In the Stop-and-Wait flow-control protocol, what best describes the sender’s (S) and receiver’s (R) respective window sizes?

Answer:

The correct answers are A,C,D,E or Remove unwanted files, Back up her data. Scan for viruses, Delete the browsing history.

Explanation:

I just did the test and got it right

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;

}

Answer:

the higher order bits in AL will be true

Explanation:

We know from the truth table that in an OR operation, once one operand is true the result is true, no matter the value of the other operand. From the above AL will have the higher order bits to be true since the first four higher order bits are 1s, 1111_0000, this is regardless of the original values of AL. We cannot however ascertain if the lower bits of AL are true from this. Note 1 evaluates to true while 0 is false

Answer:

b. Lack of coordination

Explanation:

LAteral communication in human resources and organizational communication is known as the communication that exists between peers and colleagues, so that wouldn cause a lack of cirection, delegation or control because its not between bosses and employees, what it would cause would be a lack of coordination between the employees.

Answer:

b

Explanation:

b, lack of coordination

Answer:Encryption is the process

of translating plain text data into something that appears to be random and meaningless. A symmetric key is used during both the encryption and decryption process. To decrypt a particular piece of ciphertex, the key that was used to encrypt the data must be used.

Types of encryption

1.Symmetric encryption :This is also known as public key encryption. In symmetric encryption, there is only one key, and all communication patties use the same key for encryption and decryption.

2.Asymmetric encryption :This is foundational technology for SSL(TLS)

How encryption is being configured

Encryption uses algorithms to scramble your information. It is then transmitted to the receiving party, who is able to decode the message with a key.

Explanation: