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shutvik [7]
3 years ago
9

Part 1 of 4 parts for this set of problems: Given an 4777 byte IP datagram (including IP header and IP data, no options) which i

s carrying a single TCP segment (which contains no options) and network with a Maximum Transfer Unit of 1333 bytes. How many IP datagrams result, how big are each of them, how much application data is in each one of them, what is the offset value in each. For the first of four datagrams: Segment
Computers and Technology
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

Fragment 1: size (1332), offset value (0), flag (1)

Fragment 2: size (1332), offset value (164), flag (1)

Fragment 3: size (1332), offset value (328), flag (1)

Fragment 4: size (781), offset value (492), flag (1)

Explanation:

The maximum = 1333 B

the datagram contains a header of 20 bytes and a payload of 8 bits( that is 1 byte)

The data allowed = 1333 - 20 - 1 = 1312 B

The segment also has a header of 20 bytes

the data size = 4777 -20 = 4757 B

Therefore, the total datagram = 4757 / 1312 = 4

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Output a table that show the cube of the numbers 1-15<br> (C++)
Rainbow [258]

Answer:

The c++ program to display cube of numbers from 1 to 15 is given below.

#include <iostream>

using namespace std;

int main() {    

   // variables declared and initialized  

   int num = 15, k, cube;    

   cout << " Cubes of numbers from 1 to 15 are shown below " << endl;    

   for( k = 1; k <= num; k++ )

   {

       // cube is calculated for each value of k and displayed

       cube = k * k * k ;

       cout << " \t " << cube << endl;

   }

return 0;

}

 

OUTPUT

Cubes of numbers from 1 to 15 are shown below  

  1

  8

  27

  64

  125

  216

  343

  512

  729

  1000

  1331

  1728

  2197

  2744

  3375

Explanation:

The variables are declared and initialized for loop, cube and for condition in the loop – k, cube, num respectively.

Inside for loop which executes over k, beginning from 1 to 15, cube of each value of k is calculated and displayed. The loop executes 15 times. Hence, cube for numbers from 1 to 15 is displayed after it is calculated.

   for( k = 1; k <= num; k++ )

   {

      cube = k * k * k ;

       cout << " \t " << cube << endl;

   }

In the first execution of the loop, k is initialized to 1 and variable cube contains cube of 1. Hence, 1 is displayed on the screen.

In the second execution, k is incremented by 1 and holds the value 2. The variable cube contains cube of 2, which is calculated, and 8 gets displayed on the screen.

After each execution of the loop, value of k is incremented.

After 15 executions, value of k is incremented by 1 and k holds the value 16. This new value is greater than num. Hence, the loop terminates.

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The answer & explanation for this question is given in the attachment below.

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Epic is fuzzy-figure8
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Answer:

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2 years ago
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What is a utility application that monitors the network path of packet data sent to a remote computer?
stepladder [879]

Answer:

"Traceroute " is the  correct answer.

Explanation:

The traceroute commands in the networking that traces the network path The  traceroute is also displaying the information about the pack delay that is sent over the internet."Traceroute is also known as tracepath it means it calculated the path between the packets in the network. Traceroute takes the one IP address of the computer machine and taking another Ip address to calculate the path between the packets.

Traceroute Is a type of utility application that monitors the network between the packet that is sent into a remote computer.

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