]=s[i];printf("\n enter the timeslice:");scanf("%d",&ts);i=1;while(flag){

for(i=1;i<=n;i++) { f(s[i]==0) ontinue; f(s[i]<=ts)

{count=count+s[i];a[i]=count-a[i];

t[i]=count; s[i]=0;

} else

{ s[i]=s[i]-ts; count=count+ts;

} } flag=0;

for(i=1;i<=n;i++) {

if(s[i]!=0) {

flag=1; break;

} }

}printf("\nprocess\t st\twt\t tat");for(i=1;i<=n;i++)printf("\n%d\t%d\t%d\t%d",i,a[i],w[i],t[i]);getch();

}

INPUT:Enter no of process: 5


enter the timeslice: 3

OUTPUT:

process st wt tat

1 6 12 18

2 4 15 19

3 3 6 9

4 8 19 27

5 7 21 28

SIMULATE ALL PAGE REPLACEMENT ALGORITHMS

a)FIRST IN FIRST OUT

#include <stdio.h>#include<conio.h>void main( ){ int r[15],ff=0,n,f[3]={0,0,0},k,l,x=0,y=0,p=0,a;

clrscr();printf(“enter the no of referencepages;”);scanf(“%d”,&n);printf(“enter the frames”);scanf(%d”,&ff);for( k=0; k<n;k++){

scanf(“%d”,&r[k]);}for(l=0;l<n;l++){

for(k=0;k<ff;k++){

if(f[k]==r[l]) x=1;

} if(x==0) { f[y]=r[l]; for(a=0;a<ff;a++) { printf(“%d\t”,f[a]) } printf(“\n”); y++; p+=1; } if(x==1) x=0; if(y==ff) y=0;

}p=p-ff;printf(“no of page faults %d”,p);getch( );

}INPUT:

Enter the no of reference paes 12Enter the frames 32 3 2 1 5 2 4 5 3 2 5

OUTPUT:2 0 02 3 02 3 1 5 3 15 2 15 2 43 2 43 5 4 3 5 2

no of page faults 6

b)LEAST RECENTLY USED

#include <stdio.h>#include<conio.h>void main( ){ int r[15],n,f[5]={0,0,0,0,0},k,l,x=0,y=0,p=0,a,s,fr,d,b[5]={0,0,0,0,0}, c[5]={0,0,0,0,0};

clrscr();printf(“enter the no of referencepages;”);scanf(“%d”,&n);printf(“enter the frames”);scanf(%d”,&fr);for( k=0; k<n;k++){

scanf(“%d”,&r[k]);}for(l=0;l<n;l++){

for(k=0;k<fr;k++){

if(f[k]==r[l]) x=1;

}if(x==0){

f[y]=r[l];for(a=0;a<fr;a++){

printf(“%d\t”,f[a]); printf(“\n”); y++; p+=1;

}if(x==1)x=0;if(y==fr)break;

}for (;l<n;l++){ for(k=0;k<fr;k++)

{ ig(f[k]==r[l])

x=1;

}if(x==0){

for(k=0;k<fr;k++){

for(d=l;d>=0;d--){

if(f[l]==r[d]&&b[k]==0){

c[k]=d; b[k]=1;

}}

}a=c[0];y=0;for(d=1;d<fr;d++){

if (a>c[d]) {

a=c[d]; y=d;

}}f[y]=r[l];for(s=0;s<fr;s++)printf(“%d”,f[s]);printf(“\n”);p++;for(d=0;d<fr;d++){

b[d]=0; c[d]=0;

}}if(x==1)x=0;

}p=p-fr;printf(“no of page faults are %d”,p);getch( );

}

INPUT:Enter the no of reference paes 12Enter the frames 32 3 2 1 5 2 4 5 3 2 5

OUTPUT:2 0 0 2 3 02 3 12 5 12 5 43 5 43 5 2no of page faults are 4

c)LEAST FREQUENTLY USED

#include<stdio.h>#include<conio.h>void main ( ){

int no,k,a,b,n,fr[30],m=0,c,p[50],pf=0,r[30],l,min,x;clrscr( );printf(“enter the no of frames”);scanf(“%d”,&n);printf(“enter the no of pages:”);scanf(“%d’,&no);printf(“enter the sequence of pages:”);for(a=0;a<no;a++)

scanf(“%d”,&p[a]);for(a=0;a<no;a++){

b=0,c=0;while (b<n){

if(p[a]!=fr[b])c++;b++;

}if(c==n){

l=0;for(k=0;k<n;k++){

c=0;for(x=a+1;x<no;x++){

if(fr[k]==p[x]) c++;

}r[l++]=c;

} min=100;for(k=0;k<n;k++){

if (min>r[k]) x=k; min=r[k];

}if(m<n)

fr[m++]=p[a];else{

fr[x]=p[I];pf++;

}display(n,fr);}else{

display(n,fr);}

}printf(“|n noof page faults are %d”,pf);getch();

}display(int n,int fr){

int a; for(a=0;a<n;a++) print(“%2d”,fr[a]); printf(‘\n”);

}

INPUT:Enter the no of frames 3Ente the pages 12Enter the sequence of pages 2 3 2 1 5 2 4 5 3 2 5 2 OUPUT2 0 02 3 02 3 02 3 12 3 52 3 54 3 5 4 3 54 3 52 3 52 3 52 3 5no of page faults are 3

SIMULATE PAGING TECHNIQUE OF MEMORY MANAGEMENT.

#include<stdio.h>#include<conio.h>#include<math.h>void main(){

int L[20],p[20],pt[20][10],a[6],b[10],i,j,m=0,n;clrscr();printf(“Enter 16 bit logical address”);for(i=0;i<16;i++)scanf(“%d”,&L[i]);for(i=0;i<16;i++)a[i]=L[i];for(i=0;i<16;i++)p[i]=L[i];for(i=5,j=0;i>=0;i--,j++)mt=a[i]*pow(2.0,j)printf(“page table address”);for(i=0;i<6;i++)printf(“%d”,a[i]);printf(“%d”,m);printf(“enter the page table size”);scanf(“%d”,&n);printf(“Enter %d bit string in page table of size of each string is 6”,n);for(i=0;i<n;i++){

b[i]=iprintf(“b[%d]=%d”,i,i);for(j=0;j<6;j++)scanf(“%d”,&pt[i][j]);

}for(i=0;i=n;i++)if(m==b[i]){

for(i=0;i<6;i++)p[i]=pt[m][i];

}printf(“the 16 bit physical address is “);for(i=0;i<16;i++)printf(“%d”,p[i]);

}INPUT:

Enter 16 bit logical address : 0000010111011110 Page table address: 000001 Enter the page table size 3

OUTPUT:3 bit string in page table of size of each string is 6 : b[0]=0 000101 b[1]=1 000110 b[2]=2 000111 The 16 bit physical address is 0001100111011110

SIMULATE BANKER’S ALGORITHM FOR DEADLOCK AVOIDANCE.

#include<stdio.h>#include<conio.h>viid main(){

int res[10],v[10],c[10][10],a[10][10],ca[10][10],tot[10],p=0;int r,I,j,n,flag,flag1,flag3[10],pos,count=0,r=0,order[10];clrscr();printf(“enter no. of processes\n”);scanf(“%d”,&n);for(i=0;i<n;i++){

flag3[i]=0;order[i]=0;

}printf(“enter the no.of resources\n”);scanf(“%d”,&r);printf(“enter the instances of each resource\n”);for(i=0;i<r;i++){

scanf(“%d”,&res[i]);}printf(“enter the max required from each process from each resorce”);for(i=0;i<n;i++){

for(j=0;j<r;j++){

scanf(“%d”,&c[i][j]);}

}printf(“\n enter the current allocation of resources “);for(i=0;i<n;i++){

for(j=0;j<r;j++){

scanf(“%d”,&a[i][j]);}

}start:

for(i=0;i<n;i++) {

for(j=0;j<r;j++){

ca[i][j]=c[i][j]-a[i][j];}

}for(i=0;i<r;i++){

for(j=0;j<n;j++){

tot[i]=tot[i]+a[j][i];}

}for(i=0;i<r;i++){

v[i]=res[i]-tot[i];}for(i=0;i<r;i++){

if(res[i]<tot[i]){

flag=0;goto out;

}else

flag=1;}out: for(i=0;i<r;i++)

{for(j=0;j<n;j++){

if(res[j][i]>tot[i])flag1=0;goto next;

}else

flag1=1;}

}next:

if((flag==1)&&(flag1==1)printf(“the system is in safe state”);elseif((flag==0)||(flag1==0)){

printf(“the system is in unsafe state”);goto next;

}

for(i=0;i<n;i++){

count=0;for(j=0;j<r;j++){

if((ca[i][j]<=v[i]&&flag3[i]==0))count++;

}if(count==r){

pos=i;flag3[pos]=1;goto cout;

}}cout:

order[p]=pos;p++;printf(“process %dis completed”,pos);for(i=0;i<r;i++){

v[i]=v[i]+a[pos][i];}for(i=0;i<r;i++){

for(j=0;j<r;j++){

c[pos][j]=0;ca[pos][j]=0;a[pos][j]=0;

}}printf(“C-A matrix “);for(i=0;i<n;i++){

printf(“\n”);for(j=0;j<r;j++){

printf(“%d”,ca[i][j]);

}for(i=0;i<n;i++){

if(flag3[i]==1)t++:if(t==n)

goto exit;elsegoto start;

}exit:

printf(“order of execution is”);for(i=0;i<n;i++)printf(“%d”,order[i]);

}getch();

}Output:

Enter the no of process4Enter the no of resources3enter the instances of each resource9 3 6enter the max required from each process for each resourece 3 2 2 6 1 3 3 1 4 4 2 2 enter the current allocation of resources0 0 6 1 2 2 1 1 0 0 2 The available matrix is 0 1 1 process 1 completed after completion of the 1 process available vecteor is 6 2 3 The available matrix is 6 2 3 process 0 completedafter completion of the 0 process available vecteor is 7 2 3 The available matrix is 7 2 3

process 2 completedafter completion of the 2 process available vecteor is 9 3 4

The available matrix is 9 3 4 process 3 completedafter completion of the 3 process available vecteor is 9 3 6

Order of execution of process 1 0 2 3

BANKERS ALGORITHM FOR DEADLOCK PREVENTION

#include<conio.h>#include<stdio.h>#include<stdlib.h>void main(){

int max = 10; int i, j, z;int P, L, N; int A[max], R[max], M[max]; int RS[max]; int LS[max]; int CP[max]={0}; int MAX[max];int lim;

do{

printf("\HOW MANY PROCESSES DO YOU WANT? ");scanf(“%d”,&P);

}while(P>max);

printf("\nINSERT THE MULTEPLICITY OF THE RESOURCE: ");scanf(“%d”,&N);

for(i=0;i<P;i++){printf("\nHOW MANY UNIT HAS THE P %d PROCESS? ",i);scanf(“%d”,&A[i]);printf("\nHOW MANY UNIT REQUEST THE P %d PROCESS? ",i);scanf(“%d”,&R[i]);printf("\nWHAT IS THE MAX NUMBER OF UNITS OF THE P %dPROCESS? ",i);scanf(“%d”,&M[i]);printf("\n\n");}

L=N;for(i=0; i<P; i++)L-=A[i];

for(i=0; i<P; i++){if(R[i]<=L)

{RS[i]=M[i]-A[i]-R[i];

LS[i]=L-R[i]; }else {LS[i]=-1;}}

for(i=0; i<P; i++)MAX[i]=-1; z=0;for(i=0; i<P; i++){if(i==0){for(j=0; j<P; j++){if(LS[j]>=MAX[z]){MAX[z]=LS[j];CP[z]=j;

lim=MAX[z];}}}else{for(j=0; j<P; j++){if((LS[j]>=MAX[z])&&(LS[j]<lim)){MAX[z]=LS[j];CP[z]=j;

lim=MAX[z];}}}

z++;}

for(i=0; i<P; i++){if(LS[i]!=-1)printf("\nTHE PROCESS NUMBER %d IS THE ",i,CP[i];elseprintf("THE PROCESS NUMBER %d IS NOT SECURE",i);}

getch();}

MULTI PROGRAMMING WITH FIXED NUMBER OF TASKS

#include<stdio.h>void main(){

int n,s,p_size,I,flag[100],k=-1,I_flag=0;char c,name[100];printf(“Enter the number of partitions”);scanf(“%d”,&n);printf(“Enter the size of each partition”);scanf(“%d”,&s);do{

printf(“Enter the process name”);scanf(“%d”,&name):printf(“Enter the size of process”);scanf(“%d”,&p_size);if(p_size<=1){

k++;flag[k]=p_size;

}else

printf(“The size of the process is too large”);if(k>=n-1){

clrscr();printf(“No more partitions are leftout free”);getch();break;

}}printf(“Any more process(y/n) : “);scanf(“%c”,&c);while(c==’y’||c==’Y’);for(I=0;I<=k;I++)I_flag=I_flag+flag[I];clrscr();printf(‘number of partitions occupied is %d”,k+1);printf(“number of partitions free”,n-k-1);printf(“Total internal fragmentation %d”,I_flag);getch();

}

OUTPUT: Enter No. of partitions 4 Enter size of each partition 100 Enter the process name p1 Enter the size of the process 80

Any more process(y/n) : nNumber of partitions occupied 1Number of partitions free 3Total internal fragmentation 20

MULTIPROGRAMMING WITH VARIABLE NO OF TASKS

#include <stdio.h>void main( ){ int I,h,holes[30],s,c,k,flag=0;

printf(“ente the no of holes”0); scanf(“%d”,&h); printf(“enter the size of the next process:”); scanf(“%d”,&s); printf(“enter the size of the each hole”); for(i=0;i<h;i++) scanf(“%d”,&holes[i]); printf(“select the algorithm from the list below”); printf(“\n 1.first fit 2.best fit 3.next fit”); scanf(‘%d”,&c); flag=0; if(c==1) {

for(i=0;i<n;i++) {

if(holes[i]>=s){ printf(“\n the process can be placed in hole :%d”,(i+1));

flag=1;break;

}}

if(flag==0) printf(“\n the process cannot be placed in any hole”);

} else if (c==2) { for (i=0;i<h;i++) if (holes[i]>s) break; k=i; for (i=0;i<k;i++) { if(holes[i]>=s&&holes[i]<holes[k]) {

k=i; flag =1;

} } if(flag==1)

printf(“\n process can be placed in hole:%d”,i+1); else

printf(“process cannot be placed in the hole :%d”,i+1); }

else if(c==3) {

printf(“\n enter the hole after previous insertion”);scanf(“%d”,&k);for(i=k-1;i<h;i++){

if(holes[i]>=s) flag=1; printf(“\n process can be placed in the hole %d”i+1); break;

}}if(flag==0)for(i=0;i<k;i++){

if(holes[i]>=s){

flag=1;printf(“\n the process can be placed in the hole :%d”,i+1);break;

}}

}if (flag ==1)printf(“ the process cannot be placed in the hole”);getch( );}

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