Conceptual Framework of Network - Assignment Example

INTERNET TECHNOLOGIES: ANSWERS Student’s Name Course Professor’s Name University City (state) Date Short Answer Questions Question 1: Advantages and Disadvantages of the Star, Bus, and Mesh Physical Topologies and Examples of Each Star Topology In the star topology, all network components are connected to a device, which is located centrally (Clark 2003). The device may be a switch, router, or hub. Consequently, all the communication passes through the central device before being routed to the right destination. The central device, which is sometimes called a hub, acts as a junction for connecting the several nodes within the topology. The star topology has several advantages. Firstly, it ensures centralized management of the network, which enhances security. Secondly, it is easy to add new devices or nodes to the network. Thirdly, as compared to other topologies, data does not have to be transmitted unnecessarily through all workstations, which improves performance (Bagad 2007). Finally, the failure of a single node does not have an adverse effect network. Likewise, the star topology has its drawbacks. Firstly, it is dependent on a central device, which if it was to fail, the whole network would fail as well (Clark 2003). Secondly, the number of devices or nodes that can be added on the network depend on the capability of the hub. Finally, using a central device increases network cost. An example of a star topology is a home network in which all devices within the household are connected to a single router for Internet connectivity. Bus Topology The bus topology is a kind of a network in which every device is connected to a single cable. Therefore, it transmits data in one given direction. It has several advantages. Firstly, it is easy to expand by simply connecting cables. Secondly, it is cost-effective as less cable is required and a central device is not necessary. Finally, it is easy to understand. One of its drawbacks is that if the cable fails, the entire network fails (Clark 2003). Secondly, cables have limited length. Thirdly, its performance can deteriorate due to heavy traffic or increased nodes. Furthermore, the bus topology is often slower as compared to other topologies. An excellent example of the bus topology is implemented in a computer in which the CPU, I/O devices, and the memory are connected on a common data and address bus. Mesh Topology In the mesh topology, devices and network nodes are interconnected. Therefore, each node sends its data and relays signals from other nodes. The main advantage of this topology is that various devices transmit data simultaneously, which means that the whole network can withstand high levels of traffic. Secondly, this topology enhances redundancy. Finally, modifying or expanding this topology can be done without affecting the flow of traffic (Clark 2003). The first disadvantage of this topology is that the cost of implementation is high because many connections are needed. Secondly, redundancy may be too high. Finally, installing and maintaining such a network is hard and expensive. The most common example of a mesh topology is the Internet. Question 2: OSI and TCP/IP The OSI model is a conceptual framework that standardizes and characterizes how applications communicate in a network. It divides a communication system into abstract layers, each with its distinct function. The goal of the OSI model is to improve compatibility among communication systems and protocols (Clark 2003). The model is made up of seven layers: physical, data link, network, transport, session, presentation, and application layers. On the other hand, the TCP/IP model consists of four layers: application, transport, Internet, and network. The OSI model is considered a better model as compared to TCP/IP because of various reasons. Firstly, it provides a similar platform for hardware and software developers, which encourages the development of networking components with increased compatibility. One of the key issues facing the computer industry during its earlier years is that every computer manufacturer developed their own protocols to support their hardware. The result was that there was no interoperability among various devices and software making collaboration difficult. Therefore, the OSI model was developed with the view of enhancing interoperability by creating a standardized framework to manage data communications (Bagad 2007). Secondly, the OSI model is considered superior because it allows data to be transmitted in small segments. Therefore, the layered approach makes the process of troubleshooting easier. Overall, the OSI model divides layers, which ensures that problems in one layer do not affect the adjacent layer (Gitlin, Hayes & Weinstein 2012). Furthermore, it explains in detail the functions of each layer, hence promoting standardization within the industry. Finally, enhanced standardization allows for multiple-vendor device and software development. Despite the fact that the OSI model is superior, it has not managed to take over from TCP/IP due to various reasons. To begin with, both these models share a similar layered architecture. Apart from sharing a common application layer, their transport and network layers are comparable. In addition, the TCP/IP model supports several routing protocols, most of which are the most commonly used. Another reason why the TCP/IP model has persisted is that it is scalable and supports the client-server architecture. The fact that it scalable ensures that it supports multi-vender development just like the OSI model (Clark 2003). Furthermore, it works independently of the operating system is use, which means that regardless of the operating system, communication would still occur without challenges. Overall, just like the OSI model, the TCP/IP model was developed as an open model to support multiple protocols. Question 3: Calculation of the Bit Rate and the Signal Levels From the question, the bandwidth is 3.5 MHz and the signal to noise ratio is 133. In data communication, it is essential to consider the data rate limits in bits per second that can be sent over a given channel. Essentially, a channel’s data rate depends on the bandwidth, the quantity of noise, and the levels of signals used (Gitlin, Hayes & Weinstein 2012). Essentially, enhancing the signal levels increases the chances of an error occurring, which also diminishes the system’s reliability. The Nyquist theorem helps us to calculate the upper limit of the bit rate. According to the theorem, C = 2 B log22n for a noiseless channel (White 2015). While B is the bandwidth, C refers to the capacity in bits per second. However, in the given problem, there is an aspect of noise. Therefore, the Shannon’s theorem is used. C = B log2 (1 + SNR). B = 3,500,000Hz and the SNR = 133 Therefore, C = 3.5MHz log2 (1+133) C = 3.5MHz log2 (134) C = 3.5MHz log2 (134) C = 3.5MHz * 7.066089 C = 24.73 Mbps From this, the highest bit rate for this channel is 24.73Mbps. To improve the capacity of this channel one can enhance the signal-to-voice ratio or expand the bandwidth The Shannon formula shows that the channel’s upper limit is 24.73Mbps. For better performance, we could select a lower value, say 21Mbps, for instance. After that, we could utilize the Nyquist formula to compute the signal levels. C = 2 B log2L Thus 21Mbps = 2 * 3.5MHz log2L 21Mbps = 7MHz log2L log2L = 3 L = 8 Thus, the approximate signal levels for the channel are eight. Question 4: IPv4 and IPv6 Private Addressing, Ranges and Sizes, and Conflicts on the Internet. An Internet Protocol (IP) address is an identifier, which is given to an interface or group of interfaces at the network layer of the TCP/IP protocol. Essentially, every IP address identifies the destination and source of IP packets (Clark 2003). The IPv4 continues to be in use but there are fears that the number of addresses it offers are running out. Therefore, the IPv6 was developed to increase the number of addresses available and simply addressing and enhance security (Gitlin, Hayes & Weinstein 2012). For IPv4, every device or node has an interface on which TCP/IP is enabled and assigned a logical address. It is important to note that the IPv4 address is logical as it is implemented at the network layer of the model, and lacks a relationship with the addresses utilized at the network interface layer. As the name suggests, the IPv4 32-bit long addresses. The IPv4 addressing syntax makes use of a string of 1s and 0s in 32 bits. A dotted decimal notation of four decimal numbers (between 0 and 255) is used to represent the strings in order to simplify the syntax. For example, 192.168.3.24 represents a 11000000101010000000001100011000 string. Therefore, IPv4 makes use of conversations from binary to decimal and vice versa. IPv4 also has prefixes that are utilized to express the range of possible addresses, routes, and subnets prefixes that an administrator assigns to subnets. To express the prefix of an IPv4 address, one has to identify the fixed high-order bits as well as their value and then express the prefix as StartingAddress/PrefixLength. For instance, 131.107.0.0/16 has a range of 65,536 addresses since 216 is equal to 65,536 probable addresses. Since the length of the prefix is 16, all addresses in the range start with similar 16 bits i.e. 131.107. The subnet mask can also be used in configuring IPv4 address prefixes. The IPv4 packet header is often made up of 20bytes of data. There are several kinds of IPv4 addresses: unicast, multicast, and broadcast. Unicast is used for one-to-one communication and is assigned to a given interface in a subnet. Multicast is implemented on one or more interfaces on many subnets to facilitate one-to-many data communication. Finally, broadcast is implemented to all interfaces in a network to ensure one-to-everyone communication. IPv4 can be public, private, or illegal. Public addresses are used to route connectivity to the Internet (Gitlin, Hayes & Weinstein 2012). Essentially, an organization may decide to use any addressing scheme it wants for its intranet. However, if the organization wants to connect to the Internet using such addresses, a conflict can occur since the addresses have already been assigned to another organization. Therefore, addresses that conflict with others are termed as being illegal. Because each interface in an IPv4 requires its own address, Internet designers reserved some IPv4 addresses to be used for private purposes in their intranets (Clark 2003). The use of private addresses for internal needs and the utilization of public addresses to connect to the Internet ensures that there are no conflicts in the Internet space. Since private addresses are not used to connect to the Internet directly, the use of the same private addresses in different locations does not lead to conflicts. Without a doubt, the differentiating factor between IPv4 and IPv6 is size. While IPv4 is only 32-bit long, IPv6 is 128 bits in length. This means that the possible addresses in IPv4 are 4,294,967,296 whereas IPv6 has 340,282,366,920,938,463,463,374,607,431,768,211,456 possible addresses. The syntax of IPv6 is known as a colon-hexadecimal. Therefore, it includes the conversation of binary to hexadecimals. IPv6 utilizes unicast, multicast, and anycast addresses in delivering packets from a single source to multiple destinations. For unicast in IPv6, one can utilize global addresses, sit-local addresses, link-local addresses, and local addresses (Bagad 2007). Generally, IPv6 identifiers are derived from IEEE EUI-64 or IEEE 802 addresses. Question 5: POP3 Session States (Closed, Authorization, Update, or Transaction) and Diagram Illustrating the Four States and the Movement of POP3 among Them POP 3 (post office protocol 3) enables the receipt e-mails without the need for an extensive system for transporting messages. It offers support for a message user agent that enables a workstation to retrieve mails being held by the mail server and forwarding outgoing mails for relaying onwards (Clark 2003). The normal operation in POP3 entails downloading the mail and deleting the copy of the mail on the server. Thus, a POP3 client establishes a connection with the server on TCP port 110 and the server responds with a greeting (Bagad 2007). While the client sends commands, the server responds to them in an offline status through the various states of authorization, transaction, and update. Finally, the TCP connection is closed. The figure below illustrates how POP3 moves among the various states. In the authorization phase, the client has to establish a connection with the server through port 110 and the server responds with a greeting. After that, the client has to identify and authenticate itself to the server. In authentication, the user has to provide a username and password. If they match, the client can successfully log into the system. However, if the credentials do not match, the server gives an error message and closes the connection (Bagad 2007). Once authorization is successful, the server locks the mailbox to prevent modifications from taking place. In the transaction state, emails are downloaded from the server to the client through a “maildrop.” Once the transfer is complete, the client issues a QUIT command to end the transaction (Bagad 2007). The next state is the update state in which the downloaded files are deleted from the server. Finally, the client or the server may close the connection. Question 6: Distributed Hash Table (DHT), its use in P2P networks, and examples The distributed hash tables were designed to address the need for efficient and effective public distributed systems. The Internet offers a possibility of exchanging data directly and without the need for having a centralized processing point (Tanenbaum & Van Steen 2007). Essentially, distributed hash tables are an extension of hash tables, which are data structured utilized in mapping keys to values, storing key/value pairs, and retrieving values by using those keys (El-Ansary et al. 2003). Hash tables are often effective when utilized in conventional applications. A distributed hash table can be defined as a class of decentralized distributed systems, which offer lookup services that are similar to the hash table: key/value pairs are stored in a table and all participating nodes that retrieve the value of any key. DHTs can be distinguished from the conventional client-server architectures in a variety of ways. Firstly, DHTs offer scalability, which means that their functionality is not affected even when there is a large traffic or node count. Secondly, they offer a decentralized operation without the need for a centralized device to control communication. As a result, DHTs are made up of a number of parts. Thirdly, DHTs provide a more efficient load balancing among nodes (Tanenbaum & Van Steen 2007). Consequently, no single node can be overburdened with traffic. Fourthly, DHTs take into account the fact that data communication within a network is always changing. Essentially, as nodes join and leave the network, the traffic changes as well. Fifthly, DHTs are developed to enhance the speed of data retrieval and routing so as it can be attained in logarithmic time. Finally, DHTs are robust, which enables them to withstand attacks. However, DHTs have various system issues including churn management, locality, heterogeneity, underlay network issues, and storage models and soft states (El-Ansary et al. 2003). DHTs find applications in P2P networks to store and retrieve data in an efficient manner. Usually, P2P applications track identities and addresses of peers and route messages between peers. Since peers do routing and naming, the IP becomes a low-level transport tool. P2P networks lack a centralized control and nodes are functionally symmetric (Tanenbaum & Van Steen 2007). In addition, P2P has many unreliable connections. However, it has numerous benefits including increased reliability through replication and distribution, enhanced capacity through parallelism, and automatic configuration (El-Ansary et al. 2003). DHTs find many applications, including in torrents where users share files. Other applications include naming services (such as Twine), file sharing, communication services, database query processing, event notification, and global file systems (such as OceanStore). References Bagad, I.D.V., 2007. Data Communication. Technical Publications. Clark, MP 2003, Data networks, IP and the Internet: protocols, design and operation, John Wiley & Sons, New Jersey. El-Ansary, S, Alima, LO, Brand, P & Haridi, S 2003, ‘Efficient broadcast in structured P2P networks’, In International workshop on Peer-to-Peer systems (pp. 304-314), Springer Berlin Heidelberg. Gitlin, RD, Hayes, J & Weinstein, SB 2012, Data communications principles, Springer Science & Business Media. Tanenbaum, AS & Van Steen, M 2007, Distributed systems, Prentice-Hall. White, C 2015, Data Communications and Computer Networks: A business user's approach, Cengage Learning, New York. Read More

On the other hand, the TCP/IP model consists of four layers: application, transport, Internet, and network. The OSI model is considered a better model as compared to TCP/IP because of various reasons. Firstly, it provides a similar platform for hardware and software developers, which encourages the development of networking components with increased compatibility. One of the key issues facing the computer industry during its earlier years is that every computer manufacturer developed their own protocols to support their hardware.

The result was that there was no interoperability among various devices and software making collaboration difficult. Therefore, the OSI model was developed with the view of enhancing interoperability by creating a standardized framework to manage data communications (Bagad 2007). Secondly, the OSI model is considered superior because it allows data to be transmitted in small segments. Therefore, the layered approach makes the process of troubleshooting easier. Overall, the OSI model divides layers, which ensures that problems in one layer do not affect the adjacent layer (Gitlin, Hayes & Weinstein 2012).

Furthermore, it explains in detail the functions of each layer, hence promoting standardization within the industry. Finally, enhanced standardization allows for multiple-vendor device and software development. Despite the fact that the OSI model is superior, it has not managed to take over from TCP/IP due to various reasons. To begin with, both these models share a similar layered architecture. Apart from sharing a common application layer, their transport and network layers are comparable.

In addition, the TCP/IP model supports several routing protocols, most of which are the most commonly used. Another reason why the TCP/IP model has persisted is that it is scalable and supports the client-server architecture. The fact that it scalable ensures that it supports multi-vender development just like the OSI model (Clark 2003). Furthermore, it works independently of the operating system is use, which means that regardless of the operating system, communication would still occur without challenges.

Overall, just like the OSI model, the TCP/IP model was developed as an open model to support multiple protocols. Question 3: Calculation of the Bit Rate and the Signal Levels From the question, the bandwidth is 3.5 MHz and the signal to noise ratio is 133. In data communication, it is essential to consider the data rate limits in bits per second that can be sent over a given channel. Essentially, a channel’s data rate depends on the bandwidth, the quantity of noise, and the levels of signals used (Gitlin, Hayes & Weinstein 2012).

Essentially, enhancing the signal levels increases the chances of an error occurring, which also diminishes the system’s reliability. The Nyquist theorem helps us to calculate the upper limit of the bit rate. According to the theorem, C = 2 B log22n for a noiseless channel (White 2015). While B is the bandwidth, C refers to the capacity in bits per second. However, in the given problem, there is an aspect of noise. Therefore, the Shannon’s theorem is used. C = B log2 (1 + SNR). B = 3,500,000Hz and the SNR = 133 Therefore, C = 3.

5MHz log2 (1+133) C = 3.5MHz log2 (134) C = 3.5MHz log2 (134) C = 3.5MHz * 7.066089 C = 24.73 Mbps From this, the highest bit rate for this channel is 24.73Mbps. To improve the capacity of this channel one can enhance the signal-to-voice ratio or expand the bandwidth The Shannon formula shows that the channel’s upper limit is 24.73Mbps. For better performance, we could select a lower value, say 21Mbps, for instance. After that, we could utilize the Nyquist formula to compute the signal levels.

C = 2 B log2L Thus 21Mbps = 2 * 3.5MHz log2L 21Mbps = 7MHz log2L log2L = 3 L = 8 Thus, the approximate signal levels for the channel are eight. Question 4: IPv4 and IPv6 Private Addressing, Ranges and Sizes, and Conflicts on the Internet.

Read More