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    Networking

    COMPUTER NETWORKING

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    COMPUTER NETWORKING


    COMPUTER NETWOKING

    Question 1 – Routing                                                                                   

    The questions

    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?

    Answer:

                150.3.0.3/16 then forward to R4

    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?

    Answer :

    Here we use interface 2

    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?

    Answer:  200.11.60.1/24 then forward to 150.3.0.2/16 then moves to 150.3.0.1/16 à 150.32.0.1/18 

    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?

    Answer:  220.10.40.0/24 à150.3.0.3/16à161.22.0.1/18 

    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?

    Answer:  from m0 to m2 port 

    6.      A packet at router R3 has a destination address of 220.10.40.5/26.   What next-hop address would R3 use for this packet?

    Answer:

    150.3.0.1/16

    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?

    Answer: 150.3.0.0/16 

    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). 

    Prefix

    Network address

    Next-hop address

    Interface

    150.3.0.0/16

    150.3.0.0

    150.3.0.4

    Interface 0

    150.32.0.0/18

    150.32.0.0

    150.32.0.1

    Interface 1

    150.32.0.1/16

    150.32.0.1

    150.3.0.1

    R3

    150.3.0.0/16

    150.3.0.0

    150.3.0.4

    R2

    default

     

     

    R4

     

    Question 2 – Fragmentation in IPv4                                                                      

    An IPinformationgram5,400 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 correct size for each fragment, and identify the starting byte and ending byte of each fragment(2.5 marks).

    Answer:

    IP informationgram5400 bytes MTU=1500 bytes, IP header =20bytes

    Hence information datagram 5400- 20 IP header =5380 bytes of information within MTU of 1500 bytes, 1500-20=1480 bytes will be send out each packets. 

    4 Packets

    1)      IP header: 20 bytes, 1480 bytes

    Correct size of information: 20+1480=1500 bytes

    Initial/Starting bytes: 1

    End bytes: 1480 bytes 

    2)      IP header: 20 bytes, 1480 bytes

    Accurate size of information: 20+1480=1500 bytes

    Initial/Starting bytes: 1481

    End bytes: 2960 bytes 

    3)      IP header: 20 bytes, 1480 bytes

    Accurate size of information: 20+1480=1500 bytes

    Initial/Starting bytes: 2961

    End bytes: 4440 bytes 

    4)      IP header: 20 bytes, 1140 bytes

    Accurate size of information: 20+1140=1160 bytes

    Initial/Starting bytes: 4441

    End bytes: 5580 bytes 

    b)      Calculate the fragmentation offset for each fragment (1.5 marks).

    Answer:

    1st packet fragment offset: 0

    2st packet fragment offset: 185

    3st packet fragment offset: 370

    2st packet fragment offset: 555 

    c)      State whether the total number of bytes from all 4 fragments leaving the router will be greater than the initial information datagram size that arrived, or less than the initial information datagram size, and the reason (1 mark).

    Answer:

    As 20 bytes is always occupied by IP header in any packet & it is similar to each packet so total size of next information datagram is automatically reduce by 20 bytes than previous information data gram .similar to number of bytes in initial packets (1500 bytes) but starting bytes of second packets is 1480 bytes because of 20 bytes of header, Hence information data gram size of second packet at starting is 20 bytes. Hence the total number of bytes should be greater than the initial information datagram size that arrived. 

     

    Question 3 – Congestion controls in TCP                                                  

    After reading both articles, answer the following questions:

    1.      Write a brief summary of the congestion controls currently available in TCP as covered in this Unit (1 mark)

    Answer 

    The characteristic over TCP (Transmission Control Protocol) is according to rule the switch on statistics so much such is dependable. The essential TCP features are association management, consistency, flow monitoring or congestion manage.  Connection administration includes connection initialization and its execution. The source & destination “TCP ports are used because creating more than one digital connection. A reliable Peer to peer transfer of hosts is performed along the numbers and repeated transmission. A repeated transmission on the TCP fragment happens after a timeout, so the ACK (acknowledgment) is now not obtained with the aid of the sender or then there is reproduction ACKs acquired. Flow power ensures as a transmitter does now not overflow an adoption host. The receiver tells a sender concerning what a whole lot of statistics it may ship besides arrival ACK from the consumer intimate regarding the receiver’s ACK point. This choice is known as the sliding or its aggregate is defined in Bytes. Recognition to the “sliding window”, a reception dynamically regulates the aggregation of records which may be obtained beyond the transmitter. The final TCP feature – prime rule ensures as the transmitter does no longer overflow the system. Comparing to the drift limit method the place the waft government method make sure up to expectation the source does now not overflow the vacation spot host, fullness power is extra worldwide. It ensures the capability of the routers along the direction does no longer emerge as overflow. 

    2.      Identify and explain two problems with current congestion controls in TCP that are pointed out in the articles (2 marks)

    Answer:

    In shallow buffers, packet loss takes place previous to blocking. With present-day high-speed link use service switches together with “shallow buffer”  &  loss-based fullness, monitoring performs result among terrible throughput due to the fact that overreact, multiplicatively reducing the sending degree above lot loss, even if the packet break comes from temporary site visitors bursts. Since on it dynamic, such is difficult according to achieve complete utilization along loss-based fulfilment limit between carrying out: supporting 10Gbps & RTT needs a packet breach dosage under 3% then at 100ms “RTT” path a more viable breach dimension as 2% execute only maintain at most three Mbps

    “Deep buffers” at bottleneck links together with flagrant buffers, congestion occurs earlier than etiquette failure. At the side of the modern network, loss-based congestion limit motives the ‘buffer bloat’ problem, by frequently permit the buffers into deep last-mile links and inflicting over to seconds about needless queue delay.

     

    3.      Summarize in your own words the difference(s) between the current TCP congestion controls and Google’s new BBR protocol 

    Answer

    TCP BBR, who attain greater bandwidths or lower latencies because of network traffic. It is the equal BBR &  powers TCP traffic beyond google.com and that improved YouTube network throughput by IV percentage of common worldwide — and by means of extra than 14 percent within the half nation.

     

    BBR allows the “Word Press” websites about our digital trip board to burden at speed. Google’s examination, BBR's throughput does reach greater than today's best loss-based fullness manage. System improvements kind of BBR is simply one of the many motives we associate together with GCP.

    TCP utilizes a congestion eyelet on the sender side to operate fulfilment avoidance. The congestion eyelet shows the most aggregate regarding data to that amount does send oversee of an attachment without being approved. TCP identify fulfilment now such fails to acquire an acknowledgment because of a lot within the predictable break. In certain a situation, that reduces the fulfilment window according to some maximum segment size, or below other instances, it will increase the congestion window via certain maximum segment size. It additionally exists a completeness window threshold, as is set after partly the fullness window altar at the day then a repeat broadcast used to be requisite. 

    4.      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)

    Answer: 

    This network conduct is now not totally beneficial if such as you necessity is a dependable statistics circulate to attack via the system. The approach adopted by means of the IP structure is to pass by the accountability because of managing the records flow, consisting of detecting & repairing packet loss as like well namely running the statistics go with the flow rate, to an end-to-end conduct protocol. We use the TCP because of that task. TCP’s meant dye regarding action is to rate its statistics movement such so it is sending records as quick as possible, however now not then fast to that amount such constantly saturate the cuddy queues, &  loses packets. This desire by using TCP to lead at a most appropriate speed is attempting according to chart a direction within joining objectives; initially, to makes use of whole handy community carrying resources, then portion the community sources fairly throughout whole simultaneous TCP data run.

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