COIT20261 Network Routing and Switching - Free Samples to Students
The questions (1 mark each except q.8):
1.From the point of view of router R4, what is the next-hop address for a packet addressed to host 161.22.0.15/18?
2.From the point of view of router R1, which of its interfaces would it choose for a packet being sent to network 161.22.0.0/18?
3.A host with an IP address of 200.11.60.36/24 has just sent a packet to a host with address 150.32.0.240/18. How many hops is required between source and destination?
4.A packet originating from network 220.10.40.0/24 arrives at router R1, however, R1 determines that the destination network is not in its routing table. What does R1 do with the packet?
5.A packet arrives at router R2 with a destination address of 140.21.0.10/22. Which interface port does R2 forward the packet out of?
6.A packet at router R3 has a destination address of 220.10.40.5/24. What next-hop address would R3 use for this packet?
7.A packet is waiting at router R4 for forwarding. If the next-hop was a “direct delivery”, which of these three networks is the destination network? 150.3.0.0/16, or 150.32.0.0/18, or 220.10.40.0/24?
8.Complete the information in the routing table for router R2 as shown in the Answer Template for networks 150.3.0.0/16, 150.32.0.0/18, and the Default network. Show the masks in longest mask order using CIDR format (3 marks).
Question 2 – Fragmentation in IPv4
An IP datagram 5,400 bytes long with no options arrives at a router, which determines that the next destination has an MTU of 1,500 bytes. Use the Answer Template to complete the following questions, showing your calculations and reasoning.
a)Assuming that the router decides to fragment the packet into 4 fragments, determine a correct size for each fragment, and identify the starting byte and ending byte of each fragment
b)Calculate the fragmentation offset for each fragment
c)State whether the total number of bytes from all 4 fragments leaving the router will be greater than the initial datagram size that arrived, or less than the initial datagram size, and the reason
Question 3 – Congestion controls in TCP
This question affords you the opportunity to extend your thinking about congestion controls in TCP beyond the textbook to observe what a real-world technology company, Google, is doing in this space.
Answer:
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1. 2. 3. 4. 5. 6. 7. 8. |
The next hop address for a packet routed to 161.22.0.15/18 is 150.3.0.2 M2 interface is chosen for sending an information packet to the system with address 161.22.0.0/18 In the event that 200.11.60.36/24 is the IP address of the host and it sends an information parcel with goal address 150.32.0.240/18 it requires least 2 hop address for effectively conveyance the data packet. On the off chance that a data packet is produced from the source address 220.10.40.0/24, the router R1 and the switch does not finds the goal address in its routing table it disposes of or drops the data packet quickly. In the event that the goal address of an information parcel is 140.21.0.10/22 and it reaches router R2 the M0 interface is utilized for sending the data packet to its goal. Router R3 would utilize 150.3.0.3/16 as the next hop address for a data packet having goal address 220.10.40.5/26 reaching the router. On the off chance that the next hop address is an immediate conveyance for the R4 router the 150.30.0.0/16 is the goal address of the network. q.8 Routing table of router R2:
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1-7 1 mark each, q.8 3 marks | |||||||||||||||||||||||||||||
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Question 2: (5 marks) | |||||||||||||||||||||||||||||||
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a) |
IP datagram = 5400 bytes MTU = 1500 bytes IP header = 20 bytes Therefore, 5400 - 20 = 5380 MTU of 1500 bytes, 1500 - 20 = 1480 bytes of data is transmitted for each of the data packets Ceiling (5380 /1480) = 4 packets are required 1st packet: 20 bytes: IP header, 1480 bytes: Data == Correct size: 20 +1480 = 1500 bytes, starting byte :1 , ending bytes : 1480 2nd packet: 20 bytes: IP header, 1480 bytes: Data == Correct Size: 20 + 148 = 1500 bytes, starting bytes : 1481, ending bytes : 2960 3rd packet: 20 bytes: IP Header, 1480 bytes: data == Correct Size: 20 + 1480 = 1500 bytes, starting bytes: 2961, ending bytes: 4440 4th packet: 20 bytes: IP Header, 1345 bytes: Data == Correct size: 20 + 1345 = 1365 bytes, starting bytes: 4441, ending bytes: 5805 |
2.5 | |||||||||||||||||||||||||||||
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b) |
Calculation of fragmentation offset For the 1st packet = 0 For the 2nd packet = 185 For the 3rd packet = 370 For the 4th packet = 555 |
1.5 | |||||||||||||||||||||||||||||
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c) |
The aggregate number of bytes from all the 4 fragments that leaves the router is greater than the initial size of the datagram this is on the grounds that the header document of an information bundle takes 20 bytes of information and the accompanying datagram measure is decreased by 20 bytes from the past ones. For instance an information parcel with 1500 bytes, 1481 is the begin byte of the second information bundle. |
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Question 3: (10 marks) | |||||||||||||||||||||||||||||||
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1. |
For controlling the clogs in the network Reno and CUBIC congestion control mechanism can be applied and its working can be learned from the unit. The unit additionally examines about the Google's new BBR convention that can be connected for expanding the speed of web upto 14%. The use of the clog control helps in expanding the throughput of the system and decreases the bottleneck and the blockage issue. The BBR calculation can be utilized for the investigation of the system way and count of the round trip delay for controlling the blockage. |
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2. |
The two issues recognized for the present blockage control in TCP IP brought up in the article is present in the deep buffer and the shallow buffer. The application of shallow buffer for controlling the congestion is dependent on the loss of the data packets. The utilization of shallow buffer can bring about terrible outcome because of its multiplicative and overcompensation nature and the rate of sending the information bundle in the system can be diminished. It is hard to completely use the shallow buffer because of its dynamic nature on the misfortune based blockage. The utilization of deep buffer can cause bluffer boat issue and causing a deferral by filling the buffer over and again with the last mile route and making an unnecessary queue in the system. |
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3. |
There are a few contrast between the Google's new BBR convention and the TCP clog control. The TCP clog is controlled with the use of various cradles to be specific shallow and profound support for putting away the accessible way in the system. The Google's BBR calculation controls the information movement send over the system with the end goal that it doesn't makes clog in the network path. An estimation is made for the roundtrip and various courses that can be utilized to achieve the goal arranged and the information activity that are sent at the speed that can be taken care of by the system. |
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4. |
There are some trouble for the new convention being acknowledged as a worldwide TCP/IP standard and it has been brought up in the Network World Article. For the institutionalization of the convention it requires to be tried for discovering its similarity issues with the other systems administration conventions working in the diverse layers of the system. There are distinctive calculations created and IETF is in charge of institutionalizing the convention since every one of the conventions pretty much takes after a similar guideline and some extra features is included for settling the issues looked in the TCP. For instance the BBR calculation depends on the Reno and CUBIC convention however it furthermore utilizes the system model and timing of the data packet for finding the congestion in the different routes of the network. |
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Total marks awarded |
25 (max) | ||||||||||||||||||||||||||||||
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Less late penalties if applicable | |||||||||||||||||||||||||||||||
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Less plagiarism penalties if applicable | |||||||||||||||||||||||||||||||
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Total marks earned | |||||||||||||||||||||||||||||||
Markers comments:
Bibliography
Bernstein, G., Lee, Y., Li, D., Imajuku, W. and Han, J., 2015. Routing and wavelength assignment information encoding for wavelength switched optical networks (No. RFC 7581).
Butler, B. (2018). How Google is speeding up the Internet . [online] Network World. Available at: https://www.networkworld.com/article/3218084/lan-wan/how-google-is-speeding-up-the-internet.html?idg_eid=f32fc7aec843db7ef67d0a4f08e3322d&email_SHA1_lc=&cid=nww_nlt_networkworld_daily_news_alert_2017-08-22&utm_source=Sailthru&utm_medium=email&utm_campaign=NWW%20Daily%20AM%20Alert%202017-08-22&utm_term=networkworld_daily_news_alert [Accessed 15 May 2018].
Medhi, D. and Ramasamy, K., 2017. Network routing: algorithms, protocols, and architectures. Morgan Kaufmann.
Tools.ietf.org. (2018). BBR Congestion Control. [online] Available at: https://tools.ietf.org/id/draft-cardwell-iccrg-bbr-congestion-control-00.html [Accessed 15 May 2018].
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