Part A One of satellite system provides the maximum data rat

Part A

One of satellite system provides the maximum data rate service of 2Mbps and it uses the packet size of 512 bits. If this satellite is running at the altitude of 700 km. Satellite communication uses the wireless technology, therefore, its medium speed is the speed of light, 3x108 m/s.

1. Find the transmission time, propagation time, and the propagation delay.

2. Stop-and-wait flow control is good for this system? Explain

Part B

In Wireless Sensor Network, two stationary sensor nodes are communicating each other with the distance between two nodes of 240m and the data rate of 20Mbps. Assume 512-byte of frame size is used in this network and the speed of light is 3x108 m/s. Ignore the transmission time.

1. Find the transmission time, propagation time, and the propagation delay.

2. Stop-and-wait flow control is good for this system? Explain

Solution

Ans)

Part A:
1. Transmission time: Transmission time is the total time taken to put entire data on a wire or time taken to leave all the packets from a sender.
              bit rate = 2mbps == 2 * 10^6 ==2000000 bits/satellite
              packet size = 512 bits
              distance = 700km == 700 * 1000 = 700000 m
              speed of light = 3 * 10^8 m/s
          
              transmission time = Packet size / Bit rate
                               = 512 / 2000000
                               =0.000256 seconds.
           So, the total transmission time is 0.000256 seconds.
                              
Propagation time:
                   Propagation time is the total time taken to move the packets from sender to receiver.
                   Propagation time = Distance / propagation speed
                                   = 700000 / 3 * 10^8
                                   =700000 / 300000000
                                   =0.002333 seconds

                   So, the total Propagation time is 0.002333 seconds
                  
Propagation delay:   The propagation delay, is the time it takes a bit to propagate from one router to the next.
Propagation time and the Propagation delay are the one and the same.

2. Here Stop-and-wait flow control is good for this system because in the stop and wait the receiver get ready to receive the data and sender waits for an acknowledgement after every frame for specified time.The receiver in this case waits for the data from a Satellite.The receiver let it be mobile phones waits for the sender to get a message and after getting a packect an acknowledgement is sent for each packet.So, for this type of transmission stop and wait is best method.


Part B:
1.   Transmission time: Transmission time is the total time taken to put entire data on a wire or time taken to leave all the packets from a sender.
       bit rate = 20Mbps = 20 * 10^6 = 20000000
       packet size = 512-bytes = 512 * 8 == 4096 bits
       distance = 240m
       speed of light = 3 * 10^8 m/s
      
       transmission time = Packet size / Bit rate
                           = 4096 / 20000000
                           = 0.0002048 seconds
               So, the total transmission time is 0.0002048 seconds
              
   Propagation time:
                   Propagation time is the total time taken to move the packets from sender to receiver.
                   Propagation time = Distance / propagation speed
                                   = 240 / 3 * 10^8
                                   = 240 / 300000000
                                   = 0.0000008
                       So, the total Propagation time is 0.0000008 seconds          
  
   Propagation delay:   The propagation delay, is the time it takes a bit to propagate from one router to the next.
   Propagation time and the Propagation delay are the one and the same.
  
2. In this case also stop and wait is the best control method because here the communication takes between two stationary sensor nodes .The communication takes place between two connected stations. By using the stop and wait ARQ is the best method because it ensures that no data will be lost.It also decides that data will be come in the ordered manner by using the automatic repeat request.

Part A One of satellite system provides the maximum data rate service of 2Mbps and it uses the packet size of 512 bits. If this satellite is running at the alti
Part A One of satellite system provides the maximum data rate service of 2Mbps and it uses the packet size of 512 bits. If this satellite is running at the alti

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