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shutvik [7]
2 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]2 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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Differentiate between refraction of light and reflection of light
lapo4ka [179]

Refraction represents a change in the direction of propagation when beams of light encounter a medium with a different density.

Reflection is the return of light to the medium it came from when it encounters a mirror.

3 0
3 years ago
Read 2 more answers
ppose we have a Rectangle class that includes length and width attributes, of type int, both set by the constructor. Define an e
vichka [17]

Answer:

Check the explanation

Explanation:

#include <bits/stdc++.h>

using namespace std;

class Rectangle{

  public:

      int length;

      int breadth;

      Rectangle(int l,int b){

          length = l;

          breadth = b;

      }

      int area(){

          return length*breadth;

      }

      int perimeter(){

          return 2*(length+breadth);

      }

      bool equals(Rectangle* r){

          // They have the exact same length and width.

          if (r->length == length && r->breadth == breadth)

              return true;

          // They have the same area

          if (r->area() == area())

              return true;

          // They have the same perimeter

          if (r->perimeter() == perimeter())

              return true;

          // They have the same shape-that is, they are similar.

          if (r->length/length == r->breadth/breadth)

              return true;

          return false;

      }

};

int main(){

  Rectangle *r_1 = new Rectangle(6,3);

  Rectangle *r_2 = new Rectangle(3,6);

  cout << r_1->equals(r_2) << endl;

  return 0;

}

8 0
3 years ago
4. In this problem, we consider sending real-time voice from Host A to Host B over a packet-switchednetwork (VoIP). Host A conve
mina [271]

Answer:

The time elapsed is 0.017224 s

Solution:

As per the question:

Analog signal to digital bit stream conversion by Host A =64 kbps

Byte packets obtained by Host A = 56 bytes

Rate of transmission = 2 Mbps

Propagation delay = 10 ms = 0.01 s

Now,

Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.

Time taken to generate and convert all the bits into digital signal is given by;

t = \frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}

t = \frac{56\times 8}{64\times 10^{3}}          (Since, 1 byte = 8 bits)

t = 7 ms = 0.007 s

Time Required for transmission of the packet, t':

t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}

t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s

Now, the time elapse between the bit creation and its decoding is given by:

t + t'  + propagation delay= 0.007 + 2.24\times 10^{- 4} s + 0.01= 0.017224 s

8 0
3 years ago
_____ is the unauthorized entry into a computer system via any means
Montano1993 [528]
Hi,

The word you are looking for is "Hacking".

"Hacking is unauthorized entry into a computer system via any means."

Hope this helps.
r3t40
4 0
3 years ago
Can i get any information on this website i'd like to know what its for ?
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Explanation: torsearch.org is a safe search engine mainly used for dark wed purposes. It does not track your location nor give any personal information.

8 0
3 years ago
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