Advantages And Disadvantages Of Star, Bus, And Mesh Physical Topologies, TCP/IP Protocol Suite, Bit Rate And Signal Level Calculation, OSI Model Vs. TCP/IP Model

Question 1: Discuss the advantages and disadvantages of star, bus, and mesh physical topologies. Provide real examples of each type.

Question 1:

Discuss the advantages and disadvantages of star, bus, and mesh physical topologies. Provide real examples of each type.

Star Topology:

Advantage: This is set with the better performance where the signals are not important to handle the transmission to the different workstations. The sending of the signals reach the destination with the passing through no more than 3 device. The easy connections are set with the new nodes or the devices which includes the addition without affecting the network. (Baolin et al., 2013).

Disadvantage: This includes the too much dependency with the forms that works on the format with use of the hub, router or the switch with the central device. It also increases the overall cost of the system network. The performance is based on the nodes number which is depending on the central device.

Example: The common carrier networks with the 3D network is set for handling the unshielded twisted pair Ethernet.

Bus Topology

Advantage: The easy implementation and extension will be important for the networks with the quick setup. The initial less expensive forms are set with the other topologies and is cheap as well.

Disadvantage: It is difficult for administer, limiting the cabling length and to handle the station number. The maintenance cost are more, with the lower security that affects the system with the star or the ring network. The proper termination is also needed which is not possible.

Example: In a circle in a family room for the office. (Edge et al., 2016).

Mesh Topology

Advantage: Here, the nodes are set for the distribution of the data and to work on the wired and wireless network. This is mainly to handle the system with the MANETs that are not restricted to the mobile adhoc networks.

Disadvantage: The issues are related to the hopping from one node to the other where there is a possibility of the mismatch of the data.

Examples: For the military market for the radio setup and to accomplish the different military missions. (Alani, 2014).

Question 2:

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

The encapsulation is for the data which moves from the layer which is above to the lower level set for the TCP/IP control stack where there is a relevant information of the header along with the actual data setup. The package of the data contains the header as well as the data that comes from the upper layer. Here, the header is mainly to handle the supplementing of the data placed at the block of the data when there is a proper transmission. (Marin et al., 2015).

The decapsulation is the reverse process which includes the data moving up from the lower to the upper layer where there is an incoming of the transmission process. It includes the unpacking of the layers that tends to correspond to the header and the use of the information that has been in the packet for the delivery of the exact network applications.

Question 2: Explain encapsulation and decapsulation in a five layer TCP/IP protocol suite. How does multiplexing and de-multiplexing differ from encapsulation and decapsulation?

The multiplexing is the process mainly set for the different forms of the multiple data streams which works on the coming from the input of the different sources and then it is combined with the transmission over the single form of the data channel or the data stream. The multiplexing is also for the time division and the frequency division form that has been set to carry out the data based on the similar form of the transmission medium by allocating it to the different forms of the frequency bands. (Baolin et al., 2013).

The multiplexing is for the transmission to the end of the communication. Whereas, at the receiving link, there is a composite signal which is then separated through the use of equipment which is called as the demultiplexer.

Question 3:

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

SNR= 20 log₁₀ Signal Level

132=20log₁₀signal level

e^6.6= signal level

C = 2B log2 M (Nyquist formula)

C= 2X 6800 Hz log₂ signal level= 13600 log₂ e^6.6

Question 4:

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 is considered to be better as it is able to cover all the generic protocol which could easily be shared over the independent network, with the communication in between the network and the end user. The setup is mainly through the vertical approach, where there is a network built up through the connection oriented and the connectionless services. TCP is less beneficial as there are issues related to the standard protocols which could be set at the communication protocol, with allowing the connection for the hosts over the network setup. In the TCP/IP it is seen that the system does not guarantee the delivery of the packages, where there is a use of the horizontal approach, and does not have any form of the presentation or the session layer. (Levesque et al., 2013).  For this, the TCP only provides the connectionless services which does not fit any protocol.

OSI

The advantage is mainly for the system as it works on the different forms of the layers like the application, presentation, session layers. This defines the services, interfaces and the protocols that vary depending upon the clear distinction but still has been a complete protocol independent.

The disadvantage is related to the forms where there is an issue in the fitting of the protocols into the model. Along with this, there is another issue related to the protocols that have been hidden in the OSI model and could replace without confirmation when there is a change in technology.

TCP

The major advantage is that it provides the best services, interface and the protocols to the users. But the major disadvantage is that the model is not able to fit the protocol where there is no easy replacement process of the protocol as well.

Question 3: Calculate the approximate bit rate and signal level(s) for a 6.8 MHz bandwidth system with a signal to noise ratio of 132.

Question 5:

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?

Propagation time= distance/speed

= 1900km/2.2*10^8=8.6ms

Transmission Time= Message Size/Bandwidth

= 5* 10^6 bits /8Mbps =0.625 seconds

Queuing Time= 10 routers * 3.5micro sec = 35 micro sec

Processing Delay= 10 routers * 1.8 micro sec = 18 micro sec

Total delay (Latency)= 8.6ms + 0.625 sec + 35 micro sec +18 micro sec

Queuing time is neglected.

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.

The authorization is mainly for the user to login, approve the password, quit and set the commands for the user, pass and quit.

The transaction is for the operations which involves the messages like the DELE, NOOP etc.

The update is for the deletion of the marked messages which includes the sessions to enter the state with the QUIT command, deleting the messages of the marked positions for the deletion and then quitting. The command used is the QUIT.

Here, the sessions of the receiving of the emails is based on handling the different forms of the dialog processing of the SMTP. (Ethier et al., 2013). It also includes the sessions through which the different states are able to set in the TCP connection with the opening of the POP3 server and working on the acquiring of the resources based on the association with the client. The session also tends to enter the system with the transaction state, where the client request for the different actions. Here, the client also issues the QUIT command where the session is based on the Updates where there is a need for the releasing of the server resources that are set at the time of transaction. It is then the connection of TCP is also closed.

Reference

Edge, C. and O’Donnell, D., 2016. Network Scanning, Intrusion Detection, and Intrusion Prevention Tools. In Enterprise Mac Security (pp. 441-457). Apress.

Wain, A., Reiff-Marganiec, S., Janicke, H. and Jones, K., 2016. Towards a Distributed Runtime Monitor for ICS/SCADA Systems.

Baolin, L., Xianbo, H. and Jin, W., 2013. The Research and implementation of Key Technologies of Deep Packet inspection based on POP3 Protocol in Router Firewall. Journal of Convergence Information Technology, 8(9), p.819.

Alani, M.M., 2014. Tcp/ip model. In Guide to OSI and TCP/IP models (pp. 19-50). Springer International Publishing.

Ethier, T. and Person, C., 2013. Session 4 Presentation-Comprehensive Software Solution for the Management of Complex Oceanographic and Meta Data.

Marín, J.M.F., Naranjo, J.Á.M. and Casado, L.G., 2015. Honeypots and honeynets: Analysis and case study. Handbook of research on digital crime, cyberspace security, and information assurance, pp.452-482.

Levesque, M., Witort, J. and Kelouwani, S., Unwired Planet, Inc., 2013. Method for activating and deactivating client-side services from a remote server. U.S. Patent 8,484,305.