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.
- 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
- Reasons for layering: reduce complexity, ease the exchange of
components, structure systems, hide implementation details, security, interworking,
Where to use: if systems grow to complex, distribute responsibilities,
price/performance reasons (e.g., seperate fast/expensive hardware from
Where not to use: small systems, if layering gives too much overhead
for the needed performance, if it would result in doubling of functions
- 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
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...
- 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.
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!
- 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.
- 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 =
- 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
- 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!
- Both, go-back-n (gbn) and selective repeat (sr), have the same worst
case behaviour. The total transmission time is 12.5 ms without the last
acknowledgements or 16 ms with the last acknowledgements. We accepted both
solutions. We also accepted a shift of 0.5 ms due to the point where you
started your transmission.
- It was important to choose the 5th (last!) packet to be dropped for
sr, any of the other packets do not represent the worst case. For
gbn it does not matter which packet you choose for dropping.
- We wanted to see you using timing-diagrams as stated in the question.
It was important to see the difference between gbn and sr in the diagrams.
Furthermore, we wanted to see if the relation between sending, transmission,
and receiving was ok. That includes 0.5 ms gaps between sending packets,
3 ms of transmission etc.
- Reading the questions thoroughly helps a lot! If we write credit =
5 we mean 5 and not 8, although maybe we used 8 in some former questions.
If we give you a packet size we want you to use it and not assuming one
- You should not carry assumption across different questions. We state
all you need in every question.
- 1 byte = 8 bit, that would have helped many in 3.2; if 1M is 1e6, then
10M is 1e7!
- Don't mix concepts! Connection oriented/Connectionless is one concept
for itself, don't mix this with acknowledgement, reliability and especially
routing. There are connectionless services which use always the same route
for all packets.
- Nobody needs a calculator for, e.g., 10e4/10e7=10e-3
- You had enough time, so we expect you to write and draw in a way we
can read it and figure out what you mean.
- Question 2 was on the list of recommended exercises.