Itc542 Internetworking With Tcp-Bus Topology Assessment Answers
Provide short answers to the following six questions. Your answers should be clear, concise and to the point. Prepare a single document (MS Word or PDF, NOT both) along with title page and submit it online using EASTS.Discuss the advantages and disadvantages of star, bus, and mesh physical topologies. Provide real examples of each type.
Explain encapsulation and decapsulation in a five layer TCP/IP protocol suite. How does multiplexing and de-multiplexing differ from encapsulation and decapsulation Calculate the approximate bit rate and signal level(s) for a 6.8 MHz bandwidth system with a signal to noise ratio of 132.
Explain why the OSI model is better than the TCP/IP model. Why hasn't it taken over from the TCP/IP model Discuss the advantages and disadvantages of both models. What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 3.5 µs and a processing time of 1.8 µs. The length of the link is 1900 km, the speed of light inside the link is 2.2 x 108 m/s, the link has a bandwidth of 8 Mbps. Which component(s) of the total delay is/are dominant Which one(s) is/are negligible.According to RFC1939, a POP3 session is one of the following states: closed, authorization, transaction or update. Draw a diagram and explain to show these four states and how POP3 moves between them.
Answer:
- The performance provided by the star topology is better than the other topologies. The content that is meant for a particular node would be delivered to that particular node.
- The expansion of the star topology can be done very easily. A number of nodes can be added to the existing model very easily.
- The server is located in the center hence, the monitoring of the whole network would be very easy.
- In addition to this the failure of a single node does not affect the whole system.
Disadvantages:
- Too much dependency on the central node. Therefore if there is a situation where the central node fails the whole network would be affected.
- This is an expensive topology. The cost of implementation of the topology is very high.
- The performances and the efficiency of the whole network depends on the efficiency of the central node.
Bus Topology
Advantages:
- The implementation and the extension processes of the bus topology is very easy. If there exist a topology comprising of 30 computers, it can easily be extended up to 70 in this topology.
- The design of the topology is very linear and also the topology is very efficient.
- The cost of the implementation of the topology is very low.
Disadvantages
- The node in the Bus topology are interconnected in a linear connection. Hence, the detection of error becomes very difficult.
- It also requires a termination which is efficient for dumping the signals.
- The maintenance of the topology also incurs great cost to the organization.
- The topology is inefficient in the networks where the traffic of the topology is very high.
- The security of the topology is also very efficient and the network can be accessed by all the nodes.
Mesh Topology
Advantages:
- The data transmissions in the topology can be done simultaneously.
- Single node failure and single link failure does not affect the efficiency of the network. Alternate nodes and links are always made available.
- Modification and expansion of the network do not require alteration of the network.
Disadvantages:
- The redundancy of the mesh topology are very high.
- The cost of implementation of the network is also very high.
- In addition to this the maintenance of the topology also involves a very complex process.
The data packets are subjected to an upward and downward movement within the network when transmission takes place from one machine to another machine. Some header files are added to the data packets every time they are transferred form one layer to another. The data packets are generated in the application layer and are encapsulated in the network layer where the data transfer takes place in between both the machines. After these process the decapsulation process takes place. The recipient receives the data in the network layer. And the decapsulation takes place as the data packet goes to the higher levels the header files are slowly removed. And the original data is received in the application layer. In this way the capsulation and the decapsulation process takes place.
The multiplexing and the demultiplexing techniques are not similar to the encapsulation and the decapsulation techniques in a network model. The multiplexing and the demultiplexing techniques are involved with the signals in the network, while the encapsulation and the decapsulation techniques are involved with the data packets in the network.
Given, B= 6.8 MHz (bandwidth)
SNR= 132 (signal to noise ratio)
C= Bit Rate.
C=B log (1+SNR) = 6.8x106 log2 (1+132) = 6.8x106 log2 133 = 48 Mbps.
Let, L be the number of signals
Therefore, C = 2 x B x log2 (L)
0r, 48= 2x6.8xlog2L
Or, log2 L=48/(6.8x2)
Or, log2 L= 3.56 0r 4 (approx)
Or, L = 24= 16.
The OSI model is almost similar to the TCP/IP model but the OSI model has more number of layers than that of the TCP/IP layers. Hence, the OSI model provides more options to the network and also the security of the OSI model is more enhanced than that of the TCP/IP model. But due to the collaboration of the model with the reputed models the TCP/IP model is the preferred one. In addition to this, The OSI model looks impressive theoretically but the TCP/IP is more preferable practically.
The OSI model does the interpretation of the functions at each level of the model. But, the implementation is very complex.
The TCP/IP model is easier to implement but the functionality of the model is slower than that of the other models.
Given, frame size (F)= 5 million bits
Propagation speed = 2.2x 108 m/s
Length of the link = 1900 km = 1900 x 103
Bandwidth = 8 x 106 bps
Queuing time= 10 x 3.5 mS = 35 mS.
Processing delay = 1.8 x 10 mS = 18 mS.
Transmission time = 5 x 106 /8 mS = 62500 =.625 s
Propagation time = 1900 x 103 / 2.2x 108 uS = 8 uS
Total delay time = 35 + 18 + .08 + 62500 = 62551.08 mS = .63 sec
The total delay time is .63 sec and the dominant component is the transmission delay and the negligible component is the propagation time.
The POP 3 protocol has four states:
Authorization: after the establishment of the connection the Authorization procedure takes place.
Transaction: After the Authorization the transactions takes place.
Update: The transactions are updated.
Closed: After the updating of the transactions the POP 3 is closed.
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