Advantages and Disadvantages of Datagram Network - Math Problem Example

NETWORKS REPORT Name Course Tutor University City and State Date Question 1 a) There are two networks in the given diagram. The networks use a star topology. Star topology involves several devices connected to a central switch. In the case of the given diagram, despite only one client computer being sown, the presence of a switch means that other devices are connected to the client side network, hence leading to the conclusion that the topology used is of the star variant (Yadav, Kaushik & Verma 2014, p. 80). Regarding the number of networks, only two networks are in the diagram given. This is due to the presence of only one router. A router is used to connect to at least two networks by serving as gateways to each of the networks involved (Cisco). In the figure given, only the router connects the client side network to the server side network, making them two discrete networks connected through one router. b) Each layer contains the following I. Client device , in this case a computer, contains the physical, data link, application, presentation and session layers. II. Client side switch contains the network and transport layers III. The router contains the network and transport layers IV. Server side switch contains network and transport layers V. The server contains the physical, data link, application, presentation and session layers The server and client devices contain the physical layer due to their connection to ethernet cables and other physical components (Kumar, Dalal & Dixit 2014, p.463). They also have the data link layer that corrects errors in the physical layer. Additionally, the server and client devices run applications such as email and browsers. These applications deal with certain file formats that have to be converted to network compatible formats by the Presentation Layer. Finally, the network connection sessions of these applications is maintained by the Session layer. The router and the switch uses the Network Layer to decide the most appropriate virtual circuits that the packets are to be transported from one point to another (Kumar, Dalal & Dixit 2014, p.464). Transport layer then kicks in to take charge of the actual transportation of the data packets. c) When data is being sent through a network, it is directed based on the IP address. The router and the switch both use the IP address to determine the destination of the data. But once the data leaves Layer 3, Network layer and enters layer 2, the data link layer, it meets 2 sub layers. The first one is called the LLC, Logical Link Control Layer, and the second one is the MAC, Media Access Control layer. The LLC uses the IP addresses but the MAC layer uses MAC addresses. Every device connected to the network has a unique MAC address. The data link layer matches the IP addresses with their respective MAC addresses. Resultantly, data is then transmitted physically to the specific MAC address of the destination. Question 2 a) Different signals have varying transmission speeds due to the interference. Wireless transmission is prone to more interference than the wired variety. Wired transmissions are subject to electromagnetic interference and radio frequency interference (Cisco Systems 2003). Wireless transmissions on the other hand have more barriers due to their unguided nature. In addition to electromagnetic and radio frequency interference, wireless systems are affected by physical barriers such as houses, hills and trees make it difficult for signals to penetrate (Harwood 2009, p. 3). Moreover, environmental factors such as lightning and fog weaken signals too. b) The potential problems that can arise from increased voltage include increase in operating temperature. The temperature may have counter effective outcomes on the network. At extreme temperatures, network devices can be damaged. c) Propagation time is defined as how long it takes for data to travel at the speed of light from source to destination (O3B n.d). To reduce this delay, an administrator can start by replacing slow propagation media with high propagation ones for example replacing coaxial cables with fiber optic cables. Transmission time refers to the time taken for data to travel through media(O3B n.d).. To reduce this time and effectively reduce latency, the media involved in the transmission should have high transmission speeds. Optical fibre cables have higher speeds than the traditional twisted cables. In the same light, wired media has higher speeds than wireless media. Queuing time is the delay packets have to experience at the router before they are transmitted (O3B n.d).. This delay contributes to latency. To tackle this issue, the administrator can utilize a router with higher processing speeds. Also, having routers with higher buffer memory will reduce the potential of packets getting lost. Loss of packets occurs if more packets keep flowing to the router while it’s buffer memory is already filled. In such a scenario, some packets are discarded hence is lost. To reduce processing delay, make sure that the network has high priority on the processor. This end can be achieved by simply terminating all processes that are utilizing the processor, this way only network activities will be given full processing priority. Significant speed improvements can be experienced once the processor load is eased. d) Manchester and Differential Manchester encoding schemes for 10011101 Start 1 2 3 4 5 6 7 8 End Data Manchester Differential Manchester ii) In Manchester decoding, synchronization is achieved by the mid bit transition while on differential Manchester decoding, synchronization is realized through middle transitions which are also used for clock information (Atmel n.d, p. 4). However, in Manchester decoding, transitions are made following the given bits unlike in differential Manchester decoding which is guided by the form of data. iii) Since data is random, the voltage provided to DC components will also vary. These voltage variations cause the baseline to stutter a situation referred to as baseline wandering (Radzyner 2003, p. 27). The two schemes solve these problem by implementing self-synchronization. e) Frequency-division multiplexing (FDM) and time- division multiplexing (TDM) are both utilized in GSM technology for cellular networks. FDM is unique in that it carries signals that are of different radio frequencies (RF wireless world 2012). The frequencies distinctly separated to avoid overlapping issues. Each signal occupies a narrow bandwidth and they are all transported at the same time. On the other hand, TDM separates its signals on the grounds of time intervals. A single transmission media is used however; the multiplexed signals are separated by time intervals (RF wireless world 2012). This separation allows the signals to use the full bandwidth. In GSM, the frequency can be divided into several channels in accordance with FDM then, each channel can be divided further into time intervals using the principles of TDM. Question 3 a) Advantages of datagram network 1. If there is a link failure, then the protocol will automatically recalculate the routing tables effectively preventing data transmission errors (Thakur 2016). 2. The autonomy of re-computing routes saves the administrator the trouble of manually altering the routing algorithm whenever a link failure occurs. Disadvantages of datagram network 1. The network sends packets meant for the same host through the different routes. This behavior sometimes results to packets arriving out of order which would go against the key requirement of the network protocol being developed by the colleague in question (Thakur 2016). 2. The fact that packets can move through any route in the network leads to congestion and flow control issues that are not present in virtual circuits which have predetermined paths. Advantages of virtual circuit network 1. Each packet in a virtual circuit network has a circuit identifier that determines the route that packet has to follow. Consequently, packets that are destined to a common destination use similar paths, this leads to packets arriving in sequence which is the desired outcome of the network protocol being developed (Thakur 2016). 2. Virtual circuits have efficient resources management since no wastage is done when the network is not transmitting any packets, just like datagram networks (Thakur 2016). However the virtual circuit networks have an added advantage since the end systems do not have to reorganize packets which further increases the efficiency of these networks. Disadvantages of virtual circuit network 1. In the occurrence of a link failure, traffic engineering is required to avoid the interruption of transmission (Thakur 2016). 2. Each packet in this network has to have a circuit identifier, an added attribute that is not needed in datagram networks. b) Reasons why a destination device might not respond to address resolution protocol (ARP) requests 1. Host busy. ARP works by converting IP addresses to their corresponding MAC addresses (Reifschneider 2013). These two corresponding addresses are cached in an ARP table for future reference. If a given IP address is missing from the ARP table, then the protocol broadcasts a message to all connected devices. The device that matches the given IP address with its MAC address should then respond with its location and only after that may the packet transmission commence. However, if the destination device is busy, it might not be able to respond to the ARP request. 2. More than one machine has the same IP address. When this happens, the machines responses will keep overlapping each other and the ARP protocol will not be able to resolve the addresses. 3. Destination device is in a different network. If the destination device which owns the IP addresses that is being broadcasted is connected to a different network, then it will not be able to respond to the ARP requests. This outcome occurs because the destination device has no direct route to communicate with the ARP protocol. Therefore, the ARP request will not reach the device and similarly, the device will not return any response. c) 11101 11001101 1011 ______ 1111 1011 ______ 1001 1011 _______ 0100 1011 1111 1011 ________ 100 remainder d) 1100 1100 1100 1100 0101 1100 0101 1100 + ____________________ 10010 0001 0010 0000 0010 0001 0010 0001 0001 0001 0001 0001+ __________________ 0011 0010 0011 0010 Complement checksum = 1100 1101 1100 1101 References Atmel n.d, Manchester Coding Basics, Atmel.com, retrieved 26th July 2016,< http://www.atmel.com/images/atmel-9164-manchester-coding-basics_application-note.pdf> Cisco n.d, CCNA: Network Media Types, Cisco Systems, retrieved 26th July 2016,< http://www.ciscopress.com/articles/article.asp?p=31276> Cisco n.d, What is a Network Switch? 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