Here are answers to some questions .... .
Note that the more recent answers are added at the
start .
As it is now 12 noon on Thursday 10 June, no more
email replies will be posted to this list.
Correction,
to question on expansion of bit stuffing.
The expansion is by 1/64, about 1.5% because we must have the bits
...011111 to force the bit stuff. This is one case out of 64.
However, there is a very improbable complication, which is what confused me
and can be ignored.
If you have a very long string of 1s, the first bit stuff is after 6 bits
(as above), but later bit stuffing occurs after each group of 5 1-bits.
The effect is absolutely negligible.
I am confused with the coverage of tcp checksum. The notes said that it covers
to whole tcp segment as well as pseudo-header. The fields included in the
pseudo header are fields from ip header. The message pass down the protocol
stack before transmission, that is TCP layer before IP layer, how can the TCP
layer knows what values the pseudo-header contains and calculate the checksum?
The answer is that TCP/IP does not use strict layering.
TCP is able to find from IP the values for the TCP header.
OK, it is messy, but TCP/IP is designed to work, rather than for elegance.
In asynchronous communication:
The format was:
Start bit, 8 data bits, parity bit, stop bit
When there is no character parity, does that mean the most significant bit of
the ascii char will always be 0? but we still have the parity bit in use in
front of the stop bit?
It depends on how many bits are specified, as the transmitted encoding is very
flexible.
Usually if there is no parity, specify an 8-bit character and no parity.
Then the most-sig bit is whatever is supplied from the computer, which is
usually 0.
The usual combinations are either -
7 data bits, with parity (odd or even)
8 data bits, no parity.
Other combinations are possible but rare, and can be ignored.
In SDLC transmission:
1: When we have no character parity does that mean the most significant bit of
the ascii char will always be 0?
Usually, but it is the value of the byte sent from the computer.
Most computers have that most-sig bit 0, but I have seen it always 1.
2: When we do have character parity, the most significant bit of the 8 ascii bit
is the parity bit and we have the FCS at the end of the frame for the entire
frame as well?
Yes
3: When we get 5 conecutive 1's we need to stuff in a 0. Which means the length
of the data field may not always be 8N bits?
The data field as seen by the user is always a multiple of 8 bits.
The number of bits actually sent/received need not be a multiple of 8.
4: for question 1 in the 01 exam,we were asked to show the transmitted bits, is
it necessary to draw the bit? or just writting out the 0s and 1s will be fine?
Whichever you like, as long as the intention is clear.
5: for this question, we only need to show the transmitted bits for the data
field for SDLC and dont need to show the flag, address and control bits?
Ignore any flag codes etc. If you feel doubtful, say that they are not shown.
6: I can only think of the flag as the symbols that violate the encoding rules,
would u give me some hints on finding out some others?
The JK 'bits' in token ring SD and ED.
I've been trying to understand how the frequency spectrum for a square wave is
drawn, and I still don't really understand the frequency spectrum. I was
wondering if you could be able to explain to me how it actually works. It would
be greatly appreciated.
This is too big to explain by email and you really just have to accept what
the notes and book say, unless you really trust your mathematics.
For questioon about the fragmentation, how should we layout our answers, can we
have a column for MTU, so that it make it easier for the marker, and do we have
to includes the fragmented packets for each networks in order?
Do whatever the question asks.
You must include what the question says, but can include anything else that is
relevant and helps you.
Questin about the questions you crossed out in the exams, what does it means,
does it means the questions is no longer applied in the exam, but the questions
on that topics can still be in there, such as the topic about satellites?
Ignore them. Either they are questions that did not work properly,
or they are topics that I do not cover.
I'm a bit confused about Q.9 from the 2001 S1 exam. Given two
BISYNC messages,
message one: STX a b c DLE STX p q r ETX
message two: DLE STX a b c DLE STX p q r DLE ETX
what is the text each message contains.
From my understanding, message one is just "a b c DLE STX p q r".
For message two, i think the first DLE STX stops the reciever from
checking for STX and ETX characters. But when it gets to the second DLE,
it switches back to checking for control bytes. So when when it gets to the
second STX, is an error generated, as another frame is starting within an
unfinished frame?
Forget it. This is a BAD question and will not be asked.
1: I am still confused with distinguishing between Ethernet frame and 802.3
frame.
My understanding is that:
IF (Ethernet frame)
IF (type/length field >0x600 ie 1536) THEN
type/length field is a type field
ELSE
type/length field is a length field
END IF
IF(802.3 frame)
IF (type/length field <1500) THEN
type/length field is a type field
ELSE
type/length field is a length field
END IF
So to tell whehter it is a type or length field, we have to first find out
whether it is an ethernet frame or 802.3 frame?
You look at the value of the length/type field and from that decide
whether it is Ethernet (large value = type),
or IEEE 802.3 (small value = length).
It is only the value of this field that matters.
2: Again for question 1 of 03Fc exam, when we are given a frame like this, they
are already in the correct order ie preambles on the left and checksum on the
right?
Yes, that is if the preamble and checksum are shown (they might not be).
Also, when we have 08 in 13th octet and 00 in 14th octet for type field, it
represent 0800 and not 0080, am I correct?
Correct. The most significant octet is to the left, so just read the whole
field as a hexadecimal number.
BUT, the bits within an octet are sent rightmost (least-significant) first.
3: ARP stands for Address RESOLUTION protocol and RARP stands for Reverse
address RECOGNITION protocol??
Yes.
I have the following questions regarding to the lecture notes:
1:
The Ethernet address is the same as MAC address?
Yes
2:
The coverage of the FCS for Ethernet frame is from destination adress onwards or
right from the preamble?
See earlier answer (below in this list)
3:
On Ethernet, each station will have an unique individual adress and 1 group
address for each multicast group that the station belongs to?
Yes, except it can sometimes have more than one unique address
4:
Can we actually distinguish an Ethernet frame and an 802.3 style frame by
looking at the type length field and the present of the SNAP headers or is it
unnecssary to do this?
Type/length field only.
1.How many bit errors cn be detected, how many can be corrected?
eg. parity bit: odd number of bits in error but not even number
Haming code:
CRC:
Checksum:
Parity : detect odd number of errors, correct none
Hamming : detect and correct single error
CRC : detect several, correct none (usually)
Checksum: as for CRC
2."Ethernet is a very indfficient network because so much data is lost
from data collision."
Criticise and comment on this statement.
(This is a complaint that was sometimes made against Ethernet.)
- Ethernet is designed for collisions, and to recover quickly.
- Collisions are detected very early, before any or much data is sent
- Transmission stops immediately after a collision,
so no transmission is wasted.
3.Two users are communicating using a protocol which uses XON/XOFF
flow control.
What is the visible effect if
a. One user , at a terminal, receives an XOFF character?
b. Each user, almost simultaneously, sends an XOFF character to the other?
a. the user cannot send until an XON is received (that terminal locks up)
b. both terminals lock up
4. A station using an XON/XOFF protocol will often send an XOFF when its
buffers are
nearly full and then an XON when the buffers are nearly empty.
a. What are the likely effects if the station waits until the buffers
are completely full or
completely empty?
b. How might you estimate a suitable ""fullness" or "emptiness" to
trigger sending the XON or XOFF.
a. full - buffers overrun; empty - wasted time during wait for new
data to arrive
b. send XOFF (buffer full) when the free space is at least twice the
data in transit,
it must say empty (send XON) while the buffer has at least data equal
to twice that in transit.
1. Is congestion occurs in Virtual circuit? How is congestion handled?
2. If deadlock does occur, gibe one way by which it may be broken...
3. one way of preventing deadlock?
1. No congestion, because the Virtual Circuit guarantees capacity through the
whole path.
2. Discard a packet,
3. Ensure that all reassembly buffers are large enough to receive their messages
I have a question about assignment 4 q3.(iv)
In the lecture note, it says that "VC number must be unique in one direction
over the one link",
but in the sample solution of assignment 4, q3(iv), port 0 (input) and port 5
(output) shared with the same VC number, which conflicts the statement in the
lecture nots, shouldn't be port 5 reassigned with another randomly generated VC
number?
The important point is "over the one link", NOT path.
Here the VC numbers are repeated on different links.
It is the pair {port, VC} that is important and must be unique.
Constellation:
If we have 2 amplitudes and 8 phases on the diagram, does that mean can have 16
combinations fo signals on each of the constellation point?
yes
Question 7:
b) "What is the smallest bufer to guarantee full use of the link?"
Is that within the time of a round trip between the stations,the buffer must be
able to accomodate the number bits that the stations can transmit
per second?
c)"Over what maximum transmission distance would a 50,000 byte buffer ensure a
link utilisation of 100%"..is it a calcualtion problem or that we cannot get a
utilisation of 100% because it is a tiop and go protocol.
b. The buffer must be able to hold the amount of data that can be sent in one
round-trip time.
c. Calculate the distance for 50 kByte taking the round-trip delay.
Question 11:
a) We were asked to compare and estimate the delay of signals by the cables and
by the repeaters.
I know that, there is are link latency caused by the cable. Is that what we were
asked for?? For the repeaters delay, I am totally lost. Would you please give
me some hints?
b) As mentioned in the last lecture, collisions are detected when the collided
signals arrived at a station that is listening to the network, in this case it
is when the signals had travelled 2km and reached the other station. So to
answer this question, we need to calcualate the link latency? After that, isn't
that we need to know the rate at which the stations transmit to see how many
bits were sent within that travel time?
a. Assume that a repeater has no delay (it is usually just 1 or 2 bits)
b. You must be working on a cable based Ethernet, because collisions such
as here cannot occur with switches. Cable Ethernet is limited to 10 Mbit/s.
I m studying the slides in TCP section. I m not quite understand, why halving
the congestion window can reduce the traffic into the network ? (TCP Flow
control, Congestion control in Internet slide 11,12)
The congestion window controls the amount of unacknowledged traffic
from this station, and therefore the amount data which may clutter up buffers
and lead to congestion.
Congestion controls do not affect the bit rate at which data is sent.
I have got really urgent questions to ask ,it is in previous final exam,well
I am not asking the exact question but the way to answer it.
Question 1:
As for spanning tree algorithm,we need to expain the picture in words in the
exam,but the containing of the words in the sample solution of assignment and
the handout are different,which way I should follow.Can pls tell me what is the
minimam infomation I should show in the exam?
In the 2003 exam I suggested that you describe each step,
as "Connect C to B, with cost to A = 12".
But see what the question asks.
Qusetion 2:
As for final exam year 2002 qustion 2 about LZW compression.You left many
columns in the table like test string,emit,make entry,comment,asciicodes....
you filled out just part of emit,comment and contents three columns.So How many
columns we need to finish in the exam under this condition?
Again, see what the question asks.
If you feel that you need more columns, then show what you think is necessary.
Qustion 2:
in you handout wide area networks-routing table page6.
The step g brings in three links all with a root cost of 11.so in the final exam
we need to state all those three link one by one or just one of them?
Last question is as for "off route" connections or "stubs",do we need to state
it out in the exam??
Give the links one by one, or state that all three are added in one step.
It doesn't matter which as long as all connections are included.
What does the question actually ask?
Mention the stubs if you think it necessary, but it might not be needed.
I am afraid I disagree with you about the approximate percentage of extra
0s needed for bit stuffing a random number of digits in HDLC (inserting a
0 after 5 1's). I believe the original handout is correct in saying the
expansion is around 1.5% - I have written a quick Java program to back up
my argument (although I agree that Math.random() is never truly random).
It looks like stuffing in a 0 after a run of four 1's would require an
expansion of around 3% using the program. It is attached - most
parameters can be changed.
I stand corrected.
In some of the past exam questions involving fragmentation/reassembly you
allowed that " The header size can be ignored".
In this case we would not have a Payload length field in our table , as we have
to deal only with the Total Length.
Therefore , all consequent calculations , such as Fragment offset Values will be
calculatedusing a Total Length.
Eg.
If a packet is 2000 octets , then an offset value will be 2000/8 = 250
I am right ?
Payload length and total length are the same.
The offset is calculated from the total payload lengths of all the
preceding fragments.
I was just going over your lecture notes on "Shannon's Channel capacity" in
part 1 of your lecture notes about Physical communication (Slide 15). I noticed
that your calculation for 1+S/N = 2^18.1 = 274,133 = 54 dB
when i punched in 2^18.1 in my calculator it gave me 280958.9826 instead of
what you had which was 274,133. I was wondering if I missed something or
whether there's an error. Either way I noticed that both give 54 dB, assumming
that S/N is the power ratio.
I am unsure where the 274,133 came from, but it is so close to 280958....
that the difference can be ignored.
But as calculators are not allowed, you will not be asked to do
calculations like this.
Having done 225, 220, I'm quite familiar w/ Djikstra's algorithm.. Im not sure
why in your notes (WAN, page 7 of 39) the link between B and C with cost 7 is
used
it is exact same cost to go through EF and then to C, but the sum of costs in
the tree would be less, Am i confusing minimum spanning tree w/ spanning tree?
the only obvious reasoning behind using B-C link is to to not overuse the link
through BEF
No reason at all.
I have one question for 2001 Q3 , you taught in class that for (a) , the
percentage of extra bit added by the bit stuffing is 1/32 = 3% , but in the
handout the pattern 011111 occurs in 1 out of 64 bit , which is about 1.5% .
Why is different ? Could you answer me please , thanks .
A zero bit is stuffed after EVERY occurrence of 11111, which has a
probability of 1/32, say 3%. The handout is wrong.
I just confused about Logic Link Control. The potocols(doing flow control and
error control etc.) in Logic Link Control layer of the Data Link Layer nowadays
are something like SDLC from IBM. As mentioned in the text book 3rd edition
page 402 Figure 9.5, there are HDLC frame(similar to SDLC I think). However,
when I used some network tool to catch ethernet frames on my computer, I found
there is no such frame head(like HDLC or SDLC) between ethernet frame head and
IP head, I have attached two pictures one shows ethernet frame head and the
other shows IP head. So I think maybe TCP has done the Error Control and Flow
Control such that no need to have Logic Link control frames.
Could you please explain it?
You are correct.
Both frames have the length/type field = 0800 hex (= 2048 dec).
As this is greater than the maximum length of 1500, it is type filed rather
than a length. The frame uses "Ethernet" conventions rather than the
old "IEEE 802.3" conventions.
The Type=0800 in fact marks it as an IP frame.
i am very confused about the collision domains, by my understanding only
the bridge can detect a collision,by after i read the text book it seems that
in collison domains it is also possible to form a collision,things like there
is no briage in collision domains as the graph shows in handout-Ethernet Page 15
how they know there is a collision occurs in collision domains,and plz fully
tell me what collison domains suppose to do for data transmission.
Collisions are detected by ANYTHING listening to a cable, whether it is a
repeater/hub, a workstation or a bridge (which really behaves just like a
station in this regard. Anything sending when it detects a collision will
force a jam sequence and then stop.
Collisions and jams are propagated through hubs and repeaters,
just as on a cable, but not through bridges and switches.
A collision anywhere within a complex of cables, hubs and repeaters will
travel throughout that complex and all are in the same "collision domain".
A collision detected while a message is being received by a bridge or
station just results in the message being ignored.
Can you please explain the diagram on Handout 5 Page 12. How to determine the
message is on which LAN(s) for the given source to destination. Thanks alot.
See the next answer
I have a question about the lecture handout on Logical Link Control.
In pg 12, for z -> x, the message is on LAN 4, but should it be, the message is
on LAN 1, and 4, since the bridge doesn't hasn't learned which LAN the station
z is on, am I right, have I misunderstood anything?
Thank
- A bridge learns the position of a station only when it
receives a message from that station.
- A bridge receiving a message for a station will forward that
message to a single port if it knows that station has been seen
from that port. Otherwise it sends the message out through all
ports.
- In this example, Z on LAN 4 is sending to X. But it already seen
a message from X also on LAN 4, so it knows that the
message need not be sent anywhere else.
In your handout note P25 about token ring frame format, it says that
Frame Check Sequence covers FC to FCS, so why SD,AC,ED and FS field need not
be checked? Is it the same for the Ethernet frame? Are preamble,SD in the
Ethernet frame covered by its checksum?
If IPv4 is replaced by IPv6,does it effect the checksumming of TCP
segment and Ethernet frame as we know IPv6 has no checksum?
-
These other 4 fields are added by the transmission hardware, rather than
supplied from higher layers. Also, SD & ED have non-data bits and cannot be
checksummed.
-
Again, the Ethernet FCS covers only the user-supplied data, not anything
added
by the physical transmission, so the preamble and SD are NOT included.
-
IPv6 certainly doesn't affect the checksumming of Ethernet because it
is just another bit pattern and Ethernet doesn't know the difference.
As far as the TCP pseudo-header is concerned, I just don't know.
It seems reasonable to assume that the pseudo-header is changed with IPv6,
but I have been quite unable to discover any details.
I am doing 314 and by looking at your handout for the tcp/ip protocol suite.
page 19 which is IPv6,it remind me you said something in the lecture that IPv6
is not that much examable,but I cant quite remember what was that.Can you
please tell me again here,that will be very kind of you.
The best thing is to look at the past questions on the web.
Past questions include the fields present in IPv6 header, and changes
to fragmentation, possibly the difference in header sizes.
From the past test question and lecture notes, the mothod for solving Hamming
code is by have the least signiciant bit on the left, how ever it the opposite
for this test, i am interested to know which is the more correct mothod for
the coming exam, and if the "most significiant bit on the left" is correct
mothod for solving the problem, can i recover my lost marks??
There is NO standard order.
I had always done Hamming code with the least significant bit on the right
(as for computers),
but then changed to fit with Shay's convention
(least significant bit on the left, as for data com transmission).
If the question doesn't say which order, use either convention,
BUT SAY WHICH ONE.
Will we be given formulas or will we have to memorise them? ie. should I
memorise Shannons equation?
I expect you to learn relatively simple formulae like that.
I got two questions about IP v6 fragmentation.
1.How does the sender know the appropriate size of packets in the links?
2.I cannot find any field about fragment info in the IP v6 header, could you
please tell me where these info will be stored?
Is the so-called fragment
header same with the IPv6 header or sth else or just part of the IPv6 header?
1. The sender must discover the MTU by probing the path using packets with
the "do not fragment" bit set.
2. IPv6 uses a "Fragmentation Header". It is one of the optional headers that
we have not discussed.
I have a quesiton about the analogue signalling, is the carrier actually the
signal that contains the actual information, so in a sense the informaiton is
encapsulated in the carrier so that it can transmit the data, but why are we do
the modulation, and can you tell me what the sidebands are?
The carrier "carries" the information by varying one of its parameters
and we recover a modulating signal by detecting those changes.
Any change from a single frequency, constant amplitude, signal MUST
produce other frequencies, known as the sidebands, which are usually
placed symmetrically around the carrier.
The information is entirely in the sidebands, and with Amplitude Modulation
the carrier (and one sideband) can be removed completely.
The modulating signal can be recovered by inserting a signal equivalent
to the carrier at the receiver.