BN204 Database Technologies

1. Bus topology uses a single common line or data path for all connecting nodes hence it requires very short length of cable and its structure is simple and easy to maintain.
2. In bus topology data passes through a single cable and all the nodes are connected to that single coil, so this topology is very simple and reliable with respect to hardware point of view.
3. Additional nodes can be connected very easily at any point in an existing bus topology.

Disadvantage of Bus topology-

1. As the bus topology consist of a simple structure, it is difficult to detect or diagnosis the fault.
2. In case of extension of nodes in the bus topology, routers need to be configured, so it requires experts consult for installation.
3. Each node in bus topology is directly connected to a central node, hence it is required to decide which node will be in use at a time. If the central node fails sometimes then the whole network will be down as all the nodes are connected to the central node. This is one of the major disadvantages of bus topology.

Advantage of Star topology-

1. In star topology failure of one node disconnects that node only not the whole network.
2. All the nodes of star topology are directly connected to a central node, which actually helps to detect the fault or failure of the system and to isolate them.
3. Access protocol of a star network is very simple as it has only one central node and one peripheral node in the whole network.

Disadvantage of Star topology-

1. In star network each nodes are directly connected to the central node which is situated in the center and requires a large length of cable.
2. Due to large length of cables it is little costly.
3. Star network is very difficult to expand as it is unknown how much length of cable is required to connect to the central node.
4. The star network system is totally central node dependent, if the central node fails the whole network stall.

Advantage of Star topology-

1. Mesh topology avoid media failure very efficiently rather than other network topologies.
2. In mesh topology every device or node is always have multiple paths between them to send or receive signals from one device to another device. If failure occurs in one path then signals can be routed to the other path.
3. Mesh topology is easy to troubleshoot because each of the nodes connected to the mesh topology network is independent of all others.

Disadvantage of Star topology-

1. Mesh topology is difficult to install and reconfigure.
2. Due to large no of cable connections its construction structure is complex.
3. It is a costly network topology against other topologies.

Real example of Bus topology is the LAN connection in an institute or office.

Real example of Star topology is the ATM service provided by the banks.

Real example of Mesh topology is in the military market.

Explain encapsulation and decapsulation in a five layer TCP/IP protocol suite. How does multiplexing and de-multiplexing differ from encapsulation and decapsulation?

Encapsulation- In TCP/IP protocol suite when data moves from upper layer to lower level outgoing transmission each layer includes a bundle of relevant information combined in one packet which contains header and the actual data. The data package containing the header and the data from the upper layer then becomes the data that is repackaged at the next lower level with lower layer's header. Header is used as a supplemental data placed at the beginning of a block. This supplemental data is used at the receiving side to extract the data from the encapsulated data packets. This packing of data at each layer is known as data encapsulation.

Decapsulation- The whole process of Encapsulation is reversely repeated when data is reached in receiver side or destination computer. As the data flows from lower to upper levels each layer unpacks the corresponding header and uses the information contained in the header to deliver the packet to the exact network application waiting for the data. This reversed process of encapsulation is called Decapsulation.

Difference between multiplexing, demultiplexing and Encapsulation, Decapsulation-

Multiplexing is the technique which allows the simultaneous transmission of multiple signals across a single data link where encapsulation is the process of wrapping header and the data part together in one block during transmission from one layer to another layer in TCP/PT protocol network.

In case of the demultiplexing the stream is fed into a demultiplexer which separates it back to its component transmission and directs them to their correspondent lines where as in the decapsulation only the reverse encapsulation is performed.

Calculate the approximate bit rate and signal level(s) for a 6.8 MHz bandwidth system with a signal to noise ratio of 132.

Shanon’s formula to find the upper limit:

Data rate or bit rate =B log₂ (1+SNR)

  =10⁶ log₂ (1+132)

 =10⁶ log₂133

  = (2123851.641)/3

  = 7Mbps

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.

OSI (Open System interconnection) model is protocol independent acts as a communication gateway between the user and the network with a vertical approach through transport layer which guarantees the delivery of pockets. OSI model consists of 7 different layers and in each layers protocols are hidden and easily changeable with the change of technologies.

Where TCP/IP are two protocols of the model namely Transmission Control Protocol/Internet protocol having only four layers instead of 7.  

• In 1990 OSI layered architecture model was the dominant data communication and networking literature. Everyone believed that the OSI model will become the ultimate standard of data communication system. But TCP/IP protocol suite becomes the dominant data communication model because it is used and tested extensively in the internet and practically the OSI model was never completely created. So practically in networks TCP/IP protocol model is implemented and used rather than OSI.
• Advantage of OSI model-

1. OSI model follows vertical approach.
2. It has a separate session and presentation layer.
3. OSI model is a reference model around which the networks are build which is generally used as a guidance tool in the network.
4. It defines al the protocols and its services very clearly.
5. It is protocol independent.
6. It is also machine independent.
7. Easy to add multiple networks in the model.

• Disadvantage of OSI model-

1. As OSI model is protocol independent, so it has a problem of fitting protocols into the model.
2. The data integrity facility availed by OSI model cannot be applicable to all the applications.
3. It is complex to implement.

• Advantage of TCP/IP model-

1. This model is OS independent.
2. It supports routing protocols and internetworking between organizations.
3. It supports client-server architecture.

• Disadvantage of OSI model-

1. It is very complex to implement the set up and manage.
2. Replacing protocols is difficult to implement.
3. It is protocol dependent and services provided by the protocols are not explained separately.

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?

Latency(Delay)= propagation time+ transmission time+ queuing time + processing delay

Here, queuing time = 3.5 µs =3.5 * 10^-6 sec,

Distance =1900 km= 1900*1000=1900000 m

Propagation time= distance / propagation speed

= 1900000 / 2.2 * 10⁸

= 8.6363 ms

Processing delay= time taken by the router to process

     =no of routers * processing time

     = 10 * 1.8 * 10^-6 sec

     = 1.8 * 10^-5 sec

Transmission time= message size / bandwidth

= 5000000 * 8 /10⁶ ms

= 40 ms

Now,

Latency (Delay)= propagation time+ transmission time+ queuing time + processing delay

    =8.6363 + 40 + 1.8*10^-5+3.5 * 10^-6

    =8.6363 + 40 + 0.00018 + 0.0000035

    =48.6364 ms

The transmission time is the most dominant section in total delay and the processing delay is negligible.       
Question 6: 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.

POP3 stands for Post Office Protocol 3 which enables a server to host communication through email and to monitor the servers, domains and mailboxes. It is mainly used by a client to access and download messages from the mail server.

POP3 has some sessions which are explained below-

POP3 Authorization State- POP3 connection is started with listening for connection port 110. Now in authorization session TCP makes a connection with the server and client provide user command with its user identification. If server responses with a positive success indicator then user issue pass command in its password section and goes to transaction session

POP3 Transaction State- In this state client issues one or more commands and server returns positive or negative responses. Servers attends some messages, if it a valid message then it is marked as deleted and send a positive response and returns the highest no of accessed messages and enters to update session.

POP3 Update State- In the update state of POP3, the server deletes all messages marked for deletion, releases resources associated with the client's mail-drop, and closes the TCP connection.

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