hwk1_570_24s_sol

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1 Problem 1 Short Questions (27 pts; 3pts each): Your answers should be precise-and-concise. A sentence or two of an answer is sufficient for most of these questions. A. Based on definitions of the edge and the core networks, identify which category does each of the following networks belong to: a. An IXP network: core b. NCSU campus WiFi: edge c. The cellular network that your T-Mobile/Verizon mobile phone directly connects to: edge d. Facebook’s wide area network that interconnects its multiple data centers: core e. The inter-continent cable under the Atlantic Ocean: core f. A road side unit which an autonomous driving vehicle is connected to: edge B. Explain what a Point-of-Presence is? It’s a set of routers / a subnetwork of the internal network of an ISP that connects to its external peering networks. C. Explain two reasons why packet switching is best suited for the Internet than circuit switching? Packet switching 1) is great for bursty data, 2) enables resource sharing without wastage (or can serve more users), 3) does not require circuit setup which incurs setup overhead, and 4) is more resilient to failures. D. Assume a packet is sent from a sender and is to be delivered to a remote destination through multiple intermediate switches or routers. The receiver however does not receive the packet in the end. What could have happened? The packet may have gone through a switch/router with insufficient buffer / memory, and hence got dropped at that switch/router. E. Are buffers / queues needed for a fully circuit switching network? No. Because resource is dedicatedly reserved in circuit switching. So as long as a user is not sending at a speed more than the reserved bandwidth, queueing will not happen. F. You went into a room with an Amazon Echo (or any other device with a voice-controlled digital personal assistant (DPA) similar to Alexa), and say “Alexa, what’s the time?” Alexa replies: “The time is 1:59pm.” Describe the above process in the realm of a protocol between you and Alexa. (Consider the three components of a protocol, and explain what each component is in this scenario.) Syntax: [Name, or Alexa] + [Query to Execute] Semantics: A sender sending the voice message asks the DPA with name [Name] to execute the [Query to Execute]. Action: Upon receiving the voice message, the DPA with [Name] executes the [Query to Execute], and plays the response as an audio message back to the sender. G. Explain how layering is implemented when a sender is sending out a packet.

2 Each layer (beside application layer) in the sender’s network stack will add a packet header, which contains the necessary information needed for the same layer at a remote network device or end host to process the packet and deliver it to the correct upper layer / next-hop entity. H. How can you distinguish different types of network devices such as a hub, a bridge, a router and a host? The layers of protocols that the device implements. Hub: physical layer only. Bridge: up to data link layer. Router: up to network layer. Host: up to application layer. I. Describe the major difference between the OSI and the TCP/IP layering models, and whether each of the different layers is end-to-end or not. TCP/IP model does not have the presentation and the session layers in the OSI model, which are both merged into the application layer. Both these layers are end-to-end. Problem 2 Protocol (9 pts; 3 pts each): Assume a protocol stack has 5 layers, with the topmost layer being the application layer. The four lower layers each adds an overhead of 30 bytes as headers and trailers, and does not change the packet payload. Consider the application layer sending application PDUs with different sizes as follows. Compute the percentage bit overhead in this stack as seen over a long period of communications. Use the application layer PDU as the basis. A. Application sends 200-byte fixed-length PDUs at all times. Average PDU size = 200 bytes Per-PDU overhead = 30 * 4 = 120 bytes Per-PDU overhead ratio = 120 / 200 = 60% Note: It is also okay to compute based on the final physical layer PDU size in this problem, which is 120/320 = 37.5%. But since application PDUs are the only meaningful data transmitted, the application PDU is commonly used as the “basis” of computation such as delay, throughput, loss ratio, etc. B. Application sends 200-byte PDUs at half of the times, and 400-byte PDUs at the other half. Average PDU size = (200+400)/2 = 300 bytes Per-PDU overhead = 30 * 4 = 120 bytes Per-PDU overhead ratio = 120 / 300 = 40% C. What might be the reason that we do not want to have very large PDU sizes (e.g., 1MB data per packet) to make the overhead of layering as low as possible? There are several reasons. Each one would be sufficient:

3 - The larger a packet, the more likely there will be transmission errors, which could lead to more packet retransmissions. - When a packet is lost due to either packet drop or network failure, recovery would be very costly as the sender needs to retransmit the entire packet again. - A huge buffer is needed to store all packets queuing up at an output port, which could be costly. Problem 3 Packet Loss (8 pts): Assume a client connects to a server via a lossy link. When an application is sending PDUs of different sizes to the server over the path, compute the expected number of bytes lost when transmitting an arbitrary PDU. A. (1pt) Assume the client does not have reliable data transfer. The link has packet loss probability of 20%. Application sends 200-byte fixed-length PDUs at all times. Average PDU size = 200 bytes Expected lost bytes = 200 * 20% = 40 bytes B. (1pt) Assume the client does not have reliable data transfer. The link has packet loss probability of 20%. Application sends 200-byte PDUs at half of the times, and 400-byte PDUs at the other half. Average PDU size = 300 bytes Expected lost bytes = 300 * 20% = 60 bytes C. (1pt) Assume the client does not have reliable data transfer. The link has packet loss probability of 20%. Application sends PDUs following this distribution: 100-byte (10%), 200-byte (30%), 400-byte (50%), 1000-byte (10%). Average PDU size = 370 bytes Expected lost bytes = 370 * 20% = 74 bytes D. (2.5pts) Assume that the client transmits 1 million bytes of data to the server. The PDU goes through a 5-layer network stack with 25 bytes of overhead for each layer except the application layer. During the transmission, exactly 1 bit is corrupted during the transmission, and the corresponding packet must be entirely retransmitted to ensure reliable data transfer. How many bytes in total, including both data and overhead, need to be transmitted, if the PDU size is 100, 1000, 2000, 10000, and 200000 bytes? The 1 million bytes must be transmitted all for once, which does not depend on the PDU size. The overhead depends on the PDU size. Let the PDU size be denoted by p . The number of packets needed to transmit, excluding the retransmission, is 10 6 / p. Each packet has a 4*25=100 byte overhead. Exactly one packet fails, and hence an additional PDU with (p + 100) bytes of data and overhead needs to be retransmitted. So the total (retransmission + overhead) size is: f(p) = (10 6 / p) * 100 + (p + 100). For a PDU size of 100 bytes: (10 6 / 100) * 100 + (100+100) = 1000200 bytes For a PDU size of 1000 bytes: (10 6 / 1000) * 100 + (1000+100) = 101100 bytes For a PDU size of 2000 bytes: (10 6 / 2000) * 100 + (2000+100) = 52100 bytes For a PDU size of 10000 bytes: (10 6 / 10000) * 100 + (10000+100) = 20100 bytes For a PDU size of 200000 bytes: (10 6 / 200000) * 100 + (200000+100) = 200600 bytes

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4 E. (2.5pts) What is the optimal PDU size for the above network scenario in D, i.e., requiring the fewest number of bytes to be transmitted in total? Why? Obviously between the PDU sizes given, p = 10000 bytes results in the minimum number of bytes to be transmitted. However, to prove this is indeed the smallest number, we need to show that p=10000 minimizes the function f(p) = 10 6 / p * 100 + (p + 100). To see this, we can take the derivative of f(p) as f’(p) = - 10 6 * 100 / p 2 + 1. The derivative f’(p) is equals to 0 when p = 10000. You can further verify that when p < 10000, f’(p) < 0; when p > 10000, f’(p) > 0; hence p = 10000 indeed corresponds to the minimum point of f(p), proving the result. Problem 4 Transmission Delay (31 pts): Consider the figure below, with three links, each with the specified transmission rate and link length. Assume the speed of light propagation delay on each link is v = ࠵? ∗ ࠵?࠵? ࠵? m/sec . A network layer PDU of length P = 1000 Bytes is to be transmitted from A to B starting from time 0. The PDU originates from the network layer at A, and is considered received when being processed by the network layer at B. Between the network and the DLC layers, the communications happen over a shared memory with read/write speed of 1 Gbps on all nodes. Other than memory read/write, assume the processing delay is 0 ms on all nodes. The DLC layer adds an overhead of 125 Bytes to the PDU, and the physical layer adds no additional overhead. Assume there is no queueing on any link before the transmission. Answer all time-related questions in milliseconds (ms) . Note 1: Draw a time figure of the PDU(s) through all layers/links to determine the delays. Note 2: The DLC layer must completely read a packet from shared memory before transmitting it on the bitpipe, similarly the network layer software must completely read a packet from shared memory before forwarding it, and each layer must wait until a packet is completely available in shared memory before starting to read it. A. (3 pts) Assume there is no other transmission sharing any of the links. What is the bottleneck link bandwidth for this transmission? 100 Mbps. Bottleneck link is the one with the minimum bandwidth along the transmission path. B. (3 pts) At what time T1 does the PDU leave A? A PDU leaves a router when it has been fully transmitted on to the next link. Before leaving the node, the PDU incurs: 1) processing delay on A, and 2) transmission delay onto A-R1. Delay from NET to shared memory = 1000 Bytes * 8 / 1 Gbps = 0.008 ms (NET layer write to shared memory)

5 Delay from shared memory to DLC = 1000 Bytes * 8 / 1 Gbps = 0.008 ms (DLC layer read from shared mem) DLC PDU size = 1000 + 125 = 1125 Bytes Transmission delay onto link A-R1 = 1125 Bytes * 8 / 200 Mbps = 0.045 ms T1 = 0.008 * 2 + 0.045 = 0.061 ms Note: Each communication between NET and DLC incurs 0.008 * 2 = 0.016 ms delay. We will use this number directly further on. C. (3 pts) At what time T2 does the PDU leave R1? After T1, before T2, the PDU experiences: 1) propagation delay over A-R1, 2) input processing delay at R1, 3) output processing delay at R1, and 4) transmission delay onto R1-R2. Propagation delay = 5 km / (3e8 m/sec) = 0.0167 ms Input processing delay = output processing delay = 0.016 ms Transmission delay onto link R1-R2 = 1125 Bytes * 8 / 400 Mbps = 0.0225 ms T2 = T1 + 0.0167 + 0.016 * 2 + 0.0225 = 0.1322 ms D. (3 pts) At what time T3 does the PDU leave R2? After T2, before T3, the PDU experiences: 1) propagation delay over R1-R2, 2) input+output processing delay at R2, 3) transmission delay on R2-B. Propagation delay R1-R2 = 3000 km / (3e8 m/sec) = 10 ms Input processing delay = output processing delay = 0.016 ms Transmission delay R2-B = 1125 Bytes * 8 / 100 Mbps = 0.09 ms T3 = T2 + 10 + 0.016 * 2 + 0.09 = 10.2542 ms E. (3 pts) At what time T4 does the PDU reach B? After T3, before T4, the PDU experiences: 1) propagation delay on R2-B, and 2) input processing at B. Propagation delay R2-B = 8 km / (3e8 m/sec) = 0.0267 ms T4 = T3 + 0.0267 + 0.016 = 10.2969 ms F. (4 pts) Assume that from A to B there is another PDU of the same size, which is transmitted back-to- back after the first PDU. What is the time T5 when the second PDU is ready to be sent out on link R1- R2? Is this before or after when the first PDU leaves R1? Consequently, will the second PDU experience queueing at R1? Before T5, the second PDU experiences: 1) processing delay at A, 2) queueing delay at A, 3) transmission delay on A-R1, 4) propagation delay on A-R1, and 5) input+output processing delay at R1. Queueing delay at A = transmission delay of first PDU at A = 0.045 ms Other than the queueing delay, the other delays are the same as the first PDU. T5 = 0.016 + 0.045 (queueing) + 0.045 + 0.0167 + 0.016*2 = 0.1547 ms T5 > T2. Hence the second PDU arrives at the output queue of R1 after the first PDU already leaves R1. Hence no queueing is experienced at R1.

6 G. (4 pts) Following E, what is the time T6 when the second PDU is ready to be sent out on link R2-B? Is this before or after when the first PDU leaves R2? Consequently, will the second PDU experience queueing at R2? After T5, before T6, the second PDU experiences: 1) transmission delay onto R1-R2, 2) propagation delay on R1-R2, and 3) input+output processing at R2. T6 = T5 + 0.0225 ms + 10 ms + 0.016 ms * 2 = 10.2092 ms T6 < T3. So the second PDU is ready for transmission at R2 before the first PDU is fully transmitted at R2. Hence queueing is experienced at R2. H. (3 pts) What is the total queueing delay of the second PDU before reaching B? If you consider only queueing delay in the network: (T3 – T6) = 0.045ms. If you additionally consider queueing at the host: (T3 – T6) + 0.045 = 0.09ms. Both will be regarded correct for this question. Although, it is more common for us to only consider queueing delay in the network, because the queueing delay at the host is trivial for any type of traffic: any time you send more data than a packet can carry, queueing delay is inevitable for all but the first packet, and it essentially just translates into the “transmission delay of the entire data chunk”. I. (3 pts) Assume now that there are 7 other transmissions fair sharing Link 2 (R1-R2). What is the bottleneck link bandwidth for the transmission from A to B? 50 Mbps. Note the bandwidth is the minimum bandwidth along the path that is available to this specific transmission. On Link 2, if there are in total 8 transmissions fair sharing the bandwidth, each will only get 50 Mbps. J. (2 pts) Assume that the links are not perfect. Specifically, Link 1 has a packet loss probability of p1, Link 2 has a packet loss probability of p2, and Link 3 has a packet loss probability of p3. Further assume the links are independent. What is the probability that the second PDU arrives at the destination successfully? (1 – p1) * (1 – p2) * (1 – p3). Problem 5 Packet switching queueing delay (10 pts): A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is 1/3 done being transmitted on this outbound link and 8 other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 1,500 bytes and the link rate is 10 Mbps. A. (3 pts) What is the queuing delay for the packet? (1000 + 1500 * 8) Bytes * 8 (bits-per-byte) / 10 Mbps = 10.4 ms

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7 B. (3 pts) More generally, what is the queuing delay when all packets have length L, the transmission rate is R, n bits of the currently-being-transmitted packet have been transmitted, and x packets are already in the queue? (L - n + L * x) / R C. (4 pts) Following question B, suppose X other packets arrive at the same outbound link simultaneously and immediately after that packet. What will be the average queueing delay for the X arriving packets? (L - n + L * x) / R + (( 1 + 2 + … + X ) * L / R) / X = (L - n + (x + (X + 1) / 2) * L) / R Problem 6 Traceroute Practical (10 pts): Read the online documentation of the traceroute utility in your computer’s OS (Linux/Mac: traceroute, Windows: tracert). Perform a traceroute from your residential/home network to each of these three destinations: www.unige.ch (University of Geneva), sydney.edu.au (University of Sidney) and vt.edu (Virginia Tech) (3 pts) Add screenshots of the three traceroutes. Answer A and B for each of the three traceroutes: www.unige.ch sydney.edu.au

8 vt.edu A. (3 pts) What is the average RTT delay? www.unige.ch: 123.33ms sydney.edu.au: 228.33ms vt.edu: 41.67ms B. (3 pts) How many routers are in the path? www.unige.ch: 14 sydney.edu.au: 21 vt.edu: 12 C. (1 pt) Compare the three traceroutes, what are your observations? There can be many different observations. Here are some examples: a. The further the distance, the longer the RTT. b. The further the distance, the more hops (routers). c. All communications start with the same two routers, which are likely the home gateway router, and the access ISP gateway router, respectively.

9 Problem 7 RFC Lookup (5 pts): The purpose of this question is to give you a quick look at an Internet RFC. You are not expected to read the RFCs closely or in its entirety (of course you are welcome to do so, but the RFCs are pretty long!). Go to www.rfc-editor.org and answer the following questions. A. (2 pt) Who wrote RFC 8774, and which institution is she/he from? On what day was it published? M. Welzl from University of Oslo, on Apr. 1, 2020. This is actually one of the “April Fools’ Day Request for Comments”, published each year featuring some funny (but quite rigorous) RFCs. B. (3 pt) What is RFC 5321 about? What is the general format of a MAIL command sent by a sender? What is the response when the server accepts this command? The SMTP protocol. MAIL FROM:<reverse-path> [SP <mail-parameters> ] <CRLF> 250 OK

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