Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program 1 : Character Stuffing
Program :
#include<stdio.h>#include<conio.h>#include<string.h>void main() {int n,i;long a;char str[100]=" ",temp[20]=" ",ch;FILE *fp1,*fp2,*fp3;clrscr();fp1=fopen("input1.txt","r");if(fp1==NULL){printf("not created");}fp2=fopen("z:/input2.txt","w");printf("\n\t **Character Stuffing** \t\n");printf("The Original Text is :");fseek(fp1,0,SEEK_END);a=ftell(fp1);rewind(fp1);while((ch=fgetc(fp1))!=EOF) printf("%c",ch);rewind(fp1);printf("\n\t The Encoded Text is :");while(!feof(fp1)){ if(a==ftell(fp1)) break; else { fscanf(fp1,"%s",str); fputs("stxdle ",fp2); printf("stxdle "); if(strstr(str,"dle")!=0) { fputs(str,fp2); printf("%s",str); fputs("dle",fp2); printf("dle"); }
Page 1CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
else { fputs(str,fp2); printf("%s",str);} fputs(" etxdle ",fp2); printf(" etxdle "); }}fcloseall();fp2=fopen("z:/input2.txt","r");fp3=fopen("z:/input3.txt","w");printf("\n\t The Decoded Text is :") ;while(!feof(fp2)){ fscanf(fp2,"%s",str); if(strcmp(str,"stxdle")==0) fputs("",fp3); else if(strcmp(str,"etxdle")==0) fputs(" ",fp3); else if(strstr(str,"dledle")!=0) { n=strlen(str); n=n-3; strcpy(temp,str+n); fputs(str,fp3); } else { fputs(str,fp3); }} //for(i=0;i<n;i++) //temp[i]='\0'; fcloseall(); fp3=fopen("z:/input3.txt","r"); while((ch=fgetc(fp3))!=EOF) printf("%c",ch); fcloseall(); getch();}
Page 2CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Result:
Input File: (Input1.txt)
This Is CN Lab.The Program dle is about idle character stuffing.
Output:
** CHARACTER STUFFING **
The Original Text is :
This Is CN Lab.The Program dle is about idle character stuffing.
The Encoded Text is:
stxdle This etxdle stxdle Is etxdle stxdle CN etxdle stxdle Lab. etxdle stxdle The etxdle stxdle Program etxdle stxdle dledle etxdle stxdle is etxdle stxdle about etxdle stxdle idledle etxdle stxdle character etxdle stxdle stuffing. etxdle
The Decoded Text is:
This Is CN Lab. The Program dle is about idle character stuffing.
Page 3CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program 2: Bit Stuffing
Program:
#include<stdio.h>#include<conio.h>#include<math.h>main(){int n,r,j,a=0,b=0,l=0,i=0,arr[200],bs[200];char ch;FILE *fp1;clrscr();fp1=fopen("input1.txt","r");printf("\t\t *** Bit Stuffing ** \n\n");printf("The Original Text is:\n\n ");while((n=ch=fgetc(fp1))!=EOF){printf("%c",ch);for(i=0;n!=0;i++){r=n%2;bs[i]=r;n/=2;}for(;i<8;i++)bs[i]=0;for(j=i-1;j>=0;j--,a++)arr[a]=bs[j];b=l=a;}getch();printf("\n\n The Text after Convertion to binary is:\n\n");for(i=0;i<b;i++)printf("%d",arr[i]);getch();printf("\n\n The Encoded Text is:\n\n");for(i=0;i<=7;i++){if(i==0||i==7)bs[i]=0;elsebs[i]=1;}for(i=0,j=8;i<=l;i++){
Page 4CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
if(arr[i]==1){n++;bs[j]=arr[i];if(n==5){j++;bs[j]=0;}j++;}else{n=0;bs[j]=arr[i];j++;}}for(i=0;i<=7;i++){if(i==0||i==7)bs[i+j]=0;elsebs[i+j]=1;}j=j+7;for(i=0;i<=j;i++)printf("%d",bs[i]);getch();printf("\n\n The Decoded Text is:\n ");for(i=8,r=0;i<j-7;i++,r++)bs[r]=bs[i];for(i=r=l=n=0;i<=j-16;i++){if(bs[i]==1)n++;elsen=0;r=i;if(n==5){r++;l++;for(a=r+1;a<=j-16;a++,r++)bs[r]=bs[a];}}
Page 5CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
for(i=0;i<b;i++)
printf("%d",bs[i]);getch();printf("\n The Text aft Conversion from binary text is:\n\n");for(i=0;i<b; ){for(a=0,n=7;n>=0;n--,i++)a=a+bs[i]*pow(2,n);printf("%c",a);
}fclose(fp1);getch();return (0);}
Page 6CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Result:
Input File: (Input1.txt)
This is Computer_networks Lab.
Output:
** BIT STUFFING **
The Original Text is:
This is Computer_networks Lab.
The Text after Conversion to binary is:
010101000110100001101001011100110010000001101001011100110010000001100011011010110110101110000011101010111010001100101011100100101111101101110011001010111001110111011011110111001001101011011100110010000001101100011000010110001000101
The Encoded Text is:
01111110010101000110100001101001011100110010000001101001011100110010000001100011011110110110101110000011101010111010001100101011100100101111100110111001101011101000111011101101111011100100110101101110011001000000110110001100001011000010111001111110
The Decoded Text is:
010101000110100001101001011100110010000001101001011100110010000001100011011011110110110101110000011101010111010001100101011100100010111110110111001100100111010001110111011011110111001001101011011100110010000001101100011000010110001000101110
The Text after Conversion from binary to Text is:
This is Computer_networks Lab.
Page 7CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program 3: Cyclic Redundancy Check,CRC-12
Program:
#include<stdio.h>#include<conio.h>#include<string.h>#include<dos.h>#include<stdlib.h>int copy();int check();int i,j,k,t,count=0,num=0;int gen[10],frame[30],rem[30],temp[30];void main(){char c,plym[50];char ch[]={'0','1','2','3','4','5','6','7','8','9'};clrscr();printf("\t\t\t***CYCLIC REDUNDANCY CHECK-12***\n\n\n");for(i=0;i<50;i++)plym[i]='\0';for(i=0;i<30;i++)temp[i]=rem[i]=frame[i]='\0';for(i=0;i<10;i++)gen[i]='\0';printf("enter the polynomial: ");gets(plym);plym[strlen(plym)]='+';for(i=0;i<strlen(plym);i++){if(plym[i]=='x'){i++;for(;plym[i]!='+';count++,i++){}if(count==3){for(i=i-1,j=3;j<=9;j++)if(plym[i]==ch[j]){printf("\Enter the polynomial's");printf("degree is high");getch(); exit(0);}for(j=0,num=10;j<=2;j++)if(plym[i]==ch[j])
Page 8CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
{num=num+j;frame[num]=1;}}if(count==2){for(i=i-1,j=1,num=0;j<=9;j++)if(plym[i]==ch[j])num=j;frame[num]=1;}if(count==0)frame[1]=1;count=0;}else if(plym[i]=='1')frame[0]=1;}printf("FRAME is: ");for(i=12,j=0;i>=0;i--,j++){temp[j]=frame[i];printf("%d",frame[i]);}printf("\n\n\n>>>>both high & low orders");printf("bits of GENERATOR must be 1<<<<");printf("\n enter the generator: ");for(num=i=0;(c=getchar())!='\n';i++,num++){if(c=='1')gen[i]=1;else if(c=='0')gen[i]=0;else{printf("\n Enter the GENERATOR");printf("is other then 0 or 1");getch(); exit(0);}}for(j=13,i=i-1;i>0;i--,j++)temp[j]=0;printf("\n\n FRAME after appending 0's: ");copy(); check();printf("\n The REMAINDER is: ");for(i=13;i<j;i++)
Page 9CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
{temp[i]=rem[i];printf("%d",rem[i]);}printf("\n\n\n Transmitting FRAME......");delay(10000);printf("\n\n\n Transmitted FRAME is: ");copy(); check();printf("\n frame recieved");printf("\n\n\n checking for errors......");delay(10000);printf("\n\n\n recieved frame is: ");copy(); check();printf("\n the remainder is: ");for(i=13;i<j;i++)printf("%d",rem[i]);printf("\n DATA SENT SUCCESSFULLY");getch();}check(){for(i=0;i<=12;i++){if(rem[i]==0)continue;else{for(k=0,t=i;k<num;k++,t++){if(rem[t]==1 && gen[k]==1)rem[t]=0;else if(rem[t]==0 && gen[k]==0)rem[t]=0;else if(rem[t]==1 && gen[k]==0)rem[t]=1;else if(rem[t]==0 && gen[k]==1)rem[t]=1;}}}return 0;}copy(){for(i=0;i<j;i++){printf("%d",temp[i]);
Page 10CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
rem[i]=temp[i];}return 0;}
Result:
Output :
*** CYCLIC REDUNDANCY CHECK-12 ***
Enter the Polynomial: x^12+x^6+x+1FRAME is: 1000001000011
>>>> Both High & Low order bits of GENERATOR must be 1 <<<<Enter the GENERATOR: 10010
FRAME after appending 0’s: 10000010000110000The REMAINDER is: 0110
Transmitting FRAME ………
The Transmitted FRAME is: 10000010000110110FRAME RECEIVED.
Checking for ERRORS………
Received FRAME is: 10000010000110110The REMAINDER is: 0000
DATA SENT SUCCESSFULLY.
Page 11CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program 4: Cyclic Redundancy Check,CRC-16
Program:
#include<stdio.h>#include<conio.h>#include<string.h>#include<dos.h>#include<stdlib.h>int copy();int check();int i,j,k,t,count=0,num=0;int gen[10],frame[30],rem[30],temp[30];void main(){char c,plym[50];char ch[]={'0','1','2','3','4','5','6','7','8','9'};clrscr();printf("\t\t\t***CYCLIC REDUNDANCY CHECk-16***\n\n\n");for(i=0;i<50;i++)plym[i]='\0';for(i=0;i<30;i++)temp[i]=rem[i]=frame[i]='\0';for(i=0;i<10;i++)gen[i]='\0';printf("Enter the polynomial: ");gets(plym);plym[strlen(plym)]='+';for(i=0;i<strlen(plym);i++){if(plym[i]=='x'){i++;for(;plym[i]!='+';count++,i++){}if(count==3){for(i=i-1,j=7;j<=9;j++)if(plym[i]==ch[j]){printf("\entered polynomial's");printf("degree is high");getch(); exit(0);}for(j=0,num=10;j<=6;j++)
Page 12CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
if(plym[i]==ch[j]){num=num+j;frame[num]=1;}}if(count==2){for(i=i-1,j=1,num=0;j<=9;j++)if(plym[i]==ch[j])num=j;frame[num]=1;}if(count==0)frame[1]=1;count=0;}else if(plym[i]=='1')frame[0]=1;}printf("FRAME is: ");for(i=16,j=0;i>=0;i--,j++){temp[j]=frame[i];printf("%d",frame[i]);}printf("\n\n\n>>>>both high & low orders");printf("bits of GENERATOR must be 1<<<<");printf("\n Enter the generator: ");for(num=i=0;(c=getchar())!='\n';i++,num++){if(c=='1')gen[i]=1;else if(c=='0')gen[i]=0;else{printf("\n the Enter the generator");printf("is other then 0 or 1");getch(); exit(0);}}for(j=17,i=i-1;i>0;i--,j++)temp[j]=0;printf("\n\n FRAME after appending 0's: ");copy(); check();printf("\n The REMAINDER is: ");
Page 13CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
for(i=17;i<j;i++){temp[i]=rem[i];printf("%d",rem[i]);}printf("\n\n\n Transmitting frame......");delay(10000);printf("\n\n\n TRANSMITTED FRAME is: ");copy(); check();printf("\n frame RECEIVED");printf("\n\n\n Checking for ERRORS......");delay(10000);printf("\n\n\n Recieved Frame is: ");copy(); check();printf("\n The REMAINDER is: ");for(i=17;i<j;i++)printf("%d",rem[i]);printf("\n DATA SENT SUCCESSFULLY");getch();}check(){for(i=0;i<=16;i++){if(rem[i]==0)continue;else{for(k=0,t=i;k<num;k++,t++){if(rem[t]==1 && gen[k]==1)rem[t]=0;else if(rem[t]==0 && gen[k]==0)rem[t]=0;else if(rem[t]==1 && gen[k]==0)rem[t]=1;else if(rem[t]==0 && gen[k]==1)rem[t]=1;}}}return 0;}copy(){for(i=0;i<j;i++){
Page 14CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
printf("%d",temp[i]);rem[i]=temp[i];}return 0;}
Result:
Output:
*** CYCLIC REDUNDANCY CHECK-16 ***
Enter the Polynomial: x^16+x^12+x^10+x^6+x^5+x^4+x^3+x^2+x^1+1FRAME is: 10001010001111111
>>>> Both High & Low order bits of GENERATOR must be 1 <<<<Enter the GENERATOR: 100101
FRAME after appending 0’s: 1000101000111111100000The REMAINDER is: 01110
Transmitting FRAME ………
The Transmitted FRAME is: 1000101000111111101110FRAME RECEIVED.
Checking for ERRORS………
Received FRAME is: 1000101000111111101110The REMAINDER is: 0000
DATA SENT SUCCESSFULLY.
Page 15CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program- 5: Shortest Path Routing.
Program:
#include<stdio.h>#include<conio.h>main(){ int z,fe[10],fc[10],fg[10],hi[10],hf[10]; int a[10]={0,5,2,3,0,0,0,0,0,0}; int b[10]={0,0,0,0,3,0,0,0,0,0}; int c[10]={0,0,0,0,0,4,0,0,0,0}; int d[10]={0,0,0,0,0,0,4,0,0,0}; int e[10]={0,0,0,0,0,2,0,0,2,0}; int f[10]={0,0,0,0,0,0,0,5,0,0}; int g[10]={0,0,0,0,0,6,0,0,0,0}; int h[10]={0,0,0,0,0,0,0,0,0,1}; int i[10]={0,0,0,0,0,0,0,2,0,0}; int j[10]={0,0,0,0,0,0,0,0,0,0}; clrscr(); printf("\t\t***shortest path routing***\n\n"); printf("\n the shortest path routing table \n\n"); for(z=0;z<10;z++) { if(b[z]!=0) b[z]=a[1]+b[z]; else b[z]=0; if(c[z]!=0) c[z]=a[2]+c[z]; else c[z]=0; if(d[z]!=0) d[z]=a[1]+d[z]; else d[z]=0; } for(z=0;z<10;z++) { if(e[z]!=0) e[z]=e[z]+b[4]; else e[z]=0; if(g[z]!=0) g[z]=g[z]+d[6]; else g[z]=0;
Page 16CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
} for(z=0;z<10;z++) { if(f[z]!=0) { fe[z]=e[5]+f[z]; fc[z]=c[5]+f[z]; fg[z]=g[5]+f[z]; } else { fe[z]=0; fc[z]=0; fg[z]=0; } } for(z=0;z<10;z++) { if(fc[z]<fe[z]&&fc[z]<fg[z]) f[z]=fc[z]; else if(fe[z]<fg[z]) f[z]=fe[z]; else f[z]=fg[z]; } for(z=0;z<10;z++) { if(i[z]!=0) i[z]=i[z]+e[8]; else i[z]=0; if(h[z]!=0) { hi[z]=h[z]+i[7]; hf[z]=h[z]+f[7]; } else { hi[z]=0; hf[z]=0; } if(hi[z]<hf[z]) h[z]=hi[z]; else h[z]=hf[z]; } printf("\tA\tB\tC\tD\tE\n");
Page 17CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
printf("A\t"); for(z=0;z<5;z++) printf("%d \t",a[z]); printf("\nB\t"); for(z=0;z<5;z++) printf("%d \t",b[z]); printf("\nC\t"); for(z=0;z<5;z++) printf("%d \t",c[z]); printf("\nD\t"); for(z=0;z<5;z++) printf("%d \t",d[z]); printf("\nE\t"); for(z=0;z<5;z++) printf("%d \t",e[z]); printf("\nF\t"); for(z=0;z<5;z++) printf("%d \t",f[z]); printf("\nG\t"); for(z=0;z<5;z++) printf("%d \t",g[z]); printf("\nI\t"); for(z=0;z<5;z++) printf("%d \t",i[z]); printf("\nH\t"); for(z=0;z<5;z++) printf("%d \t",h[z]); printf("\n\n"); printf("\tF\tG\tH\tI\tJ\n"); printf("\nA\t"); for(z=5;z<10;z++) printf("%d \t",a[z]); printf("\nB\t"); for(z=5;z<10;z++) printf("%d \t",b[z]); printf("\nC\t"); for(z=5;z<10;z++) printf("%d \t",c[z]); printf("\nD\t"); for(z=5;z<10;z++) printf("%d \t",d[z]); printf("\nE\t"); for(z=5;z<10;z++) printf("%d \t",e[z]); printf("\nF\t"); for(z=5;z<10;z++) printf("%d \t",f[z]);
Page 18CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
printf("\nG\t"); for(z=5;z<10;z++) printf("%d \t",g[z]); printf("\nI\t"); for(z=5;z<10;z++) printf("%d \t",i[z]); printf("\nH\t"); for(z=5;z<10;z++) printf("%d \t",h[z]); printf("\nthe shortest path from source 'a' to destination 'j' is: %d",h[9]); getch();}
Page 19CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Result:
Output:
*** SHORTEST PATH ROUTING ***
THE SHORTEST PATH ROUTING TABLE
A B C D E A 0 5 2 3 0B 0 0 0 0 8C 0 0 0 0 0D 0 0 0 0 0E 0 0 0 0 0F 0 0 0 0 0G 0 0 0 0 0H 0 0 0 0 0I 0 0 0 0 0
F G H I JA 0 0 0 0 0B 0 0 0 0 0C 6 0 0 0 0D 0 7 0 0 0E 10 0 0 10 0F 0 0 11 0 0G 13 0 0 0 0I 0 0 12 0 0H 0 0 0 0 12
The Shortest Path from SOURCE ‘A’ to DESTINATION’J’ is : 12
Page 20CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program-6: Distance Vector Routing Algorithm
Program:
#include<stdio.h>#include<conio.h>Main(){
int i=0,a[5],b[5],c[5],d[5],e[5],ab[5];int ac[5],ad[5],ao[5],ba[5],bc[5],be[5];int bo[5],ca[5],cb[5],ce[5],co[5],da[5];int ec[5],eb[5],eo[5],dop[5];char node[5]={‘A’,’B’,’C’,’D’,’E’};clrscr();printf(“\t\t\t\ *** DISTANCE VECTOR ROUTING *** \n\n\n\”);printf(“\n Enter values for ‘a’:\n”);for(i=0;i<5;i++)
scanf(“%d”,&a[i]);printf(“\n Enter the values for ‘b’:\n “);for(i=0;i<5;i++)
scanf(“%d”,&b[i]);printf(“\n Enter the values for ‘c’:\n “);for(i=0;i<5;i++)
scanf(“%d”,&c[i]);printf(“\n Enter the values for ‘d’:\n “);for(i=0;i<5;i++)
scanf(“%d”,&d[i]);printf(“\n Enter the values for ‘e’:\n “);for(i=0;i<5;i++)
scanf(“%d”,&e[i]);printf(“\n\n THE ROUTERS TABLE WITH OPTIMUM VALUES:\n”);printf(“ \t ------------------------------------ \n”);printf(“\t A \t B \t C \t D\t E\n”);printf(“ \t ------------------------------------ \n”);printf(“%c\t”,node[0]);for(i=0;i<5;i++){
Ab[i]=a[1]+b[i];Ac[i]=a[2]+c[i];Ad[i]=a[3]+d[i];
}for(i=0;i<5;i++){
if(a[i]<ab[i] && a[i]<ac[i] &&a[i]<ad[i])Ao[i]=a[i];
else if(ab[i]<ac[i] && ab[i]<ad[i])
Page 21CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Ao[i]=ab[i];else if(ac[i]<ad[i])
Ao[i]=ac[i];else
Ao[i]=ad[i];printf(“%d\t”,ao[i]);}printf(“\n %c \t “,node[1]);for(i=0;i<5;i++){
ba[i]=b[0]+ao[i];bc[i]=b[2]+c[i];be[i]=b[4]+e[i];
}for(i=0;i<5;i++){
if(b[i]<ba[i] && b[i]<bc[i] && b[i]<be[i])bo[i]=b[i];
else if(ba[i]<bc[i] && ba[i]<be[i])bo[i]=ba[i];
else if(bc[i]<be[i])bo[i]=bc[i];
elsebo[i]=be[i];
printf(“%d\t”,bo[i]);}printf(“\n%c\t”,node[2]);For(i=0;i<5;i++){
ca[i]=c[0]+ao[i];cb[i]=c[1]+bo[i];ce[i]=c[4]+e[i];
}for(i=0;i<5;i++){
if(c[i]<ca[i] && c[i]<cb[i] && c[i]<ce[i])co[i]=c[i];
else if(ca[i]<cb[i] && ca[i]<ce[i])co[i]=ca[i];
else if(cb[i]<ce[i])co[i]=cb[i];
elseco[i]=ce[i];
printf(“%d\t”,co[i]);}printf(“\n %c \t”,node[3]);For(i=0;i<5;i++)
da[i]=d[0]+ao[i];
Page 22CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
for(i=0;i<5;i++){
if(d[i]<da[i])dop[i]=d[i];
elsedop[i]=da[i];
printf(“%d\t”,dop[i]);}printf(“\n%c\t”,node[4]);for(i=0;i<5;i++){
ec[i]=e[2]+co[i];eb[i]=e[1]+bo[i];
}for(i=0;i<5;i++){
if(e[i]<ec[i] && e[i]<eb[i])eo[i]=e[i];
else if(ec[i]<eb[i])eo[i]=ec[i];
elseeo[i]=eb[i];
printf(“%d\t”,eo[i]);}printf(“\n\t--------------------------- \n”);printf(“\n OPTIMUM COST FROM ROUTER-A TO ROUTER-E:”);printf(“%d”,ao[4]);getch();return();
}
Page 23CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Result:
Output:
*** DISTANCE VECTOR ROUTING ***
Enter the values for ‘a’:0 5 2 3 6
Enter the values for ‘b’:5 0 4 8 3
Enter the values for ‘c’:2 4 0 5 4
Enter the values for ‘d’:3 8 5 0 9
Enter the values for ‘e’:6 3 4 9 0
THE ROUTERS TABLE WITH OPTIMUM VALUES
------------------------------A B C D E------------------------------
A 0 5 2 3 6B 5 0 4 8 3C 2 4 0 5 4D 3 8 5 0 9E 6 3 4 9 0
------------------------------
THE OPTIMUM COST FROM ROUTER-A TO ROUTER-E: 6
Page 24CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program-7: Simplified Decryption Encryption Standard
Program:#include<stdio.h>#include<conio.h>#include<math.h>int p,q,i=0,j=0,k=0,a=0,r=0,x=0,c=0.t=0,y=0,b[2];int binary[10],binmsg[500],temp[8],e[8],t4[4],tp[4];int p[10],key[10],newkey[10],key1[8],key2[8],p8[8];int ip[8]={2,6,3,1,4,8,5,7}ip_1[8]={4,1,3,5,7,2,8,6};int ep[8]={4,1,2,3,2,3,4,1},p4[4]={2,4,3,1},binmsg1[500];int s0[4][4]={1,0,3,2,3,2,1,0,0,2,1,3,3,1,3,2};int s1[4][4]={0,1,2,3,2,0,1,3,3,0,1,0,2,1,0,3};char ch;main(){
FILE *fp;clrscr();fp=fopen(“input1.txt”,”r”);printf(“\t\t\t ***SIMPLIFIED-DES***\n”);printf(“\n Original Message:\n”);
//Coverting Text to Binarywhile((a=ch=fgetc(fp))!=EOF){
printf(“%c”,ch);for(j=0;a!=0;j++){
r=a%2;a/=2;binary[j]=r;
}for(i=8-j;i>0;i__,j++)
binary[j]=0;for(i=j-1;i>=0;i--,x++)
binmsg[x]=binary[i];}printf(“\n\n Enter the 10 bit key: “);for(i=0;i<10;i++)
scanf(“%d”,&key[i]);//P10 User Defined
printf(“\n Enter the P10 Sequence” “);for(i=0;i<10;i++)
scanf(“%d”,&p10[i]);//Generating Key according to P10
for(i=0;i<10;i++){
j=p10[i]; j--;
Page 25CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
newkey[i]=key[j];}
//LS-1 operationj=newkey[4];for(i=4,k=3;i>0;i--,k--)
newkey[i]=newkey[k];newkey[0]=j;J=newkey[9];for(i=9,k=8;i>5;i--,k--)
newkey[i]=newkey[k];newkey[5]=j;printf(“\n Enter the p8 sequence: “);for(i=0;i<8;i++)
scanf(“%d”,&p8[i]);//Generating KEY-1
printf(“\n key-1:”);for(i=0;i<8;i++){
j=p8[i]; j--;key1[i]=newkey[j];printf(“%d”,key1[i]);
}//LS-2 Operation
j=newkey[3]; a=newkey[4];for(i=4,k=2;i>=2;i--,k--)
newkey[i]=newkey[k];newkey[0]=j; a=newkey[9];for(i=9,k=7;i>=7;i--,k--)
newkey[i]=newkey[k];newkey[5]=j; newkey[6]=a;
//Generating KEY-2printf(“\n Key-2: “);for(i=0;i<8;i++){
j=p8[i];key2[i]=newkey[j];printf(“%d”,key2[i]);
}printf(“\n\n Cipher Message: “);for(i=0;i<x;){// considering 8-bit plain text
for(j=0;j<8;j++,i++)temp[j]=binmsg[i];
initp(); extp(); xor 1();fk(); switching();extp(); xor2(); fk(); ipinv();for(j=7,a=0,k=0;k<8;j--,k++,y++)
Page 26CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
{binmsg1[y]=temp[k];a=a+temp[k]*pow(2,j);
}printf(“%c”,a);
}//Decryption
printf(“\n\n Decrypted Message is:”);for(i=0;i<y;){
for(j=0;j<8;j++,i++)temp[j]=binmsg1[i];
initp(); extp();xor2();fk(); switching();extp(); xor1(); fk(); ipinv();for(j=7,a=0,k=0;k<8;j--,k++)
a=a+temp[k]*pow(2,j);print(“%c”,a);
}getch();return 0;
}INITP(){//permutating using IP Function
For(j=0;j<8;j++){
a=ip[j]; a--;binary[j]=tamp[a];
}return 0;
}ipinv(){//Inverse Initial Permutation
for(j=0;j<8;j++){
a=ip_1[j]; a--;temp[j]=binary[a];
}return 0;
}switching(){//Switching
for(k=0,j=4;k<=3;k++,j++){
Page 27CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
a=binary[k];binary[k]=binary[j];binary[j]=a;
}return 0;
}extp(){//Applying E/P Operation
for(j=0;j<8;j++){
a=ep[j]; a+=3;e[j]=binary[a];
}return 0;
}fk(){//Converting Binary to Decimal
p=e[0];q=e[3];r=p*2+q;p=e[1];q=e[2];c=p*2+q;
//Finding value in S-0 boxt=s0[r][c]a=0;compare();p=e[4];q=e[7];r=p*2+q;t=s1[r][c];compare();
//EX-OR operation between p4 & first 4bits of IPfor(j=0;j<4;j++){
a=p4[j];a--;tp[j]=t4[a];if(tp[j]==0&& binary[j]==0)
binary[j]=0;else if(tp[j]==1 && binary[j]==1)
binary[j]=0;else if(tp[j]==0 && binary[j]==1)
binary[j]=1;else if (tp[j]==1 && binary[j]==0)
binary[j]=1;}return 0;
Page 28CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
}compare(){
if(t==0){
t4[a]=0; a++;t4[a]=0; a++;
}else{
for(j=0;t!=0;j++){
k=t%2;t/=2;b[j]=k;}
for(j=1;j>=0;j--,a++)T4[a]=b[j];
}return 0;
}Xor1(){//EX-OR operation b/w ep & key-2
for(j=0;j<8;j++){
if(e[j]==0 &&key2[j]==0)e[j]=0;
else if(e[j]==1 && key2[j]==1)e[j]=0;
else if(e[j]==1&&key2[j]==0)e[j]=1;
else if(e[j]==0&&key2[j]==1)e[j]=1;
}return 0;
}
Page 29CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Result:
Input File (input1.txt):
Surendra and Pradeep
Output:
*** SIMPLIFIED –DES ***
Original Message:Suerndra and pradeep
Enter the 10 BIT KEY: 1 0 1 01 0 1 0 1 0
Enter the P10 Sequence: 4 2 1 7 8 3 9 5 10 6
Enter the P8 Sequence: 2 1 7 8 3 9 5 10
Key-1: 00110110Key-2: 11000101
Cipher Message:
Decrypted Message is: surendra and pradeep
Page 30CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Program-8: Caeser Cipher Susbstitution Technique.
Program:#include<stdio.h>#include<conio.h>#include<string.h>main(){int c,i,n=0,key;char ch,cipher[100];FILE *fp;fp=fopen("input1.txt","r");clrscr();for(i=0;i<100;i++)cipher[i]='\0';printf("\t\t\t*** ceaser cipher substitution tech***\n\n\n");printf("the original text is:");while((ch=fgetc(fp))!=EOF)printf("%c",ch);fseek(fp,0,0);y:printf("\n\nenter the key");scanf("%d",&key);if(key>25){printf("\n\n***minimum value allowed for key is 25***\n");getch(); goto y;}printf("\n cipher text is :");/*encryption*/for(i=0;(n=fgetc(fp))!=EOF;){if(n==32){cipher[i]=' ';i++;printf(" ");}else{if(n>64&&n<91){c=(n+key)%91;if(c<65)
Page 31CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
c=c+65;cipher[i]=c;i++;printf("%c",c);}else if(n>96 &&n<123){c=(n+key)%123;if(c<97)c=c+97;cipher[i]=c;i++;printf("%c",c);}}}printf("\n\nthe decrypted text is :");for(i=0;i<strlen(cipher);i++){n=cipher[i];if(n==32)printf(" ");else{if(n>64&&n<91){c=(n-key)%91;if(c<64)c=90-64+c;printf("%c",c);}else if(n>96&&n<123){c=(n-key)%123;if(c<97)c=122-96+c;printf("%c",c);}}}getch();return 0;}
Page 32CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Result:
Output:
***CAESER CIPHER SUBSTITUTION ***
The Original text is : This is Anitha
Enter the key: 2
Cipher text is: vjku ku cpkvjc
The Decrypted text is : This is Anitha
Page 33CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
LiBrary management system
SYSTEM REQUIREMENT SPECIFICATION
To model the library management system using ‘Object Oriented Analysis and Designing’ concept and the unified modelling language (UML) objective behind the development of this software is to model an object oriented system which helps the librarians to easily performs their activities. It must as a best interface between the librarian and the computer so as to perform the required activities effectively. It must cover the activites of maintaining the records of issued and received books and
Other things available in the library separately for both faculty and student when by taking the secario as a college. Uml diagrams drew below helps us to easily understand how the software is developed. The diagrams are class, usecase, sequence, collaborations, component and deployment diagrams.
Page 34CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
CLASS DIAGRAMClass diagram shows structure of the software system. The class diagram shows a set of
class, interfaces and their relationships. The components are:a) Classb) Relationship
The forms of relationship are:1. Associations 2. Aggregations3. Generalization4. Compositions5. Dependency
Data base
student id : intfaculty id : int
update data()
book details
book no : intauthor : stringcost : float
add book()del book()
mags & jornels
mid : intjid : int
add()del()
publisher
pid : intpname : stringpaddras : string
add publisher()del publisher()
faculty
name : stringid : int
issue()renewal()return()cancel()enquiry()
issued books
<<->> book name : string<<->> book id : num<<->> issued date : date
<<+>> add books()
librarian
name : stringid : int
check details()check availability()check database()
Reservation
student idbook nameauthor name
priority()
book bank
studentname : stringid : int
issuing()returning()
library
coll name : stringcoll id : int
digital library
id : stringpassword : stringsys no : int
accessing()download()
student
name : srtingid : intlib card
issue of book bank()issue()renewal()returning()cancel()enquiry()
Page 35CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
USE CSE DIAGRAMUse case diagram is created to visualize the interaction of our system with the outside
world. The components of use case diagrams are:
Use Case: Scenarios of the system
Actors: some or something who is interacting with the system
Relationship: semantic link between use case and actor. The forms of relationship are:
a) Associationb) Dependencyc) Generalization
Database
impossing fines
updates
librarian
publisher
reservation
recieve books
digital library
student
issue books
return books
faculty
reading books
Page 36CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
INTERACTION DIAGRAM
An interaction diagram models the dynamic aspects of the system by showing the relationship among the objects and message they may dispatch. There are two types of interaction diagrams:
SEQUENCE DIAGRAM
Sequence diagram shows the steps to steps what much happen to accomplish a piece of functionality provided by the system. The components are:
a) Actors b) Objectsc) Messaged) Lifelinee) Focus of Control
COLLABORATION DIAGRAM
Collaboration diagrams displays object interactions organized around object and their links to one another. The components are:
a) Actorb) Objectc) Link
Page 37CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence diagram for issue and return of books
:student :books :Librarian :Faculty
refers
refer books
issue books
issue books
return books
update books
Page 38CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Collaboration diagram for issue and return of books
:books:student
:Librarian
:Faculty
1: refers
2: refer books
3: issue books
4: issue books
5: return books
6: update books
Page 39CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence diagram for book bank
:student:librarian :books
refers
request for book bank
update records
issues book bank
Collaboration diagram for book bank
:librarian :books
:student
4: update records
2: request for book bank3: issues book bank1: refers
Page 40CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence diagram for database
:librarian :books details :student details :faculty details :database
checks
checks
checks
update all details
Page 41CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Collaboration diagram for database
1: checks
:librarian
:books details
:student details
:faculty details
:database
3: checks
4: update all details
2: checks
Page 42CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence diagram for reservation
:librarian:books:student
refers
request for reservation
ask for book details
tells book details
issue book when available
Page 43CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
:books:student
:librarian
1: refers
2: request for reservation
3: sk for book details
4: tells book details
5: issue book when available
ACTIVITY DIAGRAM
Activity diagram shows the flow of events within our systems. The components are:
a) Start stateb) End statec) Transactiond) Decision Boxe) Synchronization Barf) Swim lane
Page 44CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
show id card
select operation
show lib card
issue renewal return
start operation
close operation
valid student
update database
validation/sucess
databaselibrarianstudent
Page 45CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
STATE CHART State chart diagrams shows a life cycle of a single class. The state is a condition where the object may be in. The components are:
a) Start stateb) End statec) Stated) Transition
verify student details
make/remove reservation
pay dues(if any)
take/return book
Page 46CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
calculate dues
issue book
update catalog
request for stock
manages
supply
Component diagramComponent diagrams provide a physical view of the current model. A component
diagram shows the organizations and dependencies among software components, including source code components, binary code components, and executable components.
Component diagram for library system
library
book stock details.jar
update application.dll
Page 47CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
library
book stock details.jar
update application.dll
Page 48CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Deployment DiagramDeployment diagram shows the physical architecture of the hardware and software in the system.
You can show the actual computers and devices, along with connections they have to each other and it can also show dependencies between components.
Deployment diagram for Library System
backup
library server
client-1 client-2
>> >><<<<Tcp/Ip>>
Page 49CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Placement Section
SOFTWARE REQUIREMENT SPECIFICATION
Introduction
This product is developed for the purpose of Managing and maintaining details & its work of the placement section.
DOCUMENT CONVENTIONS
AGG : AggregateACCR : AccredationPO : Placement OfficerADMIN : Administration
INTENDED AUDIENCE
Company HR Placement Officer Developers Principal
PROJECT SCOPE
The scope of the project is to maintain the entire student database of the college ie student ID, Branch, 10th &12th marks, Aggregate details.
REFERENCES
UML 2 ToolKit by WLLEY publications.
DESCRIPTION
PRODUCT PERSPECTIVE
The origin of the product is derived from the examination section of the college. The examination section of the college contains only the academic details of the student during his course of study in the college.
Page 50CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
In this product we maintain academic details, co-circular and extra circularactivites of every particular student
PRODUCT FEATURES
The main feature of this product is that it provides the entire details of the student in the format the company requires.
System should keep a track of students placed in organisation (ATLEAST ONE) and yet to be placed including the non placed student lacking areas.
BENEFITS
A well organised view of placements reducing confusion and steps to be taken for best results and reduce paper work and time of processing and saving turn over caused by them.
CLASS DIAGRAMClass diagram shows structure of the software system. The class diagram shows a set of
class, interfaces and their relationships. The components are:a) Classb) Relationship
The forms of relationship are:1. Associations 2. Aggregations3. Generalization4. Compositions5. Dependency
Page 51CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
placement section
placement officer : string
collecting()approaching()
college
college code : stringaccr : stringname : string
approaching hr()counselling student()
student
name : stringid : intbranch : stringresume
fill application()appoinment letter()
company
name
return test()gd()hr interview()select student()
hr
namecompany name
shortlist students()select students()reject students()
technical interview
shortlisted student list
conduct interview()
USE CASE DIAGRAMUse case diagram is created to visualize the interaction of our system with the outside
world. The components of use case diagrams are:
Use Case: Scenarios of the system
Actors: some or something who is interacting with the system
Relationship: semantic link between use case and actor. The forms of relationship are:
a) Associationb) Dependencyc) Generalization
Page 52CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Use-Case Diagram For Placement Section:
college management
company
visit recuriment
aptitude
gd
technical interview
personal interview
approaching hr
confermation letter
hr
counselling
student
Page 53CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
INTERACTION DIAGRAM
An interaction diagram models the dynamic aspects of the system by showing the relationship among the objects and message they may dispatch. There are two types of interaction diagrams:
SEQUENCE DIAGRAM
Sequence diagram shows the steps to steps what much happen to accomplish a piece of functionality provided by the system. The components are:
f) Actors g) Objectsh) Messagei) Lifelinej) Focus of Control
COLLABORATION DIAGRAM
Collaboration diagrams displays object interactions organized around object and their links to one another. The components are:
d) Actore) Objectf) Link
Page 54CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence Diagram of Notification:
placement college company student
request
approach
accept
confermation
notify students
Collaboration Diagram of Notification:
placement
college
company
student
1: request
2: accept
3: approach4: confermation5: notify students
Page 55CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence Diagram of Interview:
placement section
company studenttudentss
approach
fill application
eligible students
written test
shortlisted students
technical round
Collaboration Diagram Of Interview:
placement section
company
studenttudentss
1: approach
2: fill application3: eligible students4: written test
5: shortlisted students6: technical round
Page 56CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Sequence Diagram Of Appointment:
company student hr
hr round
shortlist student
appoinment letter
join
Collaboration Diagram of Appointment:
company
student hr
1: hr round2: shortlist student
3: appoinment letter
4: join
Page 57CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
ACTIVITY DIAGRAM
Activity diagram shows the flow of events within our systems. The components are:
a) Start stateb) End statec) Transactiond) Decision Boxe) Synchronization Barf) Swim lane
Page 58CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Activity Diagram For Placement Section:
request
student
fill application
company
written test G.DH.R. Interview
Technical Round
Short Listed Students
Conduct Interview
Select/Reject Students
Page 59CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
STATE CHART State chart diagrams shows a life cycle of a single class. The state is a condition where the object may be in. The components are:
a) Start stateb) End statec) Stated) Transition
State-Chart Diagram For College:
Conformation Letter
Approach Companies
Issue Applications
Page 60CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
State-Chart Diagram For Company:
G.D
Written Test
H.R.Interview
Select Students
State-Chart Diagram For Student:
Fill Application
Attend Interview
Appointment Letter
Page 61CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
State-Chart Diagram For H.R.Manager
Shortlist Students
Select Students
Reject Students
Component diagram
Component diagrams provide a physical view of the current model. A component diagram shows the organizations and dependencies among software components, including source code components, binary code components, and executable components.
Page 62CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Component Diagram For Placement Section:
placement.java
college.java
company.java
student.java
written test.dll
gd.dll
hr interview.dll
Deployment DiagramDeployment diagram shows the physical architecture of the hardware and
software in the system. You can show the actual computers and devices, along with connections they have to each other and it can also show dependencies between components
Page 63CN CASE TOOLS LAB III yr IT
Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.
Deployment diagram for Placement Section:
placement
college
company
student
Page 64CN CASE TOOLS LAB III yr IT