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Sergio [31]
3 years ago
11

Consider a channel that can lose packets but has a maximum delay that is known. Modify Protocol rdt2.1 to include sender timeout

and re-transmission. Informally argue why your revised protocol can communicate correctly over this channel

Computers and Technology
1 answer:
Bogdan [553]3 years ago
8 0

Answer:

See attached picture.

Explanation:

As the packets are being lost, so the introduction of re-transmission and timeout are used to identify the lost packets and re-transmit them. So, the revised protocol can communicate correctly over this channel.

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Write a program that prompts the user for an integer, then asks the user to enter that many values. Store these values in an arr
Svetllana [295]

Answer:

//The Scanner class is imported to allow the program receive user input

import java.util.Scanner;

//The class Solution is defined

public class Solution {

   //The main method is defined here and signify the beginning of program execution

   public static void main(String args[]) {

       

       //Scanner object 'scan' is created to receive user input

       Scanner scan = new Scanner(System.in);

       //Prompt display to the user to enter size of array

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

       //User input is assigned to arraySize

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       //userArray is initialized with arraySize as its size

       int[] userArray = new int[arraySize];

       

       //counter to count number of array element

       int count = 0;

       //while loop which continue executing till the user finish entering the array element

       while (count < arraySize){

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

           userArray[count] = scan.nextInt();

           count++;

       }

       

       //A blank line is printed for clarity

       System.out.println();

       

       //for loop to print each element of the array on straight line

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

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

       }

       

       //A blank line is printed for clarity

       System.out.println();

       

       //for loop is use to reverse the array in-place

       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;

       }

       

       //for loop to print each element of the reversed array on straight line

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

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       }

     

   }

}

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The program is commented to give detailed explanation.

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See options below

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I don't know this yet

Learn more about Accenture from

brainly.com/question/25682883

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