Wireless and Mobile
Networks
(ELEC6219)
Session 4: Efficiency of a link.
Data Link Protocols.
Adriana Wilde and Jeff Reeve
22 January 2015
Plan for this lecture
• At the end of this lecture (and related activities), students
should be able to :
– … apply a key theory of data communications
(while planning a strategy for answering exam papers)
– … consider design issues to ensure reliable delivery over
an unreliable link
– … reason why the data link layer adds control
information as an overhead in the transmission
2
Review
2nd review data encoding
(feedback from questions
at the end of last lecture)
• 4B/5B (4 bits encoded in 5 transition bits)
• These codewords are then coded using NRZ-I
4
Talking point
• Solve and discuss solution to Q2a from the past paper
shown in OHP.
– As a 7 marks’ question in a 75 marks’ exam paper, you
shouldn’t spend more than 10 mins in it.
– Do this in pairs (not in exam conditions )
Shannon’s capacity of a
communication link: C=H log (1+S/N)
2
A wireless link with carrier frequency 400MHz has S/N=7 and
uses a modulation scheme that transmits 3 bits in 32 cycles.
What is the minimum bandwidth required for the link to
operate successfully?
58%
1. 12.5 Mbps
2. 46.875 MHz
3. 133.33 MHz
4. 37.5 Mbps
5. None of the above
21%
9%
9%
3%
1
2
3
4
5
6
How confident are you on
your answer being correct?
1. Very confident!
2. Quite confident,
not sure about one
thing or two
3. Not very confident
(I guess it is OK?)
4. Help!
43%
27%
17%
13%
1
2
3
4
7
How good was your answer?
1. Very good!
2. Quite good, I need
to be careful with a
couple of things
3. Not very good
4. I had no idea!
57%
19%
19%
5%
1
2
3
4
8
Efficiency of a Link
(Case study HDLC)
HDLC Frames
• Every frame is delimited at the start and end by the
sequence 0x7E (01111110), the “flag byte”
(N.B. both sequences are part of the frame)
8
8
8
…
16/32
8
Variable
length
10
The payload
• The “Data” field is the “payload”
– i.e. the information we intend to communicate to the
other end of the point-to-point link
• All other information is discarded at the receiver
8
8
8
…
16/32
8
Variable
length
11
Talking point: What size for N?
• Let's calculate the efficiency of the transmission in the
following cases:
– Sending one character per frame (N=8bits)
– Sending a KB per frame (N=1024bytes)
8
8
8
…
16
8
Variable
length (N)
12
Sending one character per frame (N=8 bits)
• Frame overhead= 8+8+8+16+8=48 bits
• So, if a single 8-bit character is sent per frame:
Efficiency = 8 / (48 + 8) = 1/7= 14%
8
8
8
8
16
8
13
Talking point:
Sending 1024 bytes per frame
• What is the efficiency of the transmission in this case?
N
14
Talking point:
Sending 1024 bytes per frame
• Frame overhead= 1+1+1+2+1=6 bytes
• Therefore,
Efficiency = 1024 / (6 + 1024) = 99.4%
1
1
1
1024
1
1
GOOD!
So, go for a longer frame, for even higher efficiency?
15
Design considerations: Frame size
• Not necessarily…
• What if a frame must be retransmitted?
• So “frame size” is part of an overall system design
– Depends on application
– Depends on anticipated error rate
– Depends on system requirements for high link
utilization and time available for transmission
• In practice, frame sizes of 0.5KB to 4KB are typically used.
16
HDLC frames
(the other fields)
What about the other fields?
• FCS - Frame check sequence
– 16 bit CRC-CCITT (Cyclic Redundancy Check) based on
the polynomial x16 + x12 + x5 + 1
– 32 bit based on the polynomial x32+ x26 + x23 + x22 + x16 +
x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1
– Complex algorithm but hardware implementation with
shift registers allow for quick, “on the fly”, calculation
pp.196-200 (T. 4ed)
pp.219-235 (T. 5ed)
18
Frame Check Sequence (FCS)
Even a 16-bit CRC will detect all burst errors up to this
length, with only a 1/216 probability for a corrupted frame
at the receiver to have the same CRC as calculated at the
transmitter!
pp.196-200 (T. 4ed)
pp.219-235 (T. 5ed)
19
Frame Check Sequence (FCS)
Even a 16-bit CRC will detect all burst errors up to this
length, with only a 1/216 probability for a corrupted frame
at the receiver to have the same CRC as calculated at the
transmitter!
(99.998% confidence)
pp.196-200 (T. 4ed)
pp.219-235 (T. 5ed)
20
What about the other fields?
• Address and Control
– An ‘address’ is only necessary when there are multiple
receivers on the line – frequently HDLC is used for
point-to-point links and this is then not required.
– The ‘control’ information is used for sequence numbers
and acknowledgements – this is the mechanism used for
frame control (i.e. when a re-transmission is necessary).
21
Talking point
• WHEN IS A RETRANSMISSION NECESSARY?
22
HDLC frames
(the other fields)
Let’s first consider normal operation…
Consider a bi-directional* link:
time
1.The transmitter will send a FRAME
2. … after a delay this will reach
the receiver, which will process it
3. … and after a small delay
(to process the frame) will send
an acknowledgment (ACK)
4. … which after a delay will be
received by the Tx
Tx
Rx
1
2
3
4
* most HDLC links are bi-directional, to enable the receiver (Rx) to acknowledge successful
24
reception of a frame sent by the transmitter (Tx)
What happens if a frame is lost?
No problem!
time
Tx
Rx
The transmitter
does not wait “forever”
TIMEOUT
for an ACK
There is a TIMEOUT
… after which the frame is
retransmitted
25
Also works if the acknowledgment is lost!
time
Tx
Rx
The transmitter
does not wait “forever”
TIMEOUT
for an ACK
There is a TIMEOUT
… after which the frame is
retransmitted
WE NOW HAVE ANOTHER
PROBLEM!
Ideas?
26
The problem now (and solution?)
We now need to
detect that the
second frame is a
duplicate and
discard it.
time
Tx
Rx
TIMEOUT
27
The problem now (and solution!)
We now need to
detect that the
second frame is a
duplicate and
discard it.
time
Tx
Rx
TIMEOUT
Use a sequence number!
28
Sequence numbers
For the “STOP-AND-WAIT” scheme, a one-bit
sequence number is enough for FRAME and ACK
Tx sends:
FRAME '0' until ACK '1' is received, then
FRAME '1' until ACK '0' is received, and so on
What does Rx do?
pp 206-211 (T.4e)
pp 241-246 (T.5e)
Note the convention:
the ACK number gives
the NEXT frame to send
29
Sequence numbers
Note the convention:
the ACK number gives the NEXT frame to send
30
Talking point
• How long should the timeout be?
• Ideas?
31
Design consideration: Timeout
• The goldilocks solution! (Neither too short nor too long)
32
Design consideration: Timeout
• The goldilocks solution! (Neither too short nor too long)
• Too short – inefficient as frames are unnecessarily retransmitte
• Too long – inefficient when frame lost
33
What now?
• This scheme can be improved in various ways.
• Several frames could be sent simultaneously but then there is m
in the control.
34
Checking Learning Outcomes
• At the end of this lecture (and related activities), students
should be able to :
– … apply and compare several data encoding techniques
– … apply a key theory of data communications
(while planning a strategy for answering exam papers)
– … consider design issues to ensure reliable delivery over
an unreliable link
– … reason why the data link layer adds control
information as an overhead in the transmission
• I'll see you on Wednesday at 9:00am in B54/5027
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