Answers to Questions 1-3 of the Exam Datakomunikation, 961018

The following are answers for questions 1, 2, and 3. If we asked for descriptions, several answers can be right. We selected only some of them for this! So don't worry if you can't find you answer here, it might be also right. We also put just some keywords into this solutions to give you some ideas what we expected.

Question 1

  1. Connectionless: no connection set-up, data is sent as packet(s) into the network, each packet has to include at least a destination address needed to forward it, comparable to snail-mail
    Connection oriented: one must first set-up a connection before data can be transmitted, all data follows this connection, data needs at least a connection idendifier, comparable to the telephone system
  2. Reasons for layering: reduce complexity, ease the exchange of components, structure systems, hide implementation details, security, interworking, etc.
    Where to use: if systems grow to complex, distribute responsibilities, price/performance reasons (e.g., seperate fast/expensive hardware from cheaper/slower software)
    Where not to use: small systems, if layering gives too much overhead for the needed performance, if it would result in doubling of functions
  3. Common properties: layered systems, offer interfaces, use protocols between peer entities, standardized (in a different way, but they are standardized), not properties of a company, open for different applications, network/transport/application layers exist,
    Differences: for OSI the standardization came before an implementation, OSI has more layers, OSI network layer can be connection oriented, OSI is used more in the telecom environment, TCP/IP is the foundation of the Internet, and many more...
  4. Advantages: economics of scale, everyone can talk to everyone around the world, chances for smaller companies, stable systems
    Disadvantages: compromises often lead to poor standards, it takes a long time before a standard is settled and then the technology is much more advanced, very difficult to change if it is widely adopted etc.

Question 2

A noiseless channel can carry an arbitrarily large amount of information, no matter how often it is sampled or what the bandwidth is.Therefore is the maximum data rate unlimited. This can be also seen from the Shannon and Nyquist theorems without using a calculator.

Nyquist states: C = 2 W ld(M), with C the capacity of the channel in bit/s, W the bandwitdh in Hz, and M the number of levels per signaling element. And exactly this M is not limited and, therefore, C is not limited.
Shannon states: C = W ld(1 + S/N), with S/N the signal-to-noise ratio. We assumed a noiseless channel, therefore, N=0, S/N is infinite! Again, C is not limited.

If you now take the minimum you see that this minimum is not limited! Something like the first paragraph was also accepted as an answer. Nobody talked about only two signal levels, the sampling rate of 2kHz was as irrelevant as the bandwidth of 7.5kHz. The sampling rate limits only the baudrate to 1000 Baud!

Question 3

  1. You need at least 9 different sequence numbers to avoid conflicts with the acknowledgement. Think what happens if the sender has sent 8 packets (as allowed), we'd use only 8 sequence numbers, and then the sender receives an acknowledgement for packet 0. Is this for the new packet 0 (i.e., all 8 packets have been received) or for the old packet 0 (i.e., all 8 packets have been lost). Therefore, you need at least on sequence number more, which is 9, and, therefore, 4 bit to represent it.
  2. One packet has the size of 250 byte, that is 2000 bit. To send a packet we need 2000/10000000 s = 0.2 ms. To send without interruption, an ack must arrive no later than after the sending of the 5. packet to increment the credit. That results in a maximum acknowledgement time of 5x0.2ms = 1 ms.
  3. Go-back-n: If an error occurs, a sender has to resent all packets beginning from the lost packet. The receiver accepts packets only in order and needs a buffer for one packet to check it before forwarding to a higher layer. With a packet size of 250 byte this buffer must have the size of 250 byte. Simple method, but waste of bandwidth.
    Selective-repeat: If an error occurs, the sender resends only the lost packet(s). The receiver accepts packets within a certain window, reorders them if necessary. With a credit of 5, packet size of 250 byte, we need a buffer of at least 5x250byte = 1250 byte. More complicated, but saves bandwidth.
  4. For this question we accepted several answers, depending at what time you started/finished your transmission. Important results are the following, which should have been mentioned or shown in your results. Explanations and additional comments were appreciated!

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