Implementation of Subnetting

Aim:

Write a program to implement subnetting and find the subnet masks.

Algorithm :

1.Start the program.

2.Get the frame size from the user

3.To create the frame based on the user request. 4.To send frames to server from the client side.

5.If your frames reach the server it will send ACK signal to client otherwise it will send NACK signal to client.

6.Stop the program

Program

import java.util.Scanner; class Subnet

{

public static void main(String args[])

{

Scanner sc = new Scanner(System.in);

System.out.print(“Enter the ip address”);

String ip = sc.nextLine();

String split_ip[]=ip.split(“\\.”);

//SPlit the string after every . String split_bip[] = new String[4];

//split binary ip

String bip = “”; for(int i=0;i<4;i++){

split_bip[i] = appendZeros(Integer.toBinaryString(Integer.parseInt(split_ip[i])));

// “18” => 18 => 10010=>00010010

bip += split_bip[i];

}

System.out.println(“IP in binary is “+bip); System.out.print(“Enter the number of addresses: “); int n = sc.nextInt();

//Calculation of mask

int bits = (int)Math.ceil(Math.log(n)/Math.log(2));

/*eg if address = 120, log 120/log 2 gives log to the base 2 => 6.9068, ceil gives us upper integer */

System.out.println(“Number of bits required for address = “+bits); int mask = 32-bits;

System.out.println(“The subnet mask is = “ +mask);

//Calculation of first address and last

address int fbip[] = new int[32];

for(int i=0; i<32;i++) fbip[i] = (int)bip.charAt(i)-48;

//convert cahracter 0,1 to integer 0,1 for(int i=31;i>31-bits;i–)

//Get first address by ANDing last n bits with 0

fbip[i] &= 0;

String fip[] = {“”,””,””,””}; for(int i=0;i<32;i++)

fip[i/8] = new String(fip[i/8]+fbip[i]);

System.out.print(“First address is = “); for(int i=0;i<4;i++){ System.out.print(Integer.parseInt(fip[i],2)); if(i!=3) System.out.print(“.”);

}

System.out.println();

int lbip[] = new int[32];

for(int i=0; i<32;i++) lbip[i] = (int)bip.charAt(i)-48; //convert cahracter 0,1 to integer 0,1 for(int i=31;i>31-bits;i–) //Get last address by ORing last n bits with 1

lbip[i] |= 1;

String lip[] ={“”,””,””,””}; for(int i=0;i<32;i++)

lip[i/8] = new String(lip[i/8]+lbip[i]); System.out.print(“Last address is =”);

for(int i=0;i<4;i++){ System.out.print(Integer.parseInt(lip[i],2)); if(i!=3) System.out.print(“.”);

}

System.out.println();

}

static String appendZeros(String s){

String temp = new String(“00000000″); return temp.substring(s.length())+ s;

}

}

Output

Enter the ip address: 100.110.150.10

IP in binary is 01100100011011101001011000001010

Enter the number of addresses: 7

Number of bits required for address = 3

The subnet mask is = 29

First address is = 100.110.150.8

Last address is = 100.110.150.15