Explain what might happen if two stations are accidentally assigned the same hardware address?

Complete Question:

Remember that in the CSMA/CD protocol, after a collision, the adapter pauses for K. 512 bit times, where K is randomly chosen. a. For the first collision, if K=100, what is the waiting duration before sensing the channel again for a 1 Mbps broadcast channel? And for a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

With K set at 100, the duration the adapter must wait after detecting a collision is calculated using the following formula:

• Tw = K*512*bit time

The bit time is determined by the inverse of the channel's bandwidth, expressed in bits per second. Therefore, for the instances mentioned in the question, we get:

a) BW = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

Answer:

//To facilitate user input, the Scanner class is imported

import java.util.Scanner;

//The Solution class is being defined

public class Solution {

   //The main method is declared here, marking the start of program execution

   public static void main(String args[]) {

       //A Scanner object named 'scan' is instantiated to gather user input

       Scanner scan = new Scanner(System.in);

       //User is prompted to specify the size of the array

       System.out.print("Enter the range of array: ");

       //The user input is stored in arraySize

       int arraySize = scan.nextInt();

       //Initialization of userArray with the size specified by arraySize

       int[] userArray = new int[arraySize];

       //A counter is initialized to track the number of elements entered in the array

       int count = 0;

       //The following while loop will continue until the user finishes inputting the elements of the array

       while (count < arraySize){

           System.out.print("Enter each element of the array: ");

           userArray[count] = scan.nextInt();

           count++;

       }

       //A blank line is outputted for better readability

       System.out.println();

       //A for loop iterates to print all elements of the array in a single line

       for(int i =0; i <userArray.length; i++){

           System.out.print(userArray[i] + " ");

       }

       //Another blank line is printed for clarity

       System.out.println();

       //A for loop is utilized to reverse the contents of the array

       for(int i=0; i<userArray.length/2; i++){

           int temp = userArray[i];

           userArray[i] = userArray[userArray.length -i -1];

           userArray[userArray.length -i -1] = temp;

       }

       //A for loop prints each element of the reversed array in one line

       for(int i =0; i <userArray.length; i++){

           System.out.print(userArray[i] + " ");

       }

   }

}

Explanation:

The program is annotated to provide a thorough explanation.

The for-loop responsible for reversing the array operates by splitting the array into two segments and swapping elements from the first segment with those from the second segment. During each loop, an element from the first segment is temporarily stored in temp variable, and then that element is replaced with the corresponding element from the second segment. Subsequently, the element from the second segment is updated with the value held in temp.

Answer:

Below is the JAVA program:

import java.util.Scanner;   //to take input from user

public class Main {  //name of the class

  public static void main(String[] args) {  //beginning of main method

      Scanner input = new Scanner(System.in);  //creates Scanner instance

      int integer = input.nextInt();  //defines and collects an integer for the number of words

      String stringArray[] = new String[integer]; //initializes an array to hold strings

      for (int i = 0; i < integer; i++) {  //iterates through the array

          stringArray[i] = input.next();         }  //collects strings

      int frequency[] = new int[integer];  //creates an array for holding frequencies

      int count;         //initializes variable to calculate frequency of each word

      for (int i = 0; i < frequency.length; i++) {  //iterates through frequency array

          count = 0;  //sets count to zero

          for (int j = 0; j < frequency.length; j++) {  //iterates through array

              if (stringArray[i].equals(stringArray[j])) {  //checks if element at ith index matches jth index

                  count++;                 }            }  //increments count

          frequency[i] = count;      }  //stores count in the frequency array

      for (int i = 0; i < stringArray.length; i++) {  //iterates through the string array

          System.out.println(stringArray[i] + " " + frequency[i]);         }    }    } //displays each word and its frequency

Explanation:

To illustrate the program, consider:

let integer = 3

for (int i = 0; i < integer; i++) is used to input strings into stringArray.

On the first pass:

i = 0

0<3

stringArray[i] = input.next(); takes a word and saves it to the ith index of stringArray. For example, if the user types "hey", it will be in stringArray[0].

Then, i increments to i = 1

On the second pass:

i = 1

1<3

stringArray[i] = input.next(); takes another word and assigns it to stringArray[1]. If the user enters "hi", then hi will be in stringArray[1]

Next, i becomes i = 2

On the third pass:

i = 2

2<3

stringArray[i] = input.next(); captures a word and places it in stringArray[2]. If the user types "hi", it goes to stringArray[2]

Then, i increments to i = 3

The loop terminates since i<integers is false.

The next outer loop for (int i = 0; i < frequency.length; i++) and inner loop for (int j = 0; j < frequency.length; will check each word and if (stringArray[i].equals(stringArray[j])) identifies any duplicates. Words that repeat will have their count incremented and the frequency array will record these values. For example, hey appears once while hi shows up twice, thus resulting in the final outputs:

hey 1

hi 2

hi 2

The visual representation of the program and its outcome based on the example provided is attached.