PART__B
1. Write the code for
i) bit stuffing
#include<stdio.h>
int main() {
int a[15];
int i, j, k, n, c = 0, pos = 0;
// Input the number of bits
printf("\n Enter the number of bits: ");
scanf("%d", &n);
// Input the bits
printf("\n Enter the bits: ");
for (i = 0; i < n; i++)
scanf("%d", &a[i]);
// Perform bit stuffing
for (i = 0; i < n; i++) {
if (a[i] == 1) {
c++;
if (c == 5) {
pos = i + 1;
c = 0;
for (j = n; j >= pos; j--) {
k = j + 1;
a[k] = a[j];
a[pos] = 0;
n = n + 1;
} else {
c = 0;
// Print the stuffed data
printf("\n DATA AFTER STUFFING \n");
printf(" 01111110 ");
for (i = 0; i < n; i++) {
printf("%d", a[i]);
printf(" 01111110 ");
getchar(); // Pause the console window
return 0;
/*o/p:
Enter the number of bits: 9
Enter the bits: 1 0 1 1 1 1 1 0 1
DATA AFTER STUFFING
01111110 1011111001 01111110
*/
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ii) Character stuffing:
#include<stdio.h>
#include<string.h>
void main()
int i=0,j=0,n,pos;
char a[20],b[50],ch;
printf("Enter the Characters: ");
scanf("%s",a);
printf("\nOrginal Data:%s",a);
n=strlen(a);
b[0]='d';
b[1]='l';
b[2]='e';
b[3]='s';
b[4]='t';
b[5]='x';
j=6;
while(i<n)
if( a[i]=='d' && a[i+1]=='l' && a[i+2]=='e')
Computer Networks Laboratory [18ECL76]
Department of Electronics & Communication Engineering 6th Semester
b[j]='d';
b[j+1]='l';
b[j+2]='e';
j=j+3;
b[j]=a[i];
i++;
j++;
b[j]='d';
b[j+1]='l';
b[j+2]='e';
b[j+3]='e';
b[j+4]='t';
b[j+5]='x';
b[j+6]='\0';
printf("\nTransmitted Data: %s\n",b);
printf("Received Data:%s",a);
/*o/p:
Enter the Characters: abcdefg
Orginal Data:abcdefg
Transmitted Data: dlestxabcdefgdleetx
Received Data:abcdefg
*/
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2. Write a program for distance vector algorithm to find suitable path for
transmission.
#include <stdio.h>
struct node {
unsigned dist[20];
unsigned from[20];
} rt[10];
int main() {
int dmat[20][20], n, i, j, k, count = 0;
printf("\nEnter the number of nodes: ");
scanf("%d", &n);
printf("\nEnter the cost matrix:\n");
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
scanf("%d", &dmat[i][j]);
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
rt[i].dist[j] = dmat[i][j], rt[i].from[j] = j;
do {
count = 0;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
for (k = 0; k < n; k++)
if (rt[i].dist[j] > dmat[i][k] + rt[k].dist[j])
rt[i].dist[j] = rt[i].dist[k] + rt[k].dist[j], rt[i].from[j] = k, count++;
} while (count != 0);
for (i = 0; i < n; i++) {
printf("\n\nState value for router %d is \n", i + 1);
for (j = 0; j < n; j++)
printf("\t\nNode %d via %d Distance %d", j + 1, rt[i].from[j] + 1, rt[i].dist[j]);
printf("\n\n");
return 0;
/*o/p:
Enter the number of nodes: 3
Enter the cost matrix:
045
103
890
State value for router 1 is
Node 1 via 1 Distance 0
Node 2 via 2 Distance 4
Node 3 via 3 Distance 5
State value for router 2 is
Node 1 via 1 Distance 1
Node 2 via 2 Distance 0
Node 3 via 3 Distance 3
State value for router 3 is
Node 1 via 1 Distance 8
Node 2 via 2 Distance 9
Node 3 via 3 Distance 0
*/___________________________________________________________________________________
__________________
3. Implement Dijkstra’s algorithm to compute the shortest routing path.
#include <stdio.h>
#define MAX_NODES 15
int n, s, nb, nbs[MAX_NODES], snbs[MAX_NODES], delay[MAX_NODES][MAX_NODES], i, j,
temp[MAX_NODES], ze = 0;
void min();
int main() {
printf("Enter the number of nodes: ");
scanf("%d", &n);
printf("\nEnter the source node: ");
scanf("%d", &s);
printf("\nEnter the number of neighbors to %d: ", s);
scanf("%d", &nb);
printf("\nEnter the neighbors: ");
for (i = 1; i <= nb; i++)
scanf("%d", &nbs[i]);
printf("\nEnter the time delay from source to neighbors: ");
for (i = 1; i <= nb; i++)
scanf("%d", &snbs[i]);
printf("\nEnter the time delays for neighbors:\n");
for (i = 1; i <= nb; i++) {
printf("Enter the time delays from %d: ", nbs[i]);
for (j = 1; j <= n; j++)
scanf("%d", &delay[i][j]);
printf("\nThe time delays for neighbors:\n");
for (i = 1; i <= nb; i++) {
printf("Time delays from %d: ", nbs[i]);
for (j = 1; j <= n; j++)
printf("%3d ", delay[i][j]);
printf("\n");
min();
return 0;
void min() {
int sum, k, y = 1, store = 1;
printf("\n\t\t\tNew Route");
printf("\n\t\t\tTime Delay");
printf("\n");
for (i = 1; i <= n; i++) {
sum = 0;
k = 1;
for (j = 1; j <= nb; j++) {
temp[k++] = delay[j][i];
sum = temp[1] + snbs[1];
for (y = 2; y <= nb; y++) {
if (sum > temp[y] + snbs[y]) {
sum = temp[y] + snbs[y];
store = y;
}
if (s == i) {
printf("\n\t%d +\t%d =\t%d --", ze, ze, ze);
} else {
printf("\n\t%d +\t%d =\t%d\t%d", temp[store], snbs[store], sum, nbs[store]);
/*o/p:
Enter the number of nodes: 3
Enter the source node: 0
Enter the number of neighbors to 0: 2
Enter the neighbors: 1 2
Enter the time delay from source to neighbors: 10 20
Enter the time delays for neighbors:
Enter the time delays from 1: 4 8 12
Enter the time delays from 2: 3 6 9
The time delays for neighbors:
Time delays from 1: 4 8 12
Time delays from 2: 3 6 9
New Route
Time Delay
4+ 10 = 14 1
8+ 10 = 18 1
12 + 10 = 22 1
*/
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________________
4. For the given data, use CRC-CCITT polynomial to obtain CRC code. Verify the
program for the cases
a. Without error
b. With error
#include<stdio.h>
#include<string.h>
#define N strlen(g)
char t[28],cs[28],g[]="10001000000100001";
int a,i,j;
void xor()
for(j = 1;j < N; j++)
cs[j] = (( cs[j] == g[j])?'0':'1');
void crc()
for(i=0;i<N;i++)
cs[i]=t[i];
do
if(cs[0]=='1')
xor();
for(j=0;j<N-1;j++)
cs[j]=cs[j+1];
cs[j]=t[i++];
}
while(i<=a+N-1);
int main()
printf("\nEnter data : ");
scanf("%s",t);
printf("\nGeneratng polynomial : %s",g);
a=strlen(t);
for(i=a;i<a+N-1;i++)
t[i]='0';
printf("\nModified data is : %s",t);
crc();
printf("\nChecksum is : %s",cs);
for(i=a;i<a+N-1;i++) t[i]=cs[i-a];
printf("\nFinal codeword is : %s",t);
printf("\nEnter received message ");
scanf("%s",t);
crc();
for(i=0;(i<N-1) && (cs[i]!='1');i++);
if(i<N-1)
printf("\nError detected\n\n");
else
printf("\nNo error detected\n\n");
return 0;
}
/*op:
Enter data : 1011101
Generatng polynomial : 10001000000100001
Modified data is : 10111010000000000000000
Checksum is : 1000101101011000
Final codeword is : 10111011000101101011000
Enter received message 10111001100010110101000
Error detected
Enter data : 1011101
Generatng polynomial : 10001000000100001
Modified data is : 10111010000000000000000
Checksum is : 1000101101011000
Final codeword is : 10111011000101101011000
Enter received message 10111011000101101011000
No error detected*/
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_____________________
5. Implementation of Stop and Wait Protocol and Sliding Window Protocol
i) Program: Stop and Wait Protocol
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h> // For sleep function
int main() {
int i, f, frames[50];
printf("\nEnter number of frames to transmit: ");
scanf("%d", &f);
printf("\nEnter the %d frames: ", f);
for (i = 1; i <= f; i++)
scanf("%d", &frames[i]);
for (i = 1; i <= f; i++) {
if ((rand() % 2) == 1) {
printf("%d\n", frames[i]);
printf("Acknowledgement of above frames sent is received by sender\n\n");
else {
sleep(3);
printf("Negative acknowledgement: Resend %d frame\n", i);
i = i - 1; // Resend the same frame
sleep(1);
return 0;
/*Output:
Enter number of frames to transmit: 6
Enter the 6 frames: 1 4 5 7 8 9
Acknowledgement of above frames sent is received by sender
Negative acknowledgement: Resend 2 frame
4
Acknowledgement of above frames sent is received by sender
Acknowledgement of above frames sent is received by sender
Acknowledgement of above frames sent is received by sender
Acknowledgement of above frames sent is received by sender
Negative acknowledgement: Resend 6 frame
Negative acknowledgement: Resend 6 frame
Acknowledgement of above frames sent is received by sender
*/
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ii) Program: Sliding Window Protocol
#include <stdio.h>
int main() {
int w, i, f, frames[50];
printf("Enter Window Size: ");
scanf("%d", &w);
printf("\nEnter number of frames to transmit: ");
scanf("%d", &f);
printf("\nEnter the %d frames: ", f);
for (i = 1; i <= f; i++)
scanf("%d", &frames[i]);
for (i = 1; i <= f; i++) {
if (i % w == 0) {
printf("%d\n", frames[i]);
printf("Acknowledgment of above frames sent is received by sender\n\n");
else {
printf("%d\n", frames[i]);
if (f % w != 0)
printf("Acknowledgment of above frames sent is received by sender\n\n");
return 0;
/*Output:
Enter Window Size: 3
Enter number of frames to transmit: 5
Enter the 5 frames: 12 20 87 65 4
12
20
87
Acknowledgment of above frames sent is received by sender
65
Acknowledgment of above frames sent is received by sender
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6. Write a program for congestion control using leaky bucket algorithm.
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h> // For sleep function
int bucket_size;
void bucket_input(int pkt_sz, int op_rt) {
if (pkt_sz > bucket_size) {
printf("\n\n Bucket Overflow\n");
else {
sleep(1);
while (pkt_sz > op_rt) {
printf("\n %d bytes outputted", op_rt);
pkt_sz -= op_rt;
sleep(1);
if (pkt_sz > 0) {
printf("\n Last %d bytes sent\n", pkt_sz);
printf("\n Bucket output Successful \n");
int main() {
int i, op_rate, packet_size;
printf("\n Enter Bucket Size: ");
scanf("%d", &bucket_size);
printf("\n Enter Output rate: ");
scanf("%d", &op_rate);
for (i = 1; i <= 5; i++) {
sleep(1);
packet_size = rand() % 1000; // Using rand() for random packet sizes
printf("\n Packet number [%d] \t Packet size=%d", i, packet_size);
bucket_input(packet_size, op_rate);
return 0;
/*Ouput:
Enter Bucket Size: 500
Enter Output rate: 80
Packet number [1] Packet size=383
80 bytes outputted
80 bytes outputted
80 bytes outputted
80 bytes outputted
Last 63 bytes sent
Bucket output Successful
Packet number [2] Packet size=886
Bucket Overflow
Packet number [3] Packet size=777
Bucket Overflow
Packet number [4] Packet size=915
Bucket Overflow
Packet number [5] Packet size=793
Bucket Overflow
*/
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