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dedylja [7]
3 years ago
8

The IP protocol (and UDP) are called "connectionless" and "unreliable." Describe what those two words mean in the context of dat

a communications?
Computers and Technology
1 answer:
igor_vitrenko [27]3 years ago
6 0

Answer:

Explanation:

In data communications, a protocol is <em>connectionless </em>if it describes communication between two devices in a network without first establishing or verifying establishment of some link (connection) between the two devices. In a connectionless protocol, data are exchanged between processes or devices without a prior check of whether the processes are ready to transmit data or not. In this type of protocol, there's no verification put in place to remember if these processes were or are in communication. Connectionless protocols are also called stateless protocols. Because connectionless protocol cannot be relied upon due to its lack of delivery-assurance policy, it is sometimes referred to as <em>unreliable protocol</em>. In other words, delivery of data packets is not guaranteed. An example of a connectionless protocol is the UDP (User Datagram Protocol).

However, if the protocol defines some checks to ascertain that there is first a connection between two communicating processes before beginning transmission, then, the protocol is <em>connection-oriented. </em>An example of this is the TCP (Transmission Control Protocol). Connection-oriented protocols are reliable since they ensure connection be made between processes before transmitting data packets.

A typical application of a connectionless/unreliable protocol is in the streaming of real time videos or voice communication where packets of data could be dropped and not resent.

An example of the connection-oriented/reliable protocol is analogous to making a telephone call where data packets (voice messages) are not sent until the receiver on the other end makes a connection.

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\large \mathfrak{Answer : }

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A friend of Alex has gifted a movie collection, and Alex is excited to watch them all as quickly as possible. The duration of th
marishachu [46]

To find out the duration of alex's movies, you can use the code in C++ to make this calculation, where you will only need the number of movies and their duration, as well:

<h3>Writing code in C++:</h3>

<em>#include<bits/stdc++.h></em>

<em>using namespace std;</em>

<em>struct Movie {</em>

<em>   int timeBegin, duration, timeEnd;</em>

<em>   bool operator<(const Movie& another) const {</em>

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<em>   }</em>

<em>};</em>

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<em>   vector<Movie> movies;</em>

<em>};</em>

<em>Festival* initialize(int timeBegin[], int duration[], int count) {</em>

<em>   Festival* filmFestival = new Festival;</em>

<em>   filmFestival->count = count;</em>

<em>   for (int i = 0; i < count; i++) {</em>

<em>      Movie temp;</em>

<em>      temp.timeBegin = timeBegin[i];</em>

<em>      temp.duration = duration[i];</em>

<em>      temp.timeEnd = timeBegin[i] + duration[i];</em>

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<em>   }</em>

<em>   return filmFestival;</em>

<em>}</em>

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<em>   int res = 0;</em>

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<em>   for (int i = 0; i < fest->count; i++) {</em>

<em>      if (fest->movies[i].timeBegin >= timeEnd) {</em>

<em>         res++;</em>

<em>            timeEnd = fest->movies[i].timeEnd;</em>

<em>      }</em>

<em>   }</em>

<em>   return res;</em>

<em>}</em>

<em>int main(int argc, char *argv[]) {</em>

<em>int timeBegin[] = {1, 3, 0, 5, 5, 8, 8};</em>

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Answer:

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creativ13 [48]
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Create a simple main() that solves the subset sum problem for any vector ofints. Here is an example of the set-up and output. Yo
ivolga24 [154]

Answer:

The main function is given below. Appropriate comments are made where necessary. In addition to this, I added a full code for the problem to help you understand it fully

Explanation:

// Sublist.cpp : Defines the entry point for the console application.

//

#include "stdafx.h"

#include <iostream>

#include <vector>

using namespace std;

int sumInStack = 0;

int TARGET_SUM = 180;

vector<int> choices;

void subList(vector<int> data, int fromIndex, int endIndex);

void print();

int main()

{  

  vector<int> dataSet;

  int k, j, fromIndex,endIndex;

  bool foundPerfect;

  dataSet.push_back(20); dataSet.push_back(12);     dataSet.push_back(22);

  dataSet.push_back(15); dataSet.push_back(25);

  dataSet.push_back(19); dataSet.push_back(29);

  dataSet.push_back(18);

  dataSet.push_back(11); dataSet.push_back(13); dataSet.push_back(17);

  choices.clear();

  fromIndex=0;

  endIndex=dataSet.size();

  cout << "Target time: " << TARGET_SUM << endl;

  subList(dataSet, fromIndex, endIndex);

  // code provided by student

  system("pause");

  return 0;

}

void subList(vector<int> data, int fromIndex, int endIndex) {

       /*

       * Check if sum of elements stored in Stack is equal to the expected

       * target sum.

       *

       * If so, call print method to print the candidate satisfied result.

       */

       if (sumInStack == TARGET_SUM)

         {

          print();

       }

       for (int currentIndex = fromIndex; currentIndex < endIndex; currentIndex++)

         {

           if (sumInStack + data[currentIndex] <= TARGET_SUM)

             {

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               /*

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               * further.

               */

           subList(data, currentIndex + 1, endIndex);

                  sumInStack -= choices.back();

                  choices.pop_back();

           }

       }

   }

void print()

{

    cout<<TARGET_SUM<<" = ";

    for(int i=0;i<choices.size();i++)

    {

         cout<<choices.at(i)<<"+";

    }

}

8 0
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