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Transmission & Error control
ARQ
Giuseppe Bianchi
Two scenarios
network
End to End
(transport protocol issue)
Hop by Hop
(datalink protocol issue)
Giuseppe Bianchi
Coping with rx errors
Forward Error Correction
Extra overhead capable of CORRECTING errors
Large overhead depending on protection capability
Retransmission
Extra overhead capable of DETECTING errors
Small constant overhead (2-4 bytes)
Called:
» Frame Check Sequence
» Cyclic Redundance Check
Giuseppe Bianchi
Retransmission scenarios
referred to as ARQ schemes (Automatic Retransmission reQuest)
COMPONENTS: a) error checking at receiver; b) feedback to sender; c) retx
SRC
SRC
DST
DST
Error
Check:
OK
Error
Check:
corrupted
Automatic
retransmit
Basic ACK idea
SRC
Basic NACK idea
SRC
DST
DST SRC
Retx
Timeout
(RTO)
DST
Error
Check:
corrupted
Basic ACK/Timeout idea
Giuseppe Bianchi
Sequence numbers – a must
Sender side:
Receiver side:
RTO
DATALINK
or
NETWORK
(ACK lost)
rtx
Need to univocally “label” all packets circulating
in the network between two end points.
1 bit (0-1) enough for Stop-and-wait
Giuseppe Bianchi
New data?
Old data?
Link-level model
C bits/sec
In e2e scenario, C approximated by bottleneck link rate
Stop & Wait 
one frame/packet at a time
Pipelining (continuous ARQ) 
more than one frame/packet
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Stop & Wait
Giuseppe Bianchi
sender
receiver
stop-and-wait
One way delay
MSG/C
RTT

ACK/C
For simplicity all MSG of same
size; extension to != size  use
mean value

MSG
thr 
RTT  MSG / C  ACK / C  2
time
time
Giuseppe Bianchi
REMARK: throughput always lower than
Available link rate!
Upper bound
No processing
 = 0
ACK size negligible
ACK = 0
MSG
thr 
RTT  MSG / C
1
efficiency  thr / C 
1
RTT
1
MSG / C
Giuseppe Bianchi
stop-and-wait: upper bound /1
MSG = 1500 bytes
Under-utilization with: 1) high capacity links, 2) large RTT links
Giuseppe Bianchi
stop-and-wait: upper bound /2
MSG = 1500 bytes
Under-utilization with: 1) high capacity links, 2) large RTT links
Giuseppe Bianchi
Example
MSG = 1500 bytes
RTT = 10 ms
RTT=100 ms
Link rate (kbps) throughput efficiency throughput efficiency
33
32,12
97,32%
25,88
78,43%
128
115,66
90,36%
61,94
48,39%
1.000
545,45
54,55%
107,14
10,71%
10.000
1071,43
10,71%
118,58
1,19%
100.000
1185,77
1,19%
119,86
0,12%
1.000.000
1198,56
0,12%
119,99
0,01%
INFINITE
1200,00
N/A
120,00
N/A
MSG
MSG
thr  lim

C  RTT  MSG / C
RTT
Giuseppe Bianchi
Dealing with errors
sender
receiver
IMPORTANT ISSUE: setting the Time Out right!
Too short  unnecessary rtx
Too short  inconsistent protocol operation
 Too long  waste of time
TO
Ideally: TO = RTT+ACK/C

(+2 )
Easy to say, but harder to do if RTT is not
known (e.g. in e2e scenario)
Giuseppe Bianchi
Inconsistent protocol operation
sender
receiver
M=1
M=2  will be NEVER received!
M=1
Consequence: ACK MUST be also numbered
To always guarantee consistent operation
TO
M=2
M=3
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Performance with loss
(upper bound)
 No processing time, negligible ACK
 Per packet loss probability P
 Assumed indipendent
P(immediate success) = (1-P)
P(success at second tx) = P(1-P)
P(success at third tx) = P2(1-P)
size of message
MSG
thr 


E[message delivery time] E[# tx]  ( RTT  MSG / C )
MSG
MSG (1  P)


1
 ( RTT  MSG / C ) RTT  MSG / C
1 P
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Pipelining
(Continuous ARQ)
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Pipelining idea
Up to W>1 frames can be “in fly”
In fly = frames transmitted but not yet ACKed
Sliding Window
Size W
Window “slides” forward at each received
ACK
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Pipelining cases
RTT
(+1tx)
W=4
?
W=10
time
time
WINDOW SIZING that allows
UNDER-SIZED
WINDOW: THROUGHPUT INEFFICIENCY CONTINUOUS TRANSMISSION
W  MSG


thr  min  C ,

 RTT  MSG / C 
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Continuous transmission /1
Condition in which link rate is fully utilized
MSG
W
C
Time to transmit
W frames
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MSG
 RTT 
C
Time to receive
Ack of first frame
Example
2 ms
MSG=500 bytes
W=4
C= 2 Mbps
Propagation = 16 ms
16 ms
Question 1: assuming no errors, compute
time needed to transmit K=6 messages.
4 ms
MSG 
MSG
 K  1




TTX  
RTT


1

K

1
 RTT 16 ms


mod W

W
C
C



Question 2: which min message size for continuous tx?
W  MSG MSG

 RTT
C
C
 MSG 
Giuseppe Bianchi
RTT  C
W 1
 1334 bytes
Continuous transmission /2
We may elaborate:
W  MSG  RTT  C  MSG  RTT  C
Condition in which link rate is fully utilized
Wbit
 C  RTT
W x MSG =
W expressed in “bit”
Bandwidth-delay
product
Conclusion: full link utilization is possible when window size (in bits) is
greater than the bandwidth (C bit/s) delay (RTT s) product!
Giuseppe Bianchi
Bandwidth-delay product
D
C
Network: like a pipe
C [bit/s] x D [s]
number of bits “flying” in the
network
number of bits injected in the
network by the tx, before that
the first bit is rxed
64Kbps
A 15360 (64000x0.240) bits
“worm” in the air!!
bandwidth-delay product = no of bytes that saturate network pipe
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Long Fat Networks
LFNs (el-ef-an(t)s): large bandwidth-delay product
NETWORK
Ethernet
T1, transUS
T1 satellite
RTT (ms)
3
60
480
rate (kbps)
10.000
1.544
1.544
BxD (bytes)
3.750
11.580
92.640
T3 transUS
Gigabit transUS
60
60
45.000
1.000.000
337.500
7.500.000
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Throughput with pipelining
MSS = 1500 bytes
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Maximum achievable throughput
(assuming infinite speed link…)
Wbytes = 65535
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ARQ protocols
details
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Go Back N protocol
Sequence number:
b bits
N=2b values
Repeated modulo N
Sender window
Ws=N-1  number of possible sequence numbers minus 1
Receiver rules
RX: accept frames only if received in strict sequence!!
If out of order frame, reply with NACK
NACK(i) = I have received ALL frames up to
(i-1) mod N, but I haven’t received i-th frame
Giuseppe Bianchi
GBN example
4 5 6 7 0 1 2 3 4 5 6 7
4
4 5 6 7 0 1 2 3 4 5 6 7
5
4 5 6 7 0 1 2 3 4 5 6 7
6
4
4 5 6 7 0 1 2 3 4 5 6 7
7
5
4 5 6 7 0 1 2 3 4 5 6 7
0
6
4 5 6 7 0 1 2 3 4 5 6 7
1
4 5 6 7 0 1 2 3 4 5 6 7
2
4 5 6 7 0 1 2 3 4 5 6 7
3
4 5 6 7 0 1 2 3 4 5 6 7
7
4 5 6 7 0 1 2 3 4 5 6 7
0
4 5 6 7 0 1 2 3 4 5 6 7
1
4 5 6 7 0 1 2 3 4 5 6 7
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NACK 7
7
Piggybacking
Frame 1
ACK =
overhead on
return channel
Idea: embed
ACK inside a
frame
transmitted on
the opposite
direction
Frame 2
ACK 2
Important: cumulative ACK!
ACK 2 = ACK frame 2 + frame 1 + all previous
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Why Ws = N-1?
W=N=4
Frame X-1
= ACK 1
Frame 2
Frame 3
Frame 2
Frame 3
Frame 0
Frame 0
Frame 1
Frame 1
W=N=4
Frame X
= ACK 1
All successful OR all lost???????????????
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Frame X-1
= ACK 1
Frame X
= ACK 1
Selective Repeat
Key RX difference:
Received frames
buffered also if out of
order
New parameter:
Receiver Window Wr
New window sizing
rule:
GBN: Ws < N
SR: Ws + Wr ≤ N
GBN = special case  Wr=1
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N=4; Ws=3; Wr=2 Rx window
2
2 3 0 1 2 3
3
2 3 0 1 2 3
0
2 3 0 1 2 3
2 3 0 1 2 3
ACK 0
TO
2 reTX
NEW!!
Choice of windows
N=8
Wr > 4  not useful
not enough “in-fly” frames since Ws<4
Ws > 4  out of order may harm
 not enough RX window size to capture all
possible out of order cases
Most appropriate choice: Wr=Ws=4
 but remember throughput limits, though!
Giuseppe Bianchi
SR example and TX operation
4 5 6 7 0 1 2 3 4 5 6 7
4
4 5 6 7 0 1 2 3 4 5 6 7
5
4 5 6 7 0 1 2 3 4 5 6 7
6
4
4 5 6 7 0 1 2 3 4 5 6 7
7
5
4 5 6 7 0 1 2 3 4 5 6 7
0
6
4 5 6 7 0 1 2 3 4 5 6 7
1
4 5 6 7 0 1 2 3 4 5 6 7
2
4 5 6 7 0 1 2 3 4 5 6 7
3
4 5 6 7 0 1 2 3 4 5 6 7
7
4 5 6 7 0 1 2 3 4 5 6 7
4
4 5 6 7 0 1 2 3 4 5 6 7
4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3
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5
6
NACK 7
ACK 3  7,0,1,2,3