Cse3213 11 flow_errorcontrol_f2010

1

Error and Flow Control
Required reading:
Garcia 5.2

CSE 3213, Fall 2010
Instructor: N. Vlajic

Error Control
Error Control (1) Forward Error Correction (FEC)
Approaches (2) Error Detection + Automatic Retrans. Req. (ARQ)
•

not enough redundant info to enable error correction
case (a)

receiver detects no errors
an ACK packet is sent back to sender

case (b)

receiver detects errors
no ACK sent back to sender
sender retransmits frame after a ‘time-out’

error detected !!!
= round-trip time

2

Error Control (cont.)
Challenges of
ARQ-based
Error Control

•

3

send one frame at the time, wait for ACK
easy to implement, but inefficient in terms of channel
usage

•

send multiple frames at once
better channel usage, but more complex to implement sender must keep (all) sent but unACKed frame(s) in a
buffer, as such frame(s) may have to be retransmitted

frame
ACK
frame

frame
ACK
frame

buffer of
finite size

How many frames should be sent
at any point in time?
How should frames be released from
the sending buffer?

Error and Flow Control
Flow Control – set of procedures used to restrict the amount of data that
sender can send while waiting for acknowledgment

•

two main strategies
(1) Stop-and-Wait: sender waits until it receives ACK
before sending next frame
(2) Sliding Window: sender can send W frames before
waiting for ACKs

Error + Flow Control (1) Stop-and-Wait ARQ
(2) Go-Back-N ARQ
Techniques

(3) Selective Repeat ARQ

Error Detection + ARQ (error detection with retransmissions)
must be combined with methods that intelligently limit
the number of ‘outstanding’ (unACKed) frames.
Fewer unACKed frames ⇒ fewer packets buffered at sender and receiver.

4

Stop-and-Wait ARQ

6

Stop-and-Wait ARQ – simplest flow and error control mechanism
•

sender sends an information frame to receiver

•

sender, then, stops and waits for an ACK

•

if no ACK arrives within time-out, sender will
resend the frame, and again stop and wait
time-out period > roundtrip time

•

abnormalities (and how to fix them)
lost acknowledgment
delayed acknowledgment

error detected !!!
= round-trip time

Stop-and-Wait ARQ (cont.)
Lost Acknowledgment

•

7

frame received correctly, but ACK undergoes
errors / loss
after time-out period, sender resends frame
receiver receives the same frame twice

•

frames must be numbered so that receiver
can recognize and discard duplicate frames
sequence # are included in packet header

0

frame
ACK

1

frame
ACK

2

frame
ACK
retransmitted
frame
frame
How will receiver know

that this is NOT
a new packet?!

without packet numbering

2

frame 0
ACK
frame 1
ACK
frame 2
ACK
frame 2

Receiver has already
received frame 2 –
it resends an ACK and
discards the duplicate.

with packet numbering

Delayed Acknowledgment
(Premature Timeout)

•

8

ACKs can be delayed due to problems with
links or network congestion
time-out expires early, sender resends frame
when delayed ACK arrives, sender assumes
that given ACK is for the last frame sent

•

0

receiver sees this as
ACK for 2nd frame-0
and sends frame-1
receiver sees this as
ACK for frame-1
and sends frame-2

ACKs must be numbered to prevent gaps
in delivered packet sequence

0

frame 0
ACK

ACK 1

0

2

0

0

frame 0
1

frame 0

frame 0

ACK

1

frame-1 not
delivered !!!

without ACK numbering

ACK 1

0

cannot send
frame-2

with ACK numbering

How large should the packet / ACK sequence be? Only 1-bit long !!!


http://www.net-seal.net/animations.php?aid=37

9

Stop-and-Wait
Efficiency

•

10

t0 = basic Stop-and-Wait delay – from time when frame
is transmitted into channel until time when ACK arrives
back to receiver, and another frame can be sent

t 0 = 2 ⋅ t prop + 2 ⋅ t proc + t frame + t ACK = 2 ⋅ t prop + 2 ⋅ t proc +

nf n ACK
+
R
R

Reff = effective transmission (data) rate:

•

R eff =

number of info bits delivered to destination nf − nheader
=
total time required to deliver info bits
t0

t0 = total time to transmit 1 frame
A

tproc

nf bits
nACK bits

B

tprop

first frame bit
arrives at
receiver

frame
tf time

tproc

end of frame
is received

tack

tprop

first ACK bit
arrives at
receiver

end of ACK
is received

•

11

ηSW = transmission efficiency: ratio of actual and effective
efficiency
transmission (data) rate - ideally, ηSW ≈ 1
where do we lose channel efficiency, and how can ηSW → 1
be achieved ?!

ηSW

nf − nheader
R
t0
=
= eff =
R
R

(1)

nheader
nf

(2)

n ACK
nf

(3)

nheader
nf
2(t + t )R
+ prop proc
nf

1−
1+

n ACK
nf

should be
as small
as possible

- loss in efficiency due to (need for) header
- loss in efficiency due to (need for) ACKs

2(t prop + t proc )R - bandwidth-delay product
max number of bits in transit at any given time
in Stop-and-Wait ARQ delay-bandwidth product is a measure
of lost opportunity in terms of transmitted bits

Bandwidth-delay product = 2*(tprop + tproc)*R =
= capacity of the transmission pipe from the sender to the receiver and back.

12


frame size in comparison
to bandwidth-delay product

tprop + tproc > tframe

max number of bits in transit –
‘the pipe is full’

tprop + tproc < tframe

Stop-and-Wait ARQ becomes inadequate when data is fragmented
into small frames, such that nf / R = tframe is small relative to tprop .

13


14

Example [ impact of delay-bandwidth product ]
nf = 1250 bytes = 10000 bits
nACK = nheader = 25 bytes = 200 bits

ηSW =

R eff
=
R

⇒

n ACK nheader
=
= 0.02
nf
nf

nheader
0.98
nf
=
2 ⋅ (t prop + t proc )R
2 ⋅ (t prop + t proc )R
+
1.02 +
nf
nf

1−
1+

n ACK
nf

Efficiency

200 km
(tprop = 1 ms)

2000 km
(tprop = 10 ms)

20000 km
(tprop = 100 ms)

200000 km
(tprop = 1 sec)

1 Mbps

103
88%

104
49%

105
9%

106
1%

1 Gbps

106
1%

107
0.1%

108
0.01%

109
0.001%

Stop-and-Wait does NOT work well for very high speeds or long propagation delays.

Stop-and-Wait
Efficiency in
Channel with
Errors

•

Pf = probability that transmitted frame has errors and
need to be retransmitted
(1-Pf) – probability of successful transmission

1
1- Pf

R eff_error
R

and including

– average # of (re)transmission until first correct arrival

total delay per frame:

ηSW_error =

15

n f − nheader
t0
(1- Pf )
= (1- Pf ) ⋅
=
R

t 0 ⋅ (average # of retrans.) = t 0 ⋅

nheader
nf
2(t + t )R
+ prop proc
nf

1−
1+

n ACK
nf

ηSW_error = (1- Pf ) ⋅η0

Pf increases ⇒ ηSW decreases

1
1- Pf

(∗)

successful
transmission

average # of transmissions in error
before a successful transmission
time-out

time-out

16

time-out

t0

Probability that i transmission are needed to deliver frame successfully
( i-1 transmission in error and the ith transmission is error free ):

P[ # of trans. in error = i-1 ] = (1-Pf) Pfi-1
∞

∞

i=1

i=1

E[# of transmissi ons in error] = ∑ (i - 1) ⋅ P[n trans in error = i - 1] = ∑ (i - 1) ⋅ (1− Pf )Pfi−1 =
∞

= (1− Pf ) ⋅ ∑ (i - 1) ⋅ P

i−1
f

i=1

∞

∞

= (1− Pf ) ⋅ ∑ n ⋅ Pfn =
n =1

= (1− Pf ) ⋅ Pf ⋅ ∑ n ⋅ Pfn−1 = (1− Pf ) ⋅ Pf ⋅
n =1

=
Total average
delay per frame:

1
=
(1− Pf )2

Pf
1− Pf

t 0 + time - out ⋅ E[# of transmiss in error] = t 0 + time - out ⋅

Pf
1
≈
t0
1- Pf 1- Pf

Piggybacking

17

•

Stop-and-Wait discussed so far was ‘unidirectional’

•

in ‘bidirectional’ communications, both parties send &
acknowledge data, i.e. both parties implement flow control

•

piggybacking method: outstanding ACKs are placed in
the header of information frames
piggybacking can save bandwidth since the overhead from
a data frame and an ACK frame (addresses, CRC, etc) can
be combined into just one frame

•

S(0)

S(0)
0

S(1)

R(0)

R(0)
S(1)

0
1
S(2)

R(1)

R(1)
S(2)

1

without piggybacking

with piggybacking

Go-Back-N ARQ

19

Go-Back-N ARQ – overcomes inefficiency of Stop-and-Wait ARQ –

sender continues sending enough frames to keep
channel busy while waiting for ACKs

•

a window of Ws outstanding frames is allowed

•

m-bit sequence numbers are used for both - frames
and ACKs, and Ws = 2m-1
4 frames are outstanding; so go back 4

Go-Back-4:
A

fr
0

fr
1

fr
2

fr
3

fr
4

fr
5

fr
6

fr
3

fr
4

fr
5

fr
6

fr
7

fr
8

Time

fr
9

B
A
C
K
1

Rnext

0

A
C
K
2

A
C
K
3

out of sequence
frames

1

2

3

3

A
C
K
4

4

A
C
K
5

5

A
C
K
6

6

A
C
K
7

7

A
C
K
8

8

A
C
K
9

9

Assume: Ws= 4
1) sender sends frames one by one
2) frame 3 undergoes transmission error – receiver ignores frame 3 and all subsequent frames
3) sender eventually reaches max number of outstanding frames, and takes following action:
go back N=Ws frames and retransmit all frames from 3 onwards

Go-Back-N ARQ (cont.)
Sender Sliding Window

•

20

all frames are stored in a buffer, outstanding
frames are enclosed in a window
frames to the left of the window are already
ACKed and can be purged

before ACKs for frames 0 and 1 arrive

frames to the right of the window cannot be sent
until the window slides over them
whenever a new ACK arrives, the window slides
to include new unsent frames

after ACKs for frames 0 and 1 arrive
and window slides

once the window gets full (max # of outstanding
frames is reached), entire window gets resent

Receiver Sliding Window
•

the size of receiver window is always 1
receiver is always looking for a specific frame
to arrive in a specific order
any frame arriving out of order is discarded
and needs to be resent

The complexity of the receiver in Go-Back-N is the same as that of Stop-and-Wait!!!
Only the complexity of the transmitter increases.

Problems with
Go-Back-N
(Go-Back-N
with Timeout)

•

21

Go-Back-N works correctly (retransmission of
damaged frames gets triggered) as long as the
sender has an unlimited supply of packets that
need to be transmitted
but, in case when packets arrive sporadically, there
may not be Ws-1 subsequent transmissions ⇒ window
will not be exhausted, retransmissions will not be
triggered
this problem can be resolved by modifying Go-Back-N
such that:
1) set a timer for each sent frame
2) resend all outstanding frames either when window

gets full or when the timer of first frame expires

Window size
5

6

7

0

1

2

3

4

5

6

7

0

1

2

3

Example [ lost frame in Go-Back-N with time-out ]

Note:
• ACKs number always defines the number of the next expected frame !!!
• in Go-Back-N, receiver does not have to acknowledge each frame received –
it can send one cumulative ACK for several frames

22

Sequence Numbers
and Window Size

•

23

m bits allotted within a header for seq. numbers
⇒ 2m possible sequence numbers
how big should the sender window be!?
W > 2m cannot be accepted – multiple frames with
same seq. # in the window ⇒ ambiguous ACKs
W = 2m can still cause some ambiguity – see below
W = 2m – 1 acceptable !!!

ACK1
ACK1
ACK2
ACK2
ACK3
ACK3
ACK0
ACK3
ACK1

window size 2m = 4

window size 2m-1 = 3

Go-Back-N
Efficiency

•

24

completely efficient if Ws is large enough to keep channel busy,
and if channel is error free
time

time

•

in case of error-prone channel, with Pf frame loss probability,
time to deliver a frame is:
t frame

average # of
frame/window
(re)transmission
until a successful
transmission

t frame +

- if 1st transmission succeeds – prob. (1-Pf)

1
⋅ Ws ⋅ t frame - if 1st transmission does NOT succeeds –
1 − Pf
prob. Pf

• total average time required to transmit a frame:

⎞
⎛
P
1
t GBN = (1 − Pf ) ⋅ t frame + Pf ⋅ ⎜ t frame +
⋅ Ws ⎟ = t frame + f ⋅ Ws ⋅ t frame
⎟
⎜
1 − Pf
1 − Pf
⎠
⎝
• transmission efficiency

ηGBN

n f − nheader
n
1− header
t GBN
nf
=
=
(1− Pf )
R
1+ (Ws − 1)Pf

(∗∗)


25

What is total average time required to transmit a frame, assuming Pf?
successful
transmission

successful
transmission
WS

0 1 2 3 4 5 6 7 8 9 1011

tframe

WS

0 1 2 3 4 5 6 7 R R R R R R R R 8 9 10 11

tframe

ACK keeps window ‘sliding’

1st attempt successful:

t GBN = t frame

2nd attempt successful:

t GBN = t frame + WS ⋅ t frame

average case:

t GBN = t frame + E[# of transmissions in error] ⋅ WS ⋅ t frame
E[# of transmissions in error] =

t GBN = t frame +

Pf
WS ⋅ t frame
1 − Pf

⇒

ηGBN

Pf
1 − Pf

n f − nheader
n
1− header
t GBN
nf
=
=
(1− Pf )
R
1+ (Ws − 1)Pf


26

Example [ Stop-and-Wait vs. Go-Back-N ]
nf = 1250 bytes = 10000 bits
nACK = nheader = 25 bytes = 200 bits
Compare S&W with GBN efficiency for random bit errors with pb = 0, 10-6, 10-5, 10-4 and
bandwidth-delay product R*2*(tprop+tproc) = 1 Mbps * 100 ms = 100000 bits = 10 frames →
use Ws = 11.
11

Efficiency

pb=10-5

pb=10-4

8.9%

8.8%

8.0%

3.3%

GBN

•

pb=10-6

S&W

•

pb=0
98%

88.2%

45.4%

4.9%

Go-Back-N provides significant improvement over Stop-and-Wait for large delaybandwidth product
Go-Back-N becomes inefficient as error rate increases

Selective Repeat ARQ
Selective Repeat ARQ

•

28

Go-Back-N is NOT suitable for ‘noisy links’ – in
case of a lost/damaged frame a whole window
of frames need to be resent
excessive retransmissions use up the bandwidth
and slow down transmission

•

Selective Repeat ARQ overcomes the limitations
of Go-Back-N by adding 2 new features
(1) receiver window > 1 frame, so that out-of-order

but error-free frames can be accepted
(2) retransmission mechanism is modified – only

individual frames are retransmitted
•

Window size

Window size

already ACKed
sent, not yet ACKed

Selective Repeat ARQ is used in TCP !!!

usable not yet sent
not usable

sender window of size WS

out of order buffered
but already ACKed
expected, not yet received

acceptable
(within window)
not usable

receiver window of size WR

Selective Repeat ARQ (cont.)

29

Selective Repeat ARQ Operation
Receiver:
• window advances whenever next

in-order frame arrives
Rnext

• out-of-order frames are accepted only

Rnext + WS -1 if their sequence numbers satisfy

Rnext < Rframe < Rnext + Ws
• a negative ACK (NAK) with sequence

number Rnext is sent whenever an
out-of-sequence frame is observed

Sender:
• window advances whenever an ACK

arrives
• if a timer expires, the corresponding

frame is resent, and the timer is reset
• whenever a NAK arrives, Rnext frame

is resent

Window Sizes
(WS and WR)

•

30

m bits allotted within a header for sequence numbers
⇒ 2m possible sequence numbers
how big should the windows be!?
WS and WR = 2m-1 cannot be accepted due to possible
ambiguity as shown below
W = 2m/2 = 2m-1 acceptable !!!

ACK1
ACK1
ACK2
ACK2
ACK3



Selective Repeat
Efficiency

•

31

completely efficient if Ws is large enough to keep
channel busy, and if channel is error free
of course, sequence number space must be 2X sequence
sequence number space of Go-Back-N

•

in case of error-prone channel, total average time
required to transmit a frame:
t SR =

•

t frame
nf
=
1− Pf R ⋅ (1− Pf )

transmission efficiency

ηSR

n f − nheader
⎛ n
R
t SR
= ⎜1− header
= eff =
⎜
R
R
nf
⎝

⎞
⎟ ⋅ (1− Pf )
⎟
⎠

(∗∗∗)


32

What is total average time required to transmit a frame, assuming Pf?
successful
transmission

successful
transmission

on NAK or time-out
0 1 2 3 4 5 6 7 8 9 1011

0 1 2 3 4 5 R 6 7 8 9 101112

tframe
NAK 0

1st attempt successful:

t SR = t frame

2nd attempt successful:

t SR = t frame + t frame

average case:

t SR = t frame + E[# of transmissions in error] ⋅ t frame
Pf
1 − Pf

t SR = t frame +

Pf
1 nf
⋅ t frame =
⋅
1− Pf
1− Pf R

⇒

ηSR

nf − nheader
⎛ n
t SR
= ⎜1− header
=
⎜
R
nf
⎝

⎞
⎟(1− Pf )
⎟
⎠

Stop-and-Wait vs. Go-Back-N vs. Selective Repeat
Performance
Comparison

•

33

assume nACK and nheader are negligible relative to nf, and
2(t prop + t proc )R
nf

= L = Ws − 1

WS is for 1 less than the
number of frames currently in transit

size of the “pipe” in
multiples of frames

•

efficiencies of three ARQ techniques are

ηSW =

1
⋅ (1- Pf )
1+ L

ηGBN =

1
(1− Pf )
1+ LPf

ηSW < ηGBN < ηSR

ηSR = (1− Pf )

•

for 0 < Pf < 1, Selective Repeat provides best performance

•

for Pf → 0 Go-Back-N as good as Selective Repeat

34

ARQ Efficiency Com parison
Selective
Repeat

Efficiency

1.5

Go Back N 10

1

Stop and Wait
100

0.5
0

Go Back N 100
-9 -8 -7 -6 -5 -4 -3 10 10
10-9 10-8 10-7 10-6 10-5 10-4 10-3 -2-2 -1-1
p
- LOG(p)

Delay-Bandwidth product = 10, 100

Stop and Wait
10

Cse3213 11 flow_errorcontrol_f2010

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