Department of Computer Science Computer Networks Vrije Universiteit 25.05.2004 Always explain your answers concisely and be sure to be to-the-point. Part I This part covers the same material as the midterm exam. 1a What is the difference between a network service and a network protocol? Is it possible to have different implementations of the same protocol? 5pt A service describes what the network offers to its clients by means of an interface. A protocol is a set of rules describing exactly how that service is to be realized through the exchange of messages between computers. These rules may be implemented differently (e.g., using different programming languages), as long as those implementations coincide with the rules. 1b While considering traditional POTS modems, Bob claims that “...it is unusual to use a 2400 baud modem. Modems usually support 56 kbps.” What is wrong with this statement. Motivate your answer. 5pt Traditional POTS modems still support 2400 baud, which tells how many times a signal may change per second. The higher data rate is obtained through modulation schemes. 1c What type of multiplexing is used for ADSL? Explain your answer. 5pt Frequency division multiplexing: the bandwidth of the cable is partitioned into separate 4 kHz chan- nels. This is typical for FDM. 2a Explain why bit stuffing is needed, and how it works. 5pt Bit stuffing is needed when using frame delimiters such as 01111110. In order to send the delimeter as data in a frame, a zero is inserted after five consecutive 1s have been transmitted. The receiver simply removes the zero following 5 1s in row. 2b The physical layer sends data as a “continuous” stream of bits. The data link layer uses chunks of bits, called frames. Why switch to frames? 5pt Continuous bit streams make it impossible to detect errors, to synchronize transmission rates, etc. 2c Data link protocols almost always put the checksum in a trailer, rather than in a header. Why? 5pt Yes, it’s a question from the problems section, and it has been asked before. The CRC is computed during transmission and appended to the output stream as soon as the last bit goes out onto the wire. If the CRC were in the header, it would be necessary to make a pass over the frame to compute the CRC before transmitting. This would require each byte being handled twice-once for checksumming and once for transmitting. Using the trailer cuts the work in half. 3a Explain the difference between go-back-N and selective repeat. 5pt With go-back-N, whenever a receiver detects a missing or damaged frame k, all frames succeeding k are ignored and the sender is forced to start retransmission at frame k. With selective repeat, it is possible to retransmit only frame k; the receiver buffers successive frames. 3b Let W denote the receiver’s window, and M the maximal sequence number, explain why W ≤ 12 M. 5pt Choosing W and M like this ensures that numbers of outstanding frames (as perceived by the sender) can never overlap with the receiver’s previous window of acknowledged frames. In this way, when- ever an ACK is lost, we avoid that a retransmission by the sender is interpreted as new frame by the receiver. 3c Why is it useful to have a large window size when dealing with slow connections such as with satellites? 5pt Having a large window size allows to have many outstanding frames, thus increasing the utilization of the connection. Part II 4c When a router receives an IPv4 packet, it lowers the packet’s time-to-live (TTL) value by one. If the TTL value drops to zero, the packet is discarded and an ICMP message is returned to the sender. How can this mechanism be used to trace the route a packet will follow when going from A to B? 5pt This principle is applied in the traceroute program. It sends packets from A to B with increasing TTL values, starting with 1. Every time it gets an ICMP message back, it will have discovered a router. When it finally has the TTL value set such that B is reached, it will (generally) have discovered a route from A to B. 4b Peer-to-peer routing is generally efficient in terms of the number of hops, but not in terms of the distance that is travelled in the underlying network. What is meant by this statement? 5pt P2P routing takes place between nodes that can be placed at entirely different locations in the Inter- net. In other words, two neighbors in a P2P network may actually lay in completely different parts of the Internet, effectively requiring long-distance Internet communication. 5a The difference between UDP and IP packets is minimal. Why shouldn’t applications use IP directly? 5pt It is a matter of separating concerns: the transport layer should offer an interface to applications that hides the underlying network. For connectionless services, it does so by means of UDP. Furthermore, note that IP itself has no notion of ports. This type of address belongs to the abstraction provided by the transport layer and of which applications make heavy use. There is, in general, no way that an application can get an IP packet sent directly to an application. 5b Running TCP across wireless networks may lead to performance problems. Why? 5pt The problem is that the underlying network may drop packets due to bad transmission conditions. However, TCP assumes that packet loss is caused by congestion and as a consequence starts to decrease the congestion window size, in turn leading to sending off smaller segments. What it should have done is adopt a more aggressive form of retransmission (perhaps with smaller segments) instead of just merely lowering the segment size. 5c The figure below shows the state-transition diagram for setting up and tearing down a TCP connec- tion. The thick dashed line represents the normal path for a server; the thick solid line that of a client. What happens according to this diagram when the ACK sent by the client when changing state from “SYN SENT” to “ESTABLISHED,” is lost? What happens normally in TCP? 5pt 2 CLOSED LISTEN ESTABLISHED CLOSING CLOSE WAIT (Start) CONNECT/SYN LISTEN/– SYN/SYN + ACK SYN RCVD FIN WAIT 1 TIMED WAIT LAST ACK FIN WAIT 2 SYN SENT RST/– ACK/– (Active close) FIN/ACK FIN + ACK/ACK FIN/ACK ACK/– ACK/– ACK/– SEND/SYN SYN/SYN + ACK (simultaneous open) (Data transfer stake) SYN + ACK/ACK (Step 3 of the three-way handshake) CLOSE/FIN CLOSE/FIN FIN/ACK CLOSE/– CLOSE/– CLOSE/FIN CLOSED (Passive Close) (Timeout/) (Go back to start) The server blocks in state “SYN RCVD.” Normally, a timeout occurs and the server resends the “SYN+ACK” message after which the client can resend its ACK. If even that fails, then both will eventually start closing the connection again. 6a What is the role of DNS when sending mail to steen@cs.vu.nl? Be precise! 5pt The mail server (or even client) will need to look up the mail server in the cs.vu.nl domain. It does so through DNS by requesting for the MX record of cs.vu.nl. This record contains the DNS name of the mail server, which can then be resolved to the mail server’s IP address. 6b To offload servers, DNS makes extensive use of caches. When will DNS caching not be effective? 5pt When the mapping of name-to-IP address changes a lot, which may be the case with mobile comput- ers. 7a Explain how authentication with public/private key pairs can be established. 5pt Let EA denote A’s public key. One approach is that A first sends EB[RA] to B, after which B returns EA[RA,SAB,RB)]. RA is a number generated by A and SAB is a shared key generated by B. In the third step, A returns SAB[RB]. 7b Construct a certificate issued by an authority CA, telling that EA is the public key of A. Explain the content of your certificate. 5pt [EA,A,CA,KCA(EA,A)], where KCA is the private key of CA. Final grade: (1) Add, per part, the total points. (2) Let T denote the total points for the midterm exam (0 ≤ T ≤ 45); D1 the total points for part I; D2 the total points for part II. The final number of points E is equal to max{T,D1}+D2+10. 3