Assume there is a machine with the IP address 129.82.102.63 with netmask /23, and with a parent NW whose netmask is 255.255.224.0. For each answer, do not include any spaces, give full IP addresses/netmasks where these are requested, give the "/" as part of the answer for slash notation.

Answer:

#include<iostream>

using namespace std;

//main function

int main(){

   //initialization

   float a1,a2;

//display the message

cout<<"Enter the first number: ";

   cin>>a1;  //store the value

   cout<<"Enter the second number: ";

   cin>>a2;   //store the value

   //display the calculation result

   cout<<"The sum is: "<<a1+a2<<endl;

   cout<<"The Subtraction is: "<<a1-a2<<endl;

   cout<<"The product is: "<<a1*a2<<endl;

   cout<<"The Quotient is: "<<a1/a2<<endl;

}

Explanation:

Create the main function and declare the two variables of float but we can enter the integer as well.

display the message on the screen and then store the input enter by the user into the variable using the cin instruction.

the same process for second input as well, display the message and store the input.

after that, print the output of the calculation. the operator '+' is used to add the two numbers like a1+a2, it adds the number stored in the variable.

similarly, the subtraction operator is '-', product '*' and quotient operator '/'.

and finally, we get the desired output.

Answer:

a. 7 bits

b. 8 bits

c. 9 bits

d. 24 bits

Explanation:

Required

Number of bits per word

The formula to get the number of bits per word is:

Where

n = bits

N = numbers

Solving (a) 0 to 100

There are 101 digits between 0 and 100 (inclusive)

So:

Substitute 101 for N in

101 can not be expressed as an exponent of 2.

So, we substitute 101 for the closest number greater than 101 that can be expressed as an exponent of 2.

So:

Express 128 as an exponent of 2

Apply law of indices

Hence, 7 bits are required

Solving (b): 0 through 255

There are 256 digits between 0 and 255 (inclusive)

So:

Substitute 256 for N in

Express 256 as an exponent of 2

Apply law of indices

Hence, 8 bits are required

Solving (c): 0 through 256

There are 257 digits between 0 and 256 (inclusive)

So:

Substitute 257 for N in

257 can not be expressed as an exponent of 2.

So, we substitute 257 for the closest number greater than 101 that can be expressed as an exponent of 2.

So:

Express 512 as an exponent of 2

Apply law of indices

Hence, 9 bits are required

Solving (d): 0 through 10000000

There are 10000001 digits between 0 and 10000000 (inclusive)

So:

Substitute 10000000 for N in

10000000 can not be expressed as an exponent of 2.

So, we substitute 10000000 for the closest number greater than 101 that can be expressed as an exponent of 2.

So:

Express 16777216  as an exponent of 2

Apply law of indices

Hence, 24 bits are required

Answer:

A. No, it would not succeed.

B. Yes, the attack will succeed.

Explanation:

In the seven-layer OSI model of computer networking, packet strictly refers to a protocol data unit at layer 3, the network layer. So if an attacker wants to insert bogus/ fake packets this will happen at the network layer.

A. IPSec works on network layer (the IP layer) of the OSI model protecting and authenticating IP packets. It provides data integrity, authentication and confidentiality between participating peers at the IP layer.

Thus, inserting bogus packets into the communications by an attacker will not have any affect on security of a system which has IPSec security and therefore it will not succeed.

B. SSL/TLS is an application layer protocol that fits into layer 5 to 7 of the OSI model. However, an attack by inserting bogus packets into the communications will succeed as SSL only provides protection for layer 5 to 7 and not the network layer.

Answer:

Explanation:

#include <iostream>

#include <string>

#include<vector>

using namespace std;  

vector<int> permute(vector<int>, vector<int>);

string encrypt(vector<int>s1 , vector<int> t1, string p);

string decrypt(vector<int>s1, vector<int> t1, string p);

int main() {

  string plaintext = "cryptology";

  string plaintext2 = "RC4";

  vector<int> S(256);

  vector<int> T(256);

  int key[] = { 1,2,3,6 };

  int key2[] = { 5,7,8,9 };

  int tmp = 0;

  for (int i = 0; i < 256;i++) {

      S[i] = i;

      T[i] = key[( i % (sizeof(key)/sizeof(*key)) )];

  }

  S = permute(S, T);

  for (int i = 0; i < 256 ;i++) {

      cout << S[i] << " ";

      if ((i + 1) % 16 == 0)

          cout << endl;

  }

  cout << endl;

  string p = encrypt(S, T, plaintext);

  cout << "Message: " << plaintext << endl;

  cout << "Encrypted Message: " << " " << p << endl;

  cout << "Decrypted Message: " << decrypt(S, T, p) << endl << endl;

  tmp = 0;

  for (int i = 0; i < 256;i++) {

      S[i] = i;

      T[i] = key2[(i % (sizeof(key) / sizeof(*key)))];

  }

  S = permute(S, T);

  for (int i = 0; i < 256;i++) {

      cout << S[i] << " ";

      if ((i + 1) % 16 == 0)

          cout << endl;

  }  

  cout << endl;

  p = encrypt(S, T, plaintext2);

  cout << "Message: " << plaintext2 << endl;

  cout << "Encrypted Msg: " << p << endl;

  cout << "Decrypted Msg: "<<decrypt(S, T, p) << endl << endl;

  return 0;

}

string decrypt(vector<int>s1, vector<int> t1, string p) {

  int i = 0;

  int j = 0;

  int tmp = 0;

  int k = 0;

  int b;

  int c;

  int * plain = new int[p.length()];

  string plainT;

  for (int r = 0; r < p.length(); r++) {

      i = (i + 1) % 256;

      j = (j + s1[i]) % 256;

      b = s1[i];

      s1[i] = s1[j];

      s1[j] = b;

      tmp = (s1[i] + s1[j]) % 256;

      k = s1[tmp];

      c = ((int)p[r] ^ k);

      plain[r] = c;

      plainT += (char)plain[r];

  }

  return plainT;

}  

string encrypt(vector<int>s1, vector<int> t1, string p) {

  int i = 0;

  int j = 0;

  int tmp = 0;

  int k = 0;

  int b;

  int c;

  int * cipher = new int [p.length()];

  string cipherT;

  cout << "Keys Generated for plaintext: ";

  for (int r = 0; r < p.length(); r++) {

      i = (i + 1) % 256;

      j = (j + s1[i]) % 256;

      b = s1[i];

      s1[i] = s1[j];

      s1[j] = b;

      tmp = (s1[i] + s1[j]) % 256;

      k = s1[tmp];

      cout << k << " ";

      c = ((int)p[r] ^ k);

      cipher[r] = c;  

      cipherT += (char)cipher[r];

  }

  cout << endl;

  return cipherT;

}

vector<int> permute(vector<int> s1, vector<int> t1) {

  int j = 0;

  int tmp;

  for (int i = 0; i< 256; i++) {

      j = (j + s1[i] + t1[i]) % 256;

      tmp = s1[i];

      s1[i] = s1[j];

      s1[j] = tmp;

  }

  return s1;

}