Answering Routing, Fragmentation, And Congestion Control Questions

Given the following network diagram, assume that all the networks shown are aware of each other and have fully updated routing tables. Answer the questions that follow.

The questions (1 mark each except q.8):

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?
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?
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?
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?
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?
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?
A packet is waiting at router R4 for forwarding. If the next-hop was a “direct delivery”, which of these three networksis the destination network? 3.0.0/16, or 150.32.0.0/18, or 220.10.40.0/24?
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 (5 marks)

An IPdatagram5,400 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.

Assuming that the router decides to fragment the packet into 4 fragments, determinea correct size for each fragment, and identify the starting byte and ending byte of each fragment(2.5 marks).
Calculate the fragmentation offset for each fragment(1.5 marks).
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(1 mark).

Question 3 – Congestion controls in TCP (10 marks)

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.

First, read this Network World article that reports on Google’s approach to improving congestion controls in TCP:

You should also read the following more technical paper about it:

After reading both articles, answer the following questions:

Write a brief summary of the congestion controls currently available in TCP as covered in this Unit (1 mark)
Identify and explain two problems with current congestion controls in TCP that are pointed out in the articles (2 marks)
Summarize in your own words the difference(s) between the current TCP congestion controls and Google’s new BBR protocol (3 marks)
The Network World article points out that it is difficult to get a new protocol accepted as a global standard for TCP/IP. Why do you think this is the case? Give carefully thought out reasons for your answer. (4 marks)

Important: for every direct quotation you use from these two sources or any other source, you must immediately, after the quote, provideyourownexplanation of the quotation (for example, explain why are you quoting it, how does it help answer the question, how does it support what you are saying?) – marks will be deducted for failure to do so. In addition, correct referencing conventions must be used throughout your work using the Harvard referencing convention. Your answers will be marked on clarity, logic, relevance, use of own words and fully addressing all parts of each question.

Remember that quotations alone will not be accepted as your explanation of the questions. Quotations can support your explanations, but you must still provide the explanations yourself. Best way forward is to keep direct quotations to a minimum, and use your own words.

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Type your answers in this section in the spaces provided. Spaces can expand as you type.

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Question Number

Mark

allocated

Marks earned

Question 1: (10 marks)

For Router R4, next hp address for any data packet along destination address for hosting 161.22.0.15/18 is 150.3.0.2.

For Router R1, the selected interface to send data packet to network 161.22.0.0/18 is M2.

To host the Router with an IP address 200.11.60.36/24 for sending data packet to a host with IP address 150.32.0.240/18 needed minimum 2 hops for reaching at destination address.

To originate any data packet from a network 220.10.40.0/24 reaches at router R1 whereas if destination address is not available in routing table, data packet gets dropped.

Since any data packet along destination address 140.21.0.10/22 reaches at R2 router, the selected interface port to forward data by router is M0.

Hence, the data packet of R3 router have destination address 220.10.40.5/26, the next hop address accessed by R3 router is 150.3.0.3/16 which is used to send data packet to a destination address.

For any data packet coming up at R4 router to get forwarded, if the next hop address is acts like direct delivery, the destination network address will be 150.30.0.16.

q.8 Routing table of router R2:

Prefix	Network address	Next-hop address	Interface
255.255.255.0	220.10.40.0	150.3.0.3	M0
255.255.255.0	200.11.60.0	200.21.60.1	M1
255.255.252.0	140.21.0.0	140.21.0.1	M2
255.255.192.0	161.22.0.0	150.3.0.3	M0
255.255.192.0	150.32.0.0	150.3.0.1	M0
255.255.0.0	150.3.0.0	150.3.0.2	loopback

1-7 1 mark each, q.8 3 marks

Question 2: (5 marks)

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

1^st packet: 20 bytes: IP header, 1480 bytes: Data ==

Correct size: 20+1480 = 1500 bytes, starting byte :1 , ending bytes : 1480

2^nd packet: 20 bytes: IP header, 1480 bytes: Data ==

Correct Size: 20 +148 = 1500 bytes, starting bytes : 1481, ending bytes : 2960

3^rd packet: 20 bytes: IP Header, 1480 bytes: data ==

Correct Size: 20+1480 = 1500 bytes, starting bytes:2961, ending bytes: 4440

4^th packet: 20 bytes: IP Header, 1345 bytes: Data ==

Correct size: 20 +1345 = 1365 bytes, starting bytes: 4441, ending bytes: 5805

2.5

For the 1st packet:

Offset Value = 0

For the 2nd packet:

Offset Value = 185

For the 3rd packet:

Offset Value = 370

For the 4th packet:

Offset Value = 555

1.5

For any data packet to get sent in a network, 20 bytes of data get consumed by IP header file. It also gets similar for all types of header files of used data packets. However, the size of next datagram gets reduced to 20 bytes based n previous datagram and for the 1500 bytes packet, the starting byte of next packet is 1481 due to the 20 bytes length of header file.

Therefore, the total count of byte based on all 4 fragments leaving from router will become greater than initial size of used datagram which has arrived.

Question 3: (10 marks)

Various congestion control mechanisms accessible in TCP whereas unit covers BBR algorithm like CUBIC and RENO. Moreover, it also assists to increase the network throughput as well as the reduction of bottleneck along with congestion issues available in network. Apart from that, the delivery rate of connection is measured with the help of analysis of available maximum bandwidth and also the minimum range of round-trip delay.

Apart from that, the deep buffer and shallow will be used to do the congestion control whereas BBR algorithm generates network path model to avoid the available congestion in network.

The two different problems occurred in congestion control present in TCP elaborated in the article are mentioned below:

Deep Buffers: The uses of deep buffers increases the occurrence of delay by doing repeated filling of buffers with last mile link and also the creation of present needled queue in network.

Shallow Buffers: The uses of shallow buffers techniques for loss based congestion control results can be dreadful because it can overreacts as well as due to its multiplicity the data packet sending rate get decreases. However, it also follows dynamic approach due to which it becomes difficult to do the proper utilization of loss based congestion.

The recent TCP protocol is differentiated from BBR protocol formed Google in various ways. The recent TCP protocol organized a network model based on available network paths to respond the actual congestion. Apart from that, the algorithm created by Google optimizes the speed of sharing data packets by using various routes and the algorithm handle the traffics for any congested routes. However, both the throughput as well as roundtrip traffic for multiple routes are get estimated and therefore time consumed by data to cross the network will be known as well as the traffic send at high speed which can be handled by use of any network.

It has numbers of difficulties mentioned in Network World article which creates barrier for the accepted new protocol for Global Standard of TCP/IP. However, the compatibility of BBR algorithm as well as various other protocols is tested with the help of IETF to properly get standardized. Hence, both the CUBIC and Reno protocols follow similar kind of principle whereas BBR also uses packet timing technique as well as other figures for finding any congestion in route.

Total marks awarded

25 (max)

Less late penalties if applicable

Less plagiarism penalties if applicable

Total marks earned

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