Computer Networks -Development of Ethernet - Assignment Example

NETWORKING – PROBLEM SOLVING REPORT by (Student Name) Course name Professor (Tutor) Institution The Date a) In modern switched Ethernet networks, it is not unusual for a router to receive more traffic than it is able to forward due to the relative speeds of the Ethernet network versus the links from the router to other networks or the Internet. In such circumstances, packets would be randomly dropped either by the router or by the interconnecting switch when buffers are filled. Briefly discuss how polling could be used to overcome this problem, and discuss two advantages and two disadvantages of this approach. SNMP provides network managers and administrators ability to poll their devices. Polling allows regular management information collection and from it, the network administrator can always tell the Network Management system of the various thresholds that may require corrections when crossed. For instance, the Network Management System might poll the status of an interface on a router. This polling is suitable for addressing the above problem in routers. Advantages of polling: 1) Polling uses less bandwidth mainly WAN links 2) Polling is also friendly to high latencies specifically on WAN links 3) Polling requires no administrative privileges to monitor services on the remote host devices. 4) Polling is also open standard and there cost effective. Disadvantages of Polling 1) Polling does not gather the specific device performance information in relation to services such as utilization of virtual memory. 2) Polling does not offer the ability to leverage automatic re-switching on of failed services always through the Advanced Alert Manager. 3) Polling as a way of network monitoring provides security risks. (Bing, Jin and Guangsha, 2008, 377) b) Prior to the development of full-duplex Ethernet, Token Bus networks provided a means through which communication could be achieved on a shared medium without collisions. Briefly describe how devices transmitted on a network using Token Bus (ignore token management) and discuss the performance token bus as transmission speed increases. Bus is a LAN whereby the stations on the bus/tree form a sort of a logical ring. Each and every station in the ring gets assigned spaces always in an ordered sequence. The last station in the sequence is followed by the last station in the sequence. And each station is aware of the addresses of the stations on both sides (left and right) in the sequence. Token Bus is mainly applicable in those applications requiring real-time processing the minimal delay times. Given the number of collisions which are unpredictable, Ethernet is not a recommended protocol, and hence the Token Bus. Token Bus adapts both Ethernet and token ring features. It has both the physical configuration a bus topology (Ethernet) and the predictable delay (collusion-free) characteristic of the token ring. It is therefore, a physical bus with an operation as that of a logical ring. Token bus logically organizes stations into ring. A token is then passed among the various stations in the ring. In case a station wants t send data, it has to wait to capture the token before sending the data. However, the stations communicate through a common bus like in Ethernet. Each stations realizes the token once they are done communicating or in a case when a device of a higher priority needs to perform a transmission. When the speed of transmission is increased, in most cases, throughput is expected to rise with the rise in speeds. For a bus, this increase will reach a maximum beyond which no change will registered in throughput despite further speed increases. For instance, when the average frame time equalizes the minimum packet time, no additional speed will be consumed by a station in a token bus. This is due to minimum time limit restriction on the device. (Bing, Jin and Guangsha, 2008, 383) c) Collisions and broadcasts have a negative impact on the throughput of an Ethernet network. (1) Describe how collisions and broadcasts influence the throughput on an Ethernet network using a bus topology. Collision refers to the effect of two or more devices within the same Ethernet topology attempts to send data at the same time. When a collision is detected, the packets being transmitted will both be discarded and this causes delays in the data transmission hence lowering the network throughput. At the same time, both devices can transmit at the same time in which case both devices detects it, each of the devices will wait for a random amount of time and keep retrying until the transmission is successful. Broadcast refers to a type of communication where the communicating device sends a single copy data which is then delivered to every device in that specific network segment. An increase in the number of broadcasts reduces the available bandwidth. This is because all the broadcast traffic is always forwarded to all the ports. Also As the number of broadcasts increases, the processing power of the computer/devices reduces and these reduces the throughput of the network. (1) Repeaters, hubs, bridges, and switches are all devices that can be used in the construction of Ethernet networks. Briefly discuss how each of these devices would influence the impact of collisions and broadcasts within a single Ethernet network. Repeaters and hubs: These are layer one devices. While repeaters address the issues of attenuation, that is the signal loss over a distance. Hubs on the other hand are multi-port repeaters. These devices have no decision making abilities. Also referred to as dumb devices. All the devices which are attached to hub are in one collision domain. This implies that any attempt to send data at the same time will cause a collision. Repeaters and hubs, plus all the other devices attached to the hub are in one broadcasting domains. Being in one broadcasting domain implies that, any broadcast frame from one device/host is always received by all devices/hosts in the network. Bridges and switches are second layer devices: These are intelligent devices with ability to make decisions on which ports to receive which frame. These devices plays a major role in reducing collisions in a network by breaking down a given collision domain into smaller ones. After which the devices will buffer and forward frames between the small domains. With the modern switches, a configuration can be done to allow each host has its own collision domain which in turn allows simultaneous data transfer without any collision. Hosts attached to switch or even a domain, belong to one broadcast domain. (Bing, Jin and Guangsha, 2008, 421) Question 2 a) IPv4 uses a checksum to determine if any errors have occurred in the transmission of a datagram, however this checksum only checks for errors in the IPv4 header. Discuss why only the header is checked by IPv4 for errors, and discuss whether it is necessary to check for errors at all given that Ethernet already checks the entire frame using CRC. Ipv4 only checks for errors on the header because of a number of reasons: I. Almost all the higher level encapsulating data in IPv4 have checksum fields covering for the entire packets. This implies that the encapsulated data does not have to be checked for errors by the checksum for the iPv4. II. Also, it is only the header of the IPv4 packet that changes for every routing. However, the data do not change. It is therefore only logical to check errors for the dynamic section rather than the entire packet. III. Finally, it is for speed. Give that the Internet routers direct millions and millions of packets, recalculation of the checksum will therefore lower the packet processing speed. The main reasons for checking for errors in Ethernet despite CRC having checked the entire packet for error is because TCP/IP basically runs over multiple datalink protocols. In these protocols, some do not provide error checking services. Also, the mechanism employed in datalink only checks one end of the broadcast domain. On the other hand, TCP gives an end to end check and ensures reliable transmission (Nichols et al., 1998. 231). b) Given a subnet mask of 255.255.192.0, demonstrate for an IP address of your choosing the Calculation of the range of addresses that are valid from that network (host ID all 0s through to host ID of all 1s). Note: The IP address you choose must consist of four decimal numbers that are greater than or equal to 101, and must be odd numbers. 255.255.192.0 – 11111111.11111111.11000000.00000000(Mask) 192.168.101.101 – 11000000.10101000.01100101.01100101(IP Address) Bit-wise AND Operation of Mask and IP – 11000000.10101000.01000000.00000000- 192.168.64.0 Subnet address 192.168.64.0 Complement (Bit-wise NOT) of Mask – 00000000.00000000.00111111.11111111 - 0.0.63.255 Subnet Broadcast Address = (Subnet address + Complement (Bit-wise NOT) of Mask ) = 192.168.64.0 + 0.0.63.255 = 192.168.127.255 Range of addresses = 192.168.64.1 – 192.168.127.254 c) The fragmentation mechanism provided by IPv4 can contribute to network congestion. Briefly describe how fragmentation works and, given that there is no change to higher-level data, explain how this mechanism could contribute to congestion. IP fragmentation refers to the process of breaking down datagram into smaller chunks/fragments to form packets which can pass over link with the minimal MTU (maximum transmission unit) possible. These fragments are then reassembled by the receiving end. Internet Protocol layer receives a datagram, it then determines the exit interface for the datagram. The layer then compares the size of the datagram against the exit interface maximum transmission unit. The IP layer then performs the fragmentation in case that it is necessary using the sender/intermediate routers. When the packets reach the receiving end, they are reassembled. Network congestion takes place in form of router buffer overflows. In other words, it occurs when networks send packets more than the network can accommodate. Congestion might cause a high transmission rate. Fragmentation increases the number of packets transferred and this raises congestion. Fragmentation is therefore recommended for application in those instances when there is mechanism in place to suppress congestion (Nichols et al., 1998. 248). d) Consider implementing a negative acknowledgement mechanism for IPv4 that would alert a source host to a datagram being discarded due to a corruption in the path between the communicating hosts at any router in the path or the destination host. i. Briefly describe how your protocol would work. A negative acknowledgement (NACK) is a feedback from the receiver denoting that a packet sending was not successful. My protocol to provide a reliable data transfer is designing a channel that deliver segments and wait for an ACK/NACK before sending another message. The receiver will then send an appropriate packet on reception of the message or on failure after a given time span. A NACK is sent when the segments arrives at the receiving end with errors. When the sender receives back a NACK, the packet is retransmitted and waits for either ACK/NACK before sending another segment. This is a stop-and-wait protocol. It is a form of protocol where a sender will wait until a confirmation from the receiver is received by the sender on the state of the previously sent segment. ii. Illustrate the operation of your protocol using a time diagram/sequence diagram (see Session 5 Slide 39 for an example), including both successful and unsuccessful delivery of datagram. The Green arrows shows acknowledgement (ACK) The Blue arrows shows segments of data flow The Red arrows shows negative acknowledgement (NACK) iii. Discuss two advantages and two disadvantages of your protocol. Advantages: I. The main advantage of this protocol is its accuracy. The next frame is only transmitted on the confirmation of the success of the first one. II. The protocol is simple and easy to implement. Disadvantages I. This method is inefficient. This is because it slows the process of data transmission. II. The protocol is not robust. That is, in case of acknowledgement loss, the receiver may not tell whether the latter is a retransmission or a completely new packet. Question 3 a) Both RIPv2 and OSPFv2 routing protocols receive updated information about the state of the network from their neighbours. However it can be said that the decisions made by vector-distance routing protocols (such as RIPV2) are based on “second hand” information, whilst the decisions made by link state routing protocols (OSPFv2) is based on “first hand” information. Briefly describe what information is contained in the updated information routers receive, where the information comes from, and why it is considered “second hand” or “first hand”. Routers send hello packets to protocol enabled interfaces to discover its neighbors and thereafter establish adjacencies. The routers with a common datalink become the neighbours in the protocol if their hello packets contain that information that meets the adjacency formation criteria. Routers may then build link state packets depending on their local interfaces configured for prefixes leaned from the various adjacent routers. Routers generally flood LSPs to all their neighbors which are adjacent, a part from a neighbor from which the same LSP was received by the current router. The updated information received by routers consists of four processes alongside Routing information base. The routing information base consists of the link state database and also the forwarding database. The four processes includes receive – update – decision -forward. Routers adapt various routing protocols. Depending on the protocol adopted by a router, the information sent is either categorized as 'first hand' or 'second had'. Routers that listens to second hand information sends their entire rooting tables to the adjacent neighbors, while, those routers that build routing tables based on firsthand information I.e. by generating the exact topology for route computation of the entire network. (Expósito, Trujillo and Gamess, 2010. 341 ) b) Final rooting table Rooter Cost Path A 1 C-A B 3 C-E-B C 0 C D 4 C-A-F-D E 2 C-E F 2 C-A-F c) For each of the fields listed below appearing in the IPv4 and IPv6 headers, briefly describe the purpose of those fields, explain how they are related, and explain why a change has been made: i. IPv4 defines both header length (HLEN) and total length fields, whilst IPv6 defines only a payload length. IPv4 header length (HLEN) defines the length of the header of an IP packet in bits and accounts for the length, version, TOS, IHL, Flags, Identification etc. while the total length defines the total length of the packet including the header and the body. IPv6 payload length is 16bit value defining the size of the payload, always in octet. This change was necessitated to simplify the IP header and make transmission faster in IPv6. ii. IPv4 defines a protocol field, whilst IPv6 defines a next header field. The protocol field indicates the next protocol level contained within the data portion in the packets. Next header filed defines the type of the next header in IPv6 packet. The field determines the transport layer to be used by a payload. In case of extension headers, the field will specify the extensions headers which follow next. It uses the same values as the protocol filed in IPv4 but the design effect is to enable a presently defined range of next headers and easy extension. iii. IPv4 defines a time to live (TTL) field, whilst IPv6 defines a hop limit field. Time to Live (TTL) is used by IPv4 to limit the amount of time allowed for a packet to stay within a network. This field is assigned 8 bits and time is given in seconds. Hop limit field in IPv6 replaced the time to live field in the previous version (IPv4). The value is reduced by 1 for every visited node by the packet. The packet is discarded when the counter reaches zero. (Expósito, Trujillo and Gamess, 2010, 344-345 ) References Bing, S., Jin, H. and Guangsha, C., 2008. Development of Ethernet/CAN-bus NIC based on DSP. Chinese Journal of Scientific Instrument, 29(2),. Expósito, J., Trujillo, V. and Gamess, E., 2010, March. Easy-EIGRP: a didactic application for teaching and learning of the enhanced interior gateway routing protocol. In Networking and Services (ICNS), 2010 Sixth International Conference on (pp. 340-345). IEEE. Nichols, K., Blake, S., Baker, F. and Black, D., 1998. Definition of the differentiated services field (DS field) in the IPv4 and IPv6 headers (No. RFC 2474). Read More