AN exam March

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1 AN exam March Dear student This exam consists of 7 questions. The total number of points is 100. Read the questions carefully. Be precise and concise. Write in a readable way. Q1. UDP and TCP (25 points) Q.1.1 (15 points) Two nodes, node A and node B, are connected via a 100 Mbps link. The RTT between the nodes is 1ms (0.5 ms in each direction). On this connection there is no packet loss neither packet reordering. Suppose node A wants to send a file of 1.25Mbytes to node B. All segments (UDP or TCP) have a payload of 125bytes (1000bits). How much time will it take (ignoring all headers overhead of the transport and lower layer protocols): If node A uses UDP? If node A uses TCP? In this case assume the rcwn and the ssthresh have the same value and are set to 8Kbytes (64 kbits). 1.25Mbytes/125Bytes= segments. UDP: we can observe that 1.25Mbytes = 10Mbits. It takes 0.1 seconds (100ms) to send the segments at 100Mbps speed. Additionally you have to add the 0.5ms to complete the path. TCP: it will take 7 iterations to reach the rcwn and sshthresh. At that point we have sent 127 segments and 7 ms have elapsed. The cwdn is now the same as the rcwn and this allows us to send 64 segments (64kbits) per RTT. We still need to send segments= /64= 154 iterations. 154ms +17ms = 160ms (Some of you counted the times for the handshake, that is is obviously also correct). Q1.2 (5 points) Which measures could you take to improve the performance of TCP in this situation and why? We could increase the rcwn and sshthresh up to value of the BDP = 100Mbps * 1ms=100Kbits (100segments). A value higher than this will not produce any effect given the maximum sending speed is achieved. (Increasing the MSS makes no sense as at the end the limitation is caused by the rcwnd)

2 Advanced Networking 2018 Exam Q1.3 (5 points) In the above exercise we ignored the headers. Let s define the efficiency of a protocol as the ratio of payload bytes versus total bytes transmitted by the protocol. This said: Calculate the respective efficiency of UDP and TCP when they both carry 125 bytes payload. In the TCP case assume there are no options in the header. How would this affect the calculations you made in Q1.1? Data are 16 bytes, length of UDP header is 8 bytes, so the ratio is 125/ Data are 16 bytes, length of TCP header (no options) is 20 bytes, so the ratio is 125/ We would have to add 7% overhead for UDP. This would increase the time slightly. For TCP the overhead would not make a lot of difference in our calculation as we never used the full speed when being in slow start and the reaching the plateau. Q2. Acknowledgments (10 points) What is the difference between positive acknowledgments (ACKs) as used by TCP and negative acknowledgments (NACK) as used for example by QUIC? See the definition on the slides. Suppose the sender sends data only infrequently. Would a NACK-only protocol be preferable to a protocol that uses ACKs? Why? In a NAK only protocol, the loss of packet x is only detected by the receiver when packet x+1 is received. That is, the receiver receives x-1 and then x+1, only when x+1 is received does the receiver realizes that x was missed. If there is a long delay between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol. Now suppose the sender has a lot of data to send and the end-to-end connection experiences very few losses. In this second case, would a NACK-only protocol be preferable to a protocol that uses ACKs? Why? On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are always sent a significant reduction in feedback in the NAK-only case over the ACK-only case. Q3 - NDN (10 points) NDN is being proposed as a more secure and reliable network architecture. In slide 64 and slide 65 of Lecture09 we identified a couple of examples in which NDNs are more

3 resilient. Additionally to what mentioned in class, it is also expected that NDN networks are robust to reflection attacks 1. Why? NDN is resilient to reflections attacks due to the symmetric nature of the path taken by each interest and the corresponding content: the latter must follow, in reverse, the path established by the preceding interest. Although an NDN router may broadcast an incoming interest on some or all of its interfaces, even under the improbable case where each NDN hop broadcasts the interest, the maximum number of content copies a consumer can receive is bounded by the number of its interfaces. Consequently, the only effective reflection-style attack requires the adversary to be on the same physical network as the intended victim. Still NDN are not completely fail-safe. Can you describe how you could try (D)DoS a NDN network? One possible example is man-in-the-middle attacks where the malicious users sends bogus public keys to the requesters. Less originally: many of you used the example in the slides about flooding the network with diverse interest. As said in the slides throttling would help. If you could not think of anything a nice overview is provided in: Q4 Optical reach (10points) A fiber optic cable exhibits a 0.15 db/km loss. How long can a fiber optic cable be under the above loss characteristics if the total power budget is 6dB? (nb. power budget = the amount of loss that can be tolerated on the link) DB= 40 km * 0.15 db/km= 6 db How long could the same fiber cable be in the case a sender has a minimum transmitter power of -15dBm and the receiver has a minimum sensitivity of - 27dBm? The power budget in this case is PB=-15-(-27)=12dB. We could run for 80 km. Q (10 points) 1 reflection attacks involve three parties: the adversary, the victim host, and a set of secondary victims (reflectors). The goal is to use reflectors to overwhelm the victim with traffic. The adversary creates numerous forged IP packets with the source address set to the address of the intended victim. It then sends these packets to the reflectors. Responses to all such packets are routed to the victim. These attacks require some form of amplification: the amount of content sent by the adversary must be significantly smaller than that received by the victim.

4 Advanced Networking 2018 Exam A commonly discussed problem in networks is the presence of hidden terminals. A related problem is the exposed terminal problem (see slide61 lecture07). What can cause exposed terminals in a WiFi environment? We are talking of (infrastructure mde!). If two access points are on the same channel but close enough that their ranges overlap there might be clients associated with one that hear transmission from clients of the other. How could you solve the problem? First of all by carefully selecting the channels. If this is not possible, and the overlap remains one could deploy RTS/CTS mechanisms. Client A and client B would hear the respective RTS but not the CTS from the other AP and determine that their connection to the AP is safe. Still there could be interference. Q6 Programmable networks (25 points) Five nodes (node A, node B, node C, node D and node E) are in the same IPv4 subnet. Node A has restricted access: node B can only SSH into it, node C can only view webpages on port 8080, node D has full access to all the services of node A and node E has no access at all to node A. All other nodes are open to each other. How would you implement the above behavior in: o OpenFlow? o ebpf? o P4? List for all the three use cases (OpenFlow, ebpf and P4) the minimally required hardware (if any) to provide connectivity, and to support the scenario. Sketch, to the best level of detail you can, the steps and configurations needed. (Note this answer is fairly succinct, many of provided more details than this). OpenFlow will require an external controller and min. one OpenFlow switch. The controller will need to populate the flow tables to: - drop all packets from E to A - allow all packets from D to A - allow packets from C to A if port 8080 and protocol TCP otherwise drop - allow packets from B to A if port 22 and protocol TCP otherwise drop. - allow all packets from A to all - allow all packet from all to all In EBPF you would not need special hardware. Connectivity among the node could be supported with a simple L2 switch. On node A you would have to create an ebpf program with at least one sections. After this the program needs to be installed on every host. Finally, configuring the allowed flows per-host (i.e. populating a map(s) ) A P4 capable device would need to be used as the central switch for the nodes. Forwarding behavior would need to be defined in P4 based on destination MAC address. Filtering

5 behavior will also need to be implemented. This would involve defining Ethernet, IP and TCP in P4 and creating tables prepared to filter based on the egress port, destination TCP port and either the source IP or source MAC address. A system external to P4 would need to populate the tables with the correct entries for forwarding and filtering. These could be entered manually by an administrator or pushed by some control plane software. Q7 - Adaptive bit streaming (10 points) What are advantages of using adaptive bit streaming versus traditional RTP methods? HTTP-based adaptive bitrate streaming technologies yield additional benefits over traditional server-driven adaptive bitrate streaming. First, since the streaming technology is built on top of HTTP, contrary to RTP-based adaptive streaming, the packets have no difficulties traversing firewall and NAT devices. Second, since HTTP streaming is purely client-driven, all adaptation logic resides at the client. This reduces the requirement of persistent connections between server and client application. Furthermore, the server is not required to maintain session state information on each client, increasing scalability. Finally, existing HTTP delivery infrastructure, such as HTTP caches and servers can be seamlessly adopted.