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COIT20261 Network Routing and Switching Term 1

   

Added on  2021-06-17

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COIT20261 Network Routing and Switching (Term 1, 2018)Assignment item —Written Assessment-2First Name:_________________________Last Name:____________________________Student ID: __________________________Question NumberMarkallocatedMarks earnedQuestion 1: (10 marks)1.2.3.4.5.6.7.8.For the R4 router 150.3.0.2 is the next hop for the data packet to have161.22.0.15/18 For the R1 router the interface utilized for sending the packet to thenetwork 161.22.0.0/18 is M2. At the point when a packet is sent between 200.11.60.36/24 and150.32.0.240/18 2 hop address should be used for reaching the destinationnetwork. In the event that an information packet produced from the source200.11.60.36/24 achieves the R1 router and the router does not finds thegoal address in the routing table it drops the information parcel. On the off chance that with the goal address 140.21.0.10/22 achieves therouter R2 the interface M0 is utilized for sending the information parcel. The goal address of the packet at R3 router is 220.10.40.5/24 the next hopaddress for sending the information bundle would be 150.3.0.3/16 For the data packet looking out for the R4 router and if the following hopis an immediate conveyance the system 150.3.0.0/16 is the goal address.q.8 Routing table of router R2:For the analysis of the routing table the IP address assigned to the interface of the other routers and its interface is analysed and the following table is created PrefixNetwork addressNext-hop address Interface255.255.255.0220.10.40.0150.3.0.3M0255.255.192.0161.22.0.0150.3.0.3M0255.255.192.0150.32.0.0150.3.0.1M0255.255.0.0150.3.0.0150.3.0.2loopback1-7 1 mark each, q.83 marksQuestion 2: (5 marks)a)Number of packet fragments = (total size of packet) / (MaximumTransmission Unit) = 5400/ 1500 = 3.6 = 4Thus, Datagram with data 5400 byte fragmented into 4 fragments. For the 1st frame:2.5

COIT20261 Network Routing and Switching (Term 1, 2018)Assignment item —Written Assessment-2It carries bytes 0 to 1479 (MTU = 1500 bytes and header length = 20byte)Thus, the total bytes in fragments is maximum 1500-20 =1480For the 2nd frame:It carries bytes 1480 to 2959. For the 3rd fragment:It carries byte 2960 to 4439.For the 4th fragement: It carries bytes 4440 to 5400b)Offset value of the 1st packet fragmentation = 0Offset value of the 2nd packet fragmentation = 185Offset value of the 3rd packet fragmentation = 370Offset value of the 4th packet fragmentation = 5551.5c)IP addresses mainly comes up with 20 bytes of information. It mainlybundles up various kinds of header records. The extent of the providepacket is mainly decreased to 20 bytes when it is contrasted with diagramfrom the past. For proper kind of aggregation of number of informationbytes is mainly stored in 1500 bytes. It is mainly used in bytes for secondbundle is of value 1480 on grounds. It mainly comes up 20 bytes in lengthwhich is present in the header and second bundle is mainly stored in 1480.Apart from this summing byte is not more than value form the diagrammeasures which comes into diagram measures that comes into action. 1Question 3: (10 marks)1.The unit taught details of BBR algorithm created by google and isevaluated for the investigating the clog in TCP/IP protocol andaccelerating the conveyance of the information packet that are developedfor transmitting the information parcels via the web. The rate oftransmission of the frames in the network can be improvised withcalculating the round trip time required for sending and receiving the dataparcel from the source address to the destination address. 12.Diverse issues are recognized utilizing the BBR calculation, for example,in case of “long haul links” it utilizes different item switch for controllingthe clog and it causes strange throughput. This is because of theabatement in multiplicative and eruption. The data packets are optimizedfor sending the packet and the jitter and loss of data packet are needed tobe controlled for increasing the efficiency of the network. The fullutilization of the bandwidth is used for controlling the blockage andhandling the errors in the network channel. If a bottleneck condition israised in the network it should be eliminated with the implementation of“deep buffer” and “shallow buffer” The buffers are loaded using thedestination location and is adds an extra delay in the network.23.Google built up its BBR calculation to accelerate the Transmissioncontrol convention with optimizing the rate of sending and accepting theinformation movement in the system and maintaining a control on thecongestion for avoiding blockage of the current route. The present TCPclog control utilizes distinctive buffers for implementing a control on theinformation movement stream on the system and limits the quantity ofunacknowledged packets. A slow start methodology is followed in the3

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