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Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
PART I: Peer-to-Peer Protocols
5.1 Peer-to-Peer Protocols and Service Models
5.2 ARQ Protocols and Reliable Data Transfer
5.3 Other peer-to-peer Protocols:
Flow Control
Timing Recovery
TCP Reliable Stream Service and Flow Control
1
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
PART II: Data Link Controls
5.4 Framing
5.5 Point-to-Point Protocol
5.6 High-Level Data Link Control
5.7 Link Sharing Using Statistical Multiplexing
2
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
5.1 Peer-to-Peer Protocols and
Service Models
3


Peer-to-Peer Protocols
n + 1 peer process
SDU
PDU

Layer-(n+1) peer calls
layer-n and passes
Service Data Units
(SDUs) for transfer

Layer-n peers exchange
Protocol Data Units
(PDUs) to effect transfer

Layer-n delivers SDUs to
destination layer-(n+1)
peer
4
n peer process
n – 1 peer process


n – 1 peer process
Layer-n peer processes
execute protocol to
provide service to layer(n+1)
n + 1 peer process
SDU
n peer process

Service Models


The service model specifies the information transfer
service layer-n provided to layer-(n+1)
Two categories of service models:



Connection-oriented services
Connectionless services
A service offered by a given layer can include some
of the following features:





Arbitrary message size or structure
Sequencing and Reliability
Timing, Pacing, and Flow control
Multiplexing
Privacy, integrity, and authentication
5
Connection-Oriented Transfer
Service

Connection Establishment phase



Data transfer phase



Connection must be established between layer-(n+1) peers
Layer-n protocol must: Set initial parameters, e.g. sequence
numbers; and Allocate resources, e.g. buffers
Exchange of SDUs
Disconnection phase
Example: TCP, PPP
n + 1 peer process
send
SDU
n + 1 peer process
receive
Layer n connection-oriented service
SDU
6
Connectionless Transfer Service





No Connection setup is required, simply send SDU
Each block of information is transmitted independently
All address information per block must be provided
Acknowledge is not required (Simple & quick) or required
Example: UDP, IP
n + 1 peer process
send
SDU
n + 1 peer process
receive
Layer n connectionless service
7
Message Size and Structure

What message size and structure will a
service layer accept?




Different services impose restrictions on size &
structure of data it will transfer
Single bit? Block of bytes? Byte stream?
Ex: Transfer of voice mail = 1 long message
Ex: Transfer of voice call = byte stream
1 voice mail= 1 message = entire sequence of speech samples
(a)
1 call = sequence of 1-byte messages
8
(b)
Segmentation & Blocking



To accommodate arbitrary message size, a layer may
have to deal with messages that are too long or too
short for its protocol
Segmentation & Reassembly: a layer breaks long
messages into smaller blocks and reassembles these
at the destination
Blocking & Unblocking: a layer combines small
messages into bigger blocks prior to transfer
1 long message
2 or more short messages
9
2 or more blocks
1 block
Reliability & Sequencing




Reliability: Are messages or information
stream delivered error-free and without loss
or duplication?
Sequencing: Are messages or information
stream delivered in order?
ARQ protocols combine error detection,
retransmission, and sequence numbering to
provide reliability & sequencing
Examples: TCP and HDLC
10
Pacing and Flow Control




Messages can be lost if receiving system does
not have sufficient buffering to store arriving
messages (buffer overflow)
If destination layer-(n+1) does not retrieve its
information fast enough, destination layer-n
buffers may overflow
Pacing & Flow Control provide backpressure
mechanisms that control transfer according to
availability of buffers at the destination
Examples: TCP and HDLC
11
Timing





Applications involving voice and video generate
units of information that are related temporally
Destination application must reconstruct temporal
relation in voice/video units
Network transfer introduces delay & jitter
Timing Recovery protocols use timestamps &
sequence numbering to control the delay & jitter in
delivered information
Examples: RTP & associated protocols in Voice
over IP
12
Multiplexing



Multiplexing enables multiple layer-(n+1)
users to share a layer-n service
A multiplexing tag is required to identify
specific users at the destination
Examples: UDP, IP
13
Privacy, Integrity, &
Authentication





Privacy: ensuring that information transferred
cannot be read by others
Integrity: ensuring that information is not
altered during transfer
Authentication: verifying that sender and/or
receiver are who they claim to be
Security protocols provide these services and
are discussed in Chapter 11
Examples: IPSec, SSL
14
End-to-End vs. Hop-by-Hop

A service feature can be provided by implementing a
protocol



end-to-end across the network
across every hop in the network
Example:


Perform error control at every hop in the network or only
between the source and destination?
Perform flow control between every hop in the network or
only between source & destination?
15
Error Control in Data Link Layer
Packets
Packets
Data link
layer
Data link
layer
(a)
A
Frames
B
Physical
layer
Physical
layer


(b)

12
3
21
12
3
B
2
1
Medium
A
1
Physical layer entity
2
Data link layer entity
3
Network layer entity
21
Fig. 5.5

Data Link operates
over wire-like,
directly-connected
systems
Frames can be
corrupted or lost, but
arrive in order
Data link performs
error-checking &
retransmission
Ensures error-free
packet transfer
between two systems
16
Error Control in Transport Layer


Transport layer protocol (e.g. TCP) sends segments across
network and performs end-to-end error checking &
retransmission
Usually, the underlying network is assumed to be unreliable
Messages
Messages
Segments
Transport
layer
Transport
layer
Network
layer
Network
layer
Network
layer
Network
layer
Data link
layer
Data link
layer
Data link
layer
Data link
layer
layer
Physical
layer
Physical
layer
Physical
layer
End system
Physical
α
Fig. 5.6
End system
β
17
Network


Segments can experience long delays, can be lost, or
arrive out-of-order because packets can follow different
paths across network
The peer processes at the transport layer need to take
into account all the characteristics of the network
transfer service to be able to provide desired service to
its higher layer
12
C
3
21
End System
α
4 3 21
End System
β
12
3
21
Medium
A
12
3
B
2
1
21
123 4
Network
3
Network layer entity18
4
Transport layer entity
End-to-End versus hop-by-hop
Approaches
Hop-by-hop
In situations where
errors are likely, hopby-hop error
becomes necessary.
Data
1
Data
2
ACK/
NAK
Data
3
Data
4
ACK/
NAK
Faster recovery
5
ACK/
NAK
ACK/
NAK
End-to-end
ACK/NAK
1
2
Data
3
Data
5
4
Data
Data
In situations where
errors are
infrequent, end-toend mechanisms
are preferred.
19
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
5.2 ARQ Protocols and
Reliable Data Transfer
20
Automatic Repeat Request (ARQ)



Suppose that the layer n+1 process requires a reliable
data transfer service, and then layer n process would
provide an ARQ protocol service
ARQ combines with error detection code and
retransmission to ensure a sequence of information
packets being delivered in order and without errors or
duplications despite transmission errors & losses
Three basic types of ARQ protocols are investigated:




Stop-and-Wait ARQ
Go-Back N ARQ
Selective Repeat ARQ
Basic elements of ARQ:



ACKs (positive acknowledgments),
NAKs (negative acknowlegments)
Timeout Mechanism
21
5.2.1 Stop-and-Wait ARQ
Transmit a frame, wait for ACK
Packet sequence
Layer n+1,
PDU
Layer n
A timer is set
after a frame is
transmitted
Error-free
Packet
sequence
Information frame,
PDU
Receiver
(Process B)
Transmitter
(Process A)
Suppose that
the layer n+1
requires a
reliable data
transfer service
Control frame
CRC
Information Header
packet
Information frame
Header
CRC
Control frame: ACKs
22
Need for Sequence Numbers
(a) Frame 1 lost
A
B
Time-out
Time
Frame
0
ACK
(b) ACK lost
A
B





Frame
1
Frame
1
ACK
Frame
2
Time-out
Time
Frame
0
ACK
Frame
1
ACK
Frame
1
ACK
Frame
2
In cases (a) & (b) the transmitting station A acts the same way
But in case (b) the receiving station B accepts frame 1 twice
Question: How does the receiver know the second frame is also frame 1?
Answer: Add frame sequence number in header
Slast is sequence number of most recent transmitted frame
23
Sequence Numbers
(c) Premature Time-out
Time-out
A
Time
Frame
0
ACK
B





Frame
0 ACK
Frame
1
Frame
2
The transmitting station A misinterprets duplicate ACKs
Incorrectly assumes second ACK acknowledges Frame 1
Question: How does the receiver know second ACK is for frame 0?
Answer: Add frame sequence number in ACK header
Rnext is the sequence number of next frame expected by the
receiver to implicitly acknowledge the receipt of all prior frames
24
1-Bit Sequence Numbering
Suffices
0
1 0
1 0
1 0
1
0
1 0
1 0
1 0
1
Rnext
Slast
Timer
Slast
Transmitter
A
Receiver
B
Rnext
Global State:
(Slast, Rnext)
Fig. 5.12 System State
information in
Stop-and Wait ARQ
(0,0)
Error-free frame 0
arrives at receiver
ACK for
frame 1
arrives at
transmitter
(1,0)
Error-free frame 1
arrives at receiver
(0,1)
ACK for
frame 0
arrives at
transmitter
(1,1)
25
Stop-and-Wait ARQ
Transmitter
Ready state
 Await request from higher layer for
packet transfer
 When request arrives, transmit
frame with updated Slast and CRC
 Go to Wait State
Wait state
 Wait for ACK or timer to expire;
block requests from higher layer
 If timeout expires
 retransmit frame and reset
timer
 If ACK received:
 If sequence number is
incorrect or if errors detected:
ignore ACK
 If sequence number is correct
(Rnext = Slast +1): accept frame,
go to Ready state
Receiver
Always in Ready State
 Wait for arrival of new frame
 When frame arrives, check for
errors
 If no errors detected and
sequence number is correct
(Slast=Rnext), then
 accept frame,
 update Rnext,
 send ACK frame with Rnext,
 deliver packet to higher layer
 If no errors detected and wrong
sequence number
 discard frame
 send ACK frame with Rnext
 If errors detected
 discard frame
26
Stop-and-Wait ARQ Example
Slast=0 Time-out
A
Fr 0
ACK 1
B
Slast=0
Rnext: 01
Slast=1
Rnext:1
Slast:01
Rnext=1 Time-out
Ignore ACK 1
Fr 1
Fr 0
ACK 1
Fr 1
Slast=0 Error
Rnext=1 Frame
Discard Discard
Frame 0 Frame 1
Transmitter
Wait state
 If timeout expires
 retransmit frame and reset
timer
 If ACK received:
 If sequence number is
incorrect or if errors detected:
ignore ACK
Receiver
Always in Ready State
 If no errors detected and wrong
sequence number
 discard frame
 send ACK frame with Rnext
 If errors detected
27
 discard frame
Applications of
Stop-and-Wait ARQ



IBM Binary Synchronous
Communications protocol (Bisync):
character-oriented data link control
Xmodem: modem file transfer protocol
Trivial File Transfer Protocol (RFC 1350):
simple protocol for file transfer over UDP
28
Performance Issues
First frame bit
enters channel
Last frame bit
enters channel
Channel idle while
transmitter waits for ACK
ACK
arrives
t
A
B
First frame bit
arrives at
receiver
t
Last frame bit
arrives at
receiver
Receiver processes
frame and
prepares ACK
Fig. 5.13 Stop-and-Wait ARQ is inefficient when the time to receive an ACK
Is large compared to the frame transmission time



1000 bit frame @ 1.5 Mbps
If the time waiting for ACK = 40 ms, then efficiency = 1/60= 1.6%
If the time waiting for ACK = 1 ms, then efficiency =2/3 = 66.67%
29
Delay Components in
Stop-and-Wait ARQ
t0 = total time to transmit 1 frame
A
tproc
B
tprop
frame
tf time
tproc
tprop
tack
t0 = 2t prop + 2t proc + t f + t ack
= 2t prop + 2t proc
nf
bits/info frame
na
+
+
R
R
delay components
bits/ACK frame
channel transmission rate
30
Transmission Efficiency of Stopand-Wait ARQ in Error-free channel
Effective Information Transmission rate:
R
0
eff
bits for header & CRC
number of information bits delivered to destination n f - no
=
=
,
total time required to deliver the information bits
t0
Transmission efficiency η0:
n f - no
no
1Reff
nf
t0
h0 =
=
=
.
na 2(t prop + t proc ) R
R
R
1+
+
nf
nf
Effect of
ACK frame
Effect of
frame overhead
Effect of
Delay-Bandwidth Product
31
Example:
Effect of Delay-Bandwidth Product
nf=1250 bytes = 10,000 bits, na=no=25 bytes = 200 bits
Delay Bandwidth
Product
Efficiency
R=1 Mbps
R=1 Gbps
Reaction
time =
1 ms
10 ms
100 ms
1 sec
103
88%
106
1%
104
49%
107
0.1%
105
9%
108
0.01%
106
1%
109
0.001%
Stop-and-Wait does not work well for very high speeds
or long propagation delays
32
Transmission Efficiency of Stop-andWait ARQ in Channel with Errors




Let 1 – Pf = probability frame arrives without errors
Avg. number of transmissions for correct reception is 1/ (1–Pf )
If 1 – Pf = 0.1, then on average a frame will have to be
transmitted 10 times to get through
The Stop-and-Wait ARQ on average requires tsw = t0/(1 – Pf)
seconds to get a frame through
h SW =
Reff
R
=
n f - no
t0
1 - Pf
R
1=
na
1+
+
nf
no
nf
2(t prop + t proc ) R
nf
(1 - Pf )
Effect of
frame loss
33
Example: Effect of Bit Error Rate
nf=1250 bytes = 10,000 bits, na=no=25 bytes = 200 bits
Find efficiency for random bit errors with p= 0, 10-6, 10-5, 10-4
nf
1 - Pf = (1 - p) » e
Pf
-n f p
for large n f and small p
p=0
p=10-6
p=10-5
p=10-4
1
88%
0.99
86.6%
0.905
79.2%
0.368
32.2%
Efficiency
R = 1 Mbps
Reaction Time = 1 ms
Bit errors impact performance as nf p approach 1
34
Midterm Exam and Final Exam

Midterm:



Time: December 5 (Thursday) or 6, 2013
Scope: Chapters 1 – 3 and Chapter 5; all contents
given in the class before the beginning of
December
Final Exam:


Time: January 9 (Thursday) or 10, 2014
Scope: Chapter 5 – 6 and Some in Chapter 4 and
Chapter 7; all after the end of November
35
5.2.2 Go-Back-N ARQ




Improve Stop-and-Wait ARQ by allowing the transmitter to
continue sending frames to keep channel busy while the
transmitter waits for acknowledgements
In Go-Back-N ARQ pipelines, a window size Ws, the
maximum number of frames that can be outstanding without
acknowledgement, is chosen larger than the delay-bandwidth
product to ensure that the channel can be kept busy
m
Use m-bit sequence numbering from 0 to 2 -1 maximum
When will the ARQ protocol go back N? and N = ?


If ACK for oldest frame arrives before window is exhausted,
frames can continue be transmitted
If window is exhausted and the timer is out, pull back and
retransmit all outstanding frames
36
Go-Back-N ARQ
4 frames are outstanding; so go back 4
Go-Back-4:
fr
0
A
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
Pipelined
Rnext



0
A
C
K
1
01
A
C
K
2
A
C
K
3
2
3
The receiver ignores frame
3 and all subsequent
frames because of out of
sequence frames
3
A
C
K
4
3 4
A
C
K
5
5
A
C
K
6
6
A
C
K
7
A
C
K
8
7
8
A
C
K
9
Pipelined
9
Frame transmission are pipelined to keep the channel busy
Frame with errors and subsequent out-of-sequence frames are ignored
Transmitter is forced to go back when window of 4 is exhausted
37
Window size long enough to cover
round trip time
Stop-and-Wait ARQ
fr
0
A
B
B
fr
1
Time
A
C
K
1
Receiver is
looking for
Rnext=0
Go-Back-N ARQ
A
Time-out expires
fr
0
Four frames are outstanding; so go back 4
fr
0
fr fr
1 2
fr
3
fr
0
fr fr
1 2
Receiver is
Out-oflooking for
sequence
Rnext=0
fr
3
A
C
K
1
A
C
K
2
fr fr
4 5
A
C
K
3
A
C
K
4
fr
6
A
C
K
5
Time
A
C
K
6
frames
Fig. 5.16 Relationship of Stop-and-Wait ARQ and Go-Back-N ARQ
38
Go-Back-N with Timeout

Problem with Go-Back-N as presented:


If frame is lost and source does not have frame to
send, then window will not be exhausted and
recovery will not commence
Use a timeout mechanism with each frame

When timeout expires, resend all outstanding
frames
39
Go-Back-N Transmitter & Receiver
Transmitter
Send Window
...
Frames
transmitted S
last
and ACKed
Srecent
Slast
Timer
Slast+1
oldest unACKed frame
...
Timer
Srecent
most recent
transmission
...
Slast+Ws-1
Receive Window
Slast+Ws-1
Buffers
Timer
Receiver
Max Sequence
Number allowed
Frames
received
Rnext
Receiver will only accept
a frame that is error-free and
that has sequence number = Rnext
When such a frame arrives, Rnext is
incremented by one, and the receive
window slides forward by one
40
Sliding Window Operation
Transmitter
Send Window
...
Frames
transmitted S
last
and ACKed
Srecent
Slast+Ws-1
m-bit Sequence Numbering
2m –
1
0
1
2
Transmitter waits for error-free
ACK frame with sequence
number Slast
When such ACK frame arrives,
Slast is incremented by one, and
the send window slides forward
by one
Slast
send
i
window
i+1
i + Ws – 1
Ws ≦ 2m– 1
*Note that the modular 2m arithmetic is used here
41
Go-Back-N ARQ Protocol
1.
2.
3.
4.
5.
6.
Notice that the modulo 2m arithmetic is used here.
The transmitter is in ready state or block state.
If the transmitter is in the ready state waiting for a
request from its higher layer. If there is request, it sends
the frame with sequence number Srecent and sets the
timer.
If Srecent= Slast+Ws-1, then the send window becomes
empty and the transmitter goes into the blocking state.
If an error-free ACK frame is received with Rnext in the
range between Srecent and Slast, the send window slides
forward by setting Slast = Rnext and the maximum window
size is Slast+Ws-1.
If the Rnext is outside the range from Srecent to Slast, the
42
frame is discarded and no further action is taken.
Go-Back-N ARQ Protocol
1.
2.
3.
4.
The transmitter is in block state, that is Srecent= Slast+Ws1, denoting the send window becomes empty, and the
transmitter refused to accept requests for packet transfer
from the upper layer.
If a timer expires, the transmitter resends the
corresponding packet and resets the timer.
If an error-free ACK frame is received with Rnext in the
range between Srecent and Slast, the send window slides
forward by setting Slast = Rnext and the maximum window
size is Slast+Ws-1, then changing to the ready state.
If the Rnext is outside the range from Srecent to Slast, the
frame is discarded and no further action is taken.
43
Go-Back-N ARQ Protocol
1.
2.
3.
4.
The receiver is always in the ready state waiting for a
notification of an arriving frame from the lower layer.
When a frame arrives, if no errors are detected and the
sequence number is Rnext, then the frame is accepted,
the sequence number is incremented to Rnext +1, an
ACK frame is transmitted with sequence number Rnext
+1, and the packet is delivered to its higher layer.
If the received frame is correct but with wrong sequence
number, the frame is discarded and an ACK is sent with
sequence number Rnext.
If an erroneous frame is received, the frame is discarded
and no further action is taken.
44
Maximum Allowable Window Size is Ws = 2m-1
M = 22 = 4, Go-Back - 4:
A
fr
0
A
C
K
1
B
Rnext
fr
2
fr
1
0
1
fr
3
A
C
K
2
2
M = 22 = 4, Go-Back-3:
A
fr
0
fr
0
A
C
K
3
3
fr
1
A
C
K
0
fr
2
fr
3
Time
Receiver has Rnext= 0, and it regards this frame 0 is a
new frame 0. But it is actually the old frame 0. It is a
mistake. Therefore, Ws < 2m and the maximum
allowable window size is Ws = 2m-1
0
Transmitter goes back 3
fr
0
fr
2
fr
1
Transmitter goes back 4 and retransmits frame 0 - 3
fr
1
fr
2
fr
3
Time
But notice that the
maximum sequence
number can be
from 0 to 2m-1
B
Rnext
0
A
C
K
1
A
C
K
2
A
C
K
3
A
C
K
3
1
2
3
3
Receiver has Rnext= 3 , so it
rejects the old frame 0, 1, 2,
altogether.
45
ACK Piggybacking in Bidirectional GBN
SArecent RA next
Transmitter
Receiver
Transmitter
SBrecent
RB
next
“A” Receive Window
RA next
“A” Send Window
...
SA last+WA s-1
SA last
Buffers
Timer
SA last
Timer
SA last+1
...
SArecent
...
Timer
Timer
SA
last
+WA
s-1
Receiver
“B” Receive Window
Note 1: Out-ofsequence error-free
frames discarded after
Rnext examined
Note 2: If no information
frames are scheduled, the
receiver can send the
ACK timer. When the ACK
timer expires a control
frame conveying ACK is
transmitted
RB next
“B” Send Window
...
SB last
SB last+WB s-1
Buffers
Timer
SB last
Timer
SBlast+1
...
SBrecent
...
Timer
Timer
SB last+WB s-1
46
Applications of Go-Back-N ARQ


HDLC (High-Level Data Link Control): bitoriented data link control
V.42 modem: error control over telephone
modem links
47
Required Timeout & Window Size
Tproc
Tout
Tprop
Tf
Tf
Tprop
Tout = 2 Tprop + 2 Tf +Tproc

Timeout value should allow for:




Two propagation times + 1 processing time: 2 Tprop + Tproc
A frame that begins transmission at the receiver right before our
frame arrives; one Tf (as the worst case !) must be counted
Next frame carrying the piggybacked ACK takes the second Tf
Ws should be large enough to keep channel busy for Tout
48
Required Window Size for
Delay-Bandwidth Product
Frame = 1250 bytes =10,000 bits, R = 1 Mbps
2(tprop + tproc)
2 x Delay x BW
Minimum
Window size
1 ms
1000 bits
1
10 ms
10,000 bits
2
100 ms
100,000 bits
11
1 second
1,000,000 bits
101
49
Efficiency of Go-Back-N



GBN has a best possible efficiency, if Ws large enough to keep
channel busy and the channel is error-free, namely, 1-n0/nf
If the channel is erroneous and assume that the frame loss
probability is Pf , then time to deliver a frame successfully is
tf + Pf Wstf /(1-Pf), where tf =nf/R
Notice that the dela-bandwidth product determines the Ws
tGBN = t f (1 - Pf ) + Pf {t f +
n f - no
hGBN =
tGBN
R
=
Ws t f
1 - Pf
no
1nf
1 + (Ws - 1) Pf
} = t f + Pf
Ws t f
1 - Pf
and
(1 - Pf )
50
Example: Effect of Bit Error Rate
and the Delay-Bandwidth Product
nf=1250 bytes = 10000 bits, na=no=25 bytes = 200 bits
Compare S&W with GBN efficiency for random bit errors with p = 0, 10-6,
10-5, 10-4 and R = 1 Mbps & reaction time 100 ms
Delay-Bandwidth Product is 1 Mbps x 100 ms = 100000 bits = 10 frames
→ Use Ws = 11
Efficiency
0
10-6
10-5
10-4
S&W
8.9%
8.8%
8.0%
3.3%
GBN
98%
88.2%
45.4%
4.9%


Go-Back-N ARQ significant improvement over Stop-andWait for large delay-bandwidth product
Go-Back-N becomes inefficient as error rate increases 51
5.2.3 Selective Repeat ARQ


Go-Back-N ARQ protocol is inefficient because
multiple frames are retransmitted when frame errors
or losses occur
Selective Repeat ARQ retransmits only each
erroneous or lost individual frame



Timeout causes individual corresponding frame to be resent
NAK causes retransmission of the oldest un-acked frame
Receiver maintains a receive window of sequence
numbers that can be accepted


Error-free, but out-of-sequence frames with sequence
numbers within the receive window are buffered
Arrival of frame with Rnext causes window to slide forward by
1 or more
52
Send Window for
Selective Repeat ARQ
Slast
Srecent
Ws
m
1. Notice that the modulo 2 arithmetic is used here.
2. The transmitter is in ready state or block state. If Srecent= Slast+Ws-1, then the send
window becomes empty and the transmitter goes into the blocking state.
3. If an error-free ACK frame is received with Rnext in the range between Srecent and Slast,
the send window slides forward by setting Slast = Rnext and the maximum window size
is Slast+Ws-1.
4. If an error-free NAK frame is received with Rnext in the range between Srecent and Slast,
the frame with sequence number Rnext is retransmitted and the send window slides
forward by setting Slast = Rnext and the maximum window size is Slast+Ws-1.
5. If the Rnext is outside the range from Srecent to Slast, the frame is discarded and no
further action is taken.
Send Window for
Selective Repeat ARQ
1.
2.
3.
4.
5.
The transmitter is in block state, that is Srecent= Slast+Ws-1, denoting
the send window becomes empty, and the transmitter refused to
accept requests for packet transfer from the upper layer.
If a timer expires, the transmitter resends the corresponding packet
and resets the timer.
If an error-free ACK frame is received with Rnext in the range
between Srecent and Slast, the send window slides forward by setting
Slast = Rnext and the maximum window size is Slast+Ws-1, then
changing to the ready state.
If an error-free NAK frame is received with Rnext in the range
between Srecent and Slast, the frame with sequence number Rnext is
retransmitted and the send window slides forward by setting Slast =
Rnext and the maximum window size is Slast+Ws-1, then changing to
the ready state.
If the Rnext is outside the range from Srecent to Slast, the frame is
discarded and no further action is taken.
54
Receive Window for
Selective Repeat ARQ
Rnext
WR
1. The receiver is always in the ready state waiting for a notification of an arriving frame
from the lower layer.
2. If the received frame is correct and with sequence number Rnext, and if Rnext+1 up to
Rnext+k-1 have already been received and buffered, then the receive sequence
number is incremented to Rnext+k, the receive window slides forward, an ACK with
Rnext is sent, and the corresponding packets are delivered to higher layer.
3. When a frame arrives, if no errors are detected and the sequence number is not
Rnext but in the range from Rnext to Rnext+WR-1, then the frame is accepted and
buffered, and a NAK or ACK frame with sequence number Rnext is transmitted.
4. If an arriving frame has no errors but the sequence number is outside the receive
window, then the frame is discarded and an ACK with sequence number Rnext is sent.
Selective Repeat ARQ
56
Selective Repeat ARQ
A
fr
0
fr
1
fr
2
fr
3
fr
4
fr
5
fr
6
fr
2
fr
7
A
C
K
2
A
C
K
2
fr
8
fr fr fr fr
9 10 11 12
Time
B
A
C
K
1
A
C
K
2
N
A
K
2
A
C
K
2
A
C
K
7
A
C
K
8
A
C
K
9
A
C
K
1
0
A
C
K
1
1
A
C
K
1
2
Using NAK or Timeout mechanism
57
Selective Repeat ARQ Example
Selective Repeat ARQ
Receiver
Transmitter
Send Window
...
Frames
transmitted S
last
and ACKed
Timer
Timer
Srecent
Slast+ Ws -1
Receive Window
Frames
received Rnext
Buffers
Slast
Buffers
Rnext+ 1
Slast+ 1
Rnext+ 2
Rnext + Wr -1
...
Timer
Srecent
...
Slast+ Ws - 1
...
Rnext+Wr -1
The maximum
sequence number
that can be
accepted 59
The Sliding Send & Receive Windows
Transmitter
2m-1
0
Receiver
1
2m-1
0
1
2
Slast
send
i
window
i+1
i + Ws –
1
Moves k forward when ACK
arrives with Rnext = Slast + k
k = 1, …, Ws-1
2
Rnext
receive
window
j
i
j + Wr – 1
Moves forward by 1 or more
when frame arrives with
sequence number = Rnext
60
What size Ws and Wr are allowed?

Send
Window
Example: m=2, M=2m=4, Ws=2m-1=3, Wr=3
Timeout and
Frame 0 resent
{0,1,2} {1,2}
A
B
Receive
Window
fr0
{2}
fr1
{.}
fr2
ACK1
{0,1,2} {1,2,3}
fr0
ACK2
Time
ACK3
{2,3,0}
{3,0,1}
Old frame 0 accepted as a
new frame because it falls
in the receive window.
It would be an error.
61
Ws = Wr = 2m-1 is the maximum allowed
Example: m=2, M=22=4, Ws=2, Wr=2

Timeout and
Frame 0 resent
Send
Window
{0,1}
A
{1}
fr0
fr0
fr1
B
Receive
Window
{.}
ACK1
{0,1}
{1,2}
Time
ACK2
{2,3}
Old frame 0 rejected because it falls
outside the receive window
62
Applications of Selective Repeat
ARQ


TCP (Transmission Control Protocol):
transport layer protocol uses variation of
selective repeat to provide reliable stream
service
SSCOP (Service Specific Connection
Oriented Protocol): error control for signaling
messages in ATM networks
63
Efficiency of Selective Repeat

Assume Pf frame loss probability, then time required
to deliver a frame is tf /(1-Pf)
n f - no
h SR =
t f /(1 - Pf )
R
no
= (1 - )(1 - Pf )
nf
64
Example: Effect of Bit Error Rate
and Delay-Bandwidth Product
nf=1250 bytes = 10000 bits, na=no=25 bytes = 200 bits
Compare S&W, GBN & SR efficiency for random bit errors
with p=0, 10-6, 10-5, 10-4 and R= 1 Mbps & 100 ms

Efficiency
0
10-6
10-5
10-4
S&W
8.9%
8.8%
8.0%
3.3%
GBN
98%
88.2%
45.4%
4.9%
SR
98%
97%
89%
36%
Selective Repeat outperforms GBN and S&W, but its
efficiency drops significantly as the error rate is large.
65
Comparison of ARQ Efficiencies
Assume na and no are negligible relative to nf, and
L = 2(tprop+tproc)R/nf =(Ws-1) is the size of the pipe in multiples
of frames.
Selective-Repeat:
h SR
no
= (1 - Pf )(1 ) » (1 - Pf )
nf
Go-Back-N:
hGBN =
Stop-and-Wait:
h SW =
1 - Pf
1 + (WS - 1) Pf
=
1 - Pf
For Pf≈0, SR & GBN same
1 + LPf
For Pf→1, GBN & SW same
(1 - Pf )
1 - Pf
»
2(t prop + t proc ) R
na
1+ L
1+
+
nf
nf
66
ARQ Efficiencies
ARQ Efficiency Comparison
E fficie ncy
1.2
1
Selective Repeat
0.8
Go Back N 10
0.6
Stop and Wait 10
0.4
Go Back N 100
0.2
Stop and Wait 100
0
10
-9 -9 10
-8-8 10
-7-7 10
-6-6 10
-5-5 10
-4 -4 10
-3-3 10
-2 -2 10
-1 -1
- LOG(p)
p
Delay-Bandwidth product = 10, 100
67
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
5.3 Other Peer-to-Peer Protocol
68
Flow Control
buffer fill
Information frame
Transmitter
Receiver
Control frame


Receiver has limited buffering to store arriving
frames
Several situations cause buffer overflow



Mismatch between the sending rate and the rate at which
the high-layer user can retrieve data
Surges in frame arrivals
Flow control prevents buffer overflow by regulating
rate at which source is allowed to send information
69
X ON / X OFF
threshold
Information frame
Transmitter
Receiver
Transmit
X OFF
Transmit
Time
A
on
off
on
off
B
Time
2Tprop
 Use the threshold flow control method.
 When the buffer is occupied over the threshold, an OFF signal will
be activated.
The threshold would be larger than the 2 Tprop R bits which are the
still remaining free in the receive buffer.
70
Window Flow Control
tcycle
Return of permits
A
Time
B
Tcycle > Ws Tf

Sliding Window ARQ method with Ws equal to the receive buffer available



Transmitter can never send more than Ws frames
ACKs that slide window forward can be viewed as permits to transmit
more
Can also pace ACKs as shown above


Time
Return permits (ACKs) at the end of cycle to regulate the transmission
rate; Tcycle > Ws Tf
Problems using sliding window for both error and flow control



Choice of window size must take into account the delay-bandwidth
product of the channel and as well as the receive buffer size
Interplay between transmission rate and retransmissions
TCP separates error and flow control
71
5.3.2 Timing Recovery for
Synchronous Services
Network output
not periodic
Synchronous source
sends periodic
information blocks
Network



Applications that involve voice, audio, or video will generate a
synchronous information stream, meaning periodically.
Information carried by equally-spaced fixed-length packets
Network that performs multiplexing and switching will introduce random
delay



Packets experience variable transfer delay
Jitter (variation in interpacket arrival times) also introduced
Timing recovery re-establishes the synchronous nature of the stream
72
Playout Buffer
Provide playout buffer and Tmax to Perform timing Recovery
Packet Playout
Packet Arrivals
Playout
Buffer
Packet Arrivals
■Sequence numbers help order packets
Packet Playout
Tmax
 Delay first packet by maximum network delay, Tmax
 All other packets would arrive with less delay than Tmax
 Packets will play out periodically thereafter
73
Playout clock must
be synchronized to
transmitter clock
Arrival times
Time
Send
times
Playout
times
Tplayout
Time
Receiver too fast
buffer starvation
Receiver too slow;
buffer fills and
overflows
time
Receiver speed
Time just right
Many late
packets
74
Tplayout
time
Tplayout
time
Adaptive Timing Recovery
Buffer for information blocks
Timestamps inserted in
packet payloads to
indicate when information
was produced
Error signal to indicate the
difference between the transmitter
and the receiver clocks
+
t4
t3
t2
t1
Add
Smoothing
filter
Playout
command
Adjust
frequency
-
Recovered
clock
Timestamps
Counter
Fig. 5.29 Adaptive clock recovery for clock synchronization



Counter attempts to synchronize the transmitter clock
Frequency of counter (Receiver Clock) is adjusted according to arriving
timestamps
Jitter introduced by network causes fluctuations in buffer & in local clock
75
Synchronization to a Common
Clock (Network Clock using GPS)
Receiver
Transmitter
∆f
fs
∆f
Network
fr=fn-Δf
Δf=fn-fs=fn-(M/N)fn
fn/fs=N/M
fr
fn
Network clock

Clock recovery simple if a common clock is available to transmitter & receiver




E.g. SONET network clock; Global Positioning System (GPS)
Transmitter sends Δf between its frequency and network frequency
Receiver adjusts its local frequency by Δf
Packet delay jitter can be removed completely
76
Example: Real-Time Protocol


RTP (RFC 1889) designed to support realtime applications such as voice, audio, video
RTP provides means to carry:




Type of information source
Sequence numbers
Timestamps
Actual timing recovery must be done by
higher layer protocol

MPEG2 for video, MP3 for audio
77
5.3.3 TCP Reliable Stream Service
and Flow Control
Application Layer
writes bytes into send
buffer through socket
Application layer
TCP transfers byte stream
in order, without errors or Application Layer reads
duplications
bytes from receive buffer
through socket
Write 45 bytes
Write 15 bytes
Write 20 bytes
Read 40 bytes
Read 40 bytes
Transport layer
Segments
Transmitter
Receiver
Receive buffer
Send buffer
ACKs
78
TCP ARQ Method
• TCP delivers a connection-oriented service over best
effort (connectionless) service of IP
•
•
•
•
•
Packets can arrive with errors or be lost
Packets can arrive out-of-order
Packets can arrive after very long delays
Duplicate segments must be detected & discarded
Must avoid old segments from previous connections
• Sequence Numbers
• Very long sequence number (32 bits) to deal with long delays
• Initial sequence numbers negotiated during connection setup
(to deal with very old duplicates)
• Accept segments within a receive window
• TCP uses Selective Repeat ARQ
• Transfers byte stream without preserving boundaries
79
Transmitter
Receiver
Send Window
Receive Window
Slast + WA-1
...
Octets
transmitted Slast
& ACKed
...
Srecent
Rlast
Rlast + WR – 1
...
Slast + WS – 1
Slast = oldest unacknowledged byte
Srecent = highest-numbered
transmitted byte
Slast+WA-1 = highest-numbered byte
that can be transmitted
Slast+WS-1 = highest-numbered byte
that can be accepted from the
application
Rnext Rnew
Rlast = highest-numbered byte not yet
read by the application (has been
ACKed, but still in the receive buffer)
Rnext = next expected byte
Rnew = highest numbered byte
received correctly
Rlast+WR-1 = highest-numbered byte
that can be accommodated in receive
buffer
80
Flow Control



TCP separates the flow control function from the acknowledgement
function. The receiver controls the transmission rate from the sender
to prevent from buffer overflow
TCP receiver advertises a window size, WA, in the segment header,
specifying number of buffers that can be available at the receiver
WA = WR – (Rnew – Rlast)
TCP sender is obliged to keep the number of outstanding bytes below
WA, that is,
(Srecent - Slast) ≤ WA
Send Window
Receive Window
Slast + WA-1
...
...
Slast Srecent
WA
...
Slast + Ws – 1
Rlast
Rnew
Rlast + WR – 1
81
TCP Connections

TCP Connection



Connection Setup with Three-Way Handshake


Three-way exchange to negotiate initial Seq. #’s for
connections in each direction
Data Transfer


One connection each way
Identified uniquely by Send IP Address, Send TCP Port #,
Receive IP Address, Receive TCP Port #
Exchange segments carrying data
Graceful Close

Close each direction separately
82
Three Phases of TCP Connection
Host A
Host B
Three-way
Handshake
Data Transfer
Graceful
Close
83
1st Handshake: Client-Server
Connection Request
Initial Sequence number
from client to server
SYN bit set indicates request to establish
connection from client to server
84
2nd Handshake: ACK from Server
ACK Seq. no.
= Initial Seq. no. + 1
ACK bit set acknowledges connection
request; Client-to-Server connection
established
85
2nd Handshake: Server-Client
Connection Request
Initial Seq. # from server
to client
SYN bit set indicates request to establish
connection from server to client
86
3rd Handshake: ACK from Client
ACK Seq. # = Init. Seq.
#+1
ACK bit set acknowledges connection
request; Connections in both directions
established
87
TCP Data Exchange


Application Layers write bytes into buffers
TCP sender





forms segments when bytes exceed threshold or
timer expires, or upon PUSH command from
applications
Consecutive bytes from buffer inserted in payload
Sequence no. and ACK no. inserted in header
Checksum calculated and included in header
TCP receiver


Performs selective repeat ARQ functions
Writes error-free, in-sequence bytes to receive
buffer
88
Data Transfer: Server-to-Client
Segment
12 bytes of payload
Push set
12 bytes of payload carries
telnet option negotiation
89
Graceful Close: Client-to-Server
Connection
Client initiates closing of its
connection to server
90
Graceful Close: Client-to-Server
Connection
ACK Seq. no.
= Previous Seq. no. + 1
Server ACKs request; client-to-server
connection closed
91
TCP Retransmission by Timeout



TCP retransmits the segment after its corresponding timer expires
 Timeout too short: excessive number of retransmissions
 Timeout too long: recovery too slow
 Timeout value depends on round-trip time (RTT): time from when a
segment is sent to when its ACK is received
RTT in Internet is widely variable from time to time or from connection to
connection, therefore TCP adaptively sets the re-transmission time-out
value
TCP continuously estimates the RTT by
tRTT(new) = a tRTT(old) + (1 – a) tn


tn : the time elapsed from the instant the n-th segment is transmitted
until the corresponding ACK is received
a = 7/8 typical
92
RTT Variability



The time-out value tout is determined by taking into account not only
the mean tRTT but also the variability in the estimation of the round-trip
time
The variability can be obtained by the standard deviation of the roundtrip time, denoted by sRTT
The time-out value is then given by: tout = tRTT + k sRTT




If RTT highly variable, timeout value increases accordingly
If RTT nearly constant, timeout value is close to the mean RTT estimate
Approximate estimation of the standard deviation by
dRTT(new) = b dRTT(old) + (1-b) | tn - tRTT |
The typical value of b is ¾. The time-out value that has been found to
work well is then
tout = tRTT + 4 dRTT
93
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
PART II: Data Link Controls
94
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
5.4 Framing
95
Framing for Frame Synchronization



Framing involves identifying the beginning and end of
a block of information
Framing presupposes that there is enough clock
synchronization at the physical layer to at least
identify an individual bit or byte
Frame boundaries can be determined using:





Character Counts
Control Characters
Flags
CRC Checks
The data link layer will do the frame synchronization
96
Character-Oriented Framing
Data to be sent
A
DLE B
ETX DLE STX E
After stuffing and framing
DLE STX A


DLE ETX
Asynchronous transmission systems using ASCII to transmit printable
characters
Octets with HEX value <20 are nonprintable
Special 8-bit patterns used as control characters


ETX DLE DLE STX E
Frames consist of integer number of bytes


DLE DLE B
STX (start of text) = 0x02; ETX (end of text) = 0x03;
Byte used to carry non-printable characters in frame




DLE (data link escape) = 0x10
DLE STX (DLE ETX) used to indicate beginning (end) of frame
Insert extra DLE in front of occurrence of DLE STX (DLE ETX) in frame
All DLEs occur in pairs except at frame boundaries
Framing & Bit Stuffing
HDLC frame
Flag Address Control
Information
FCS
Flag
any number of bits




Frame delineated by flag character
HDLC uses bit stuffing to prevent occurrence of flag
01111110 inside the frame
Transmitter inserts extra 0 after each consecutive
five 1s inside the frame
Receiver checks for five consecutive 1s



if next bit = 0, it is removed
if next two bits are 10, then flag is detected
If next two bits are 11, then frame has errors
98
Example: Bit stuffing & destuffing
(a)
Data to be sent
0110111111111100
After stuffing and framing
01111110 01101111101111100001111110
(b)
Data received
01111110000111011111011111011001111110
After destuffing and deframing
*000111011111-11111-110*
99
PPP Frame
Flag
Address
01111110 1111111
Control
00000011
Protocol
Information
CRC
Flag
01111110
integer # of bytes
All stations are to
accept the frame



Specifies what kind of packet is contained in the
payload, e.g., LCP, NCP, IP, OSI CLNP, IPX
PPP uses similar frame structure as HDLC, except


Unnumbered
frame
Protocol type field
Payload contains an integer number of bytes
PPP uses the same flag, but uses byte stuffing
Problems with PPP byte stuffing


Size of frame varies unpredictably due to byte insertion
Malicious users can inflate bandwidth by sending 7D or 7E in the
information field
Byte-Stuffing in PPP




PPP is character-oriented version of HDLC
Flag is 0x7E (01111110)
Control escape 0x7D (01111101)
Any occurrence of flag or control escape inside of frame is
replaced with 0x7D followed by original octet XORed with
0x20 (00100000)
Data to be sent
7E
41
41
7D
42
7E
50
70
46
7D
5D
42
7D
5E
50
70
After stuffing and framing
46
7E
CRC-based Framing:
Generic Framing Procedure (GFP)
GFP payload area
2


2
2
2
0-60
PLI
cHEC
Type
tHEC
GEH
Payload
length
indicator
Core
header
error
checking
Payload
type
Type
header
error
checking
GFP
extension
headers
GFP payload
GFP
payload
GFP combines a frame length indication field with Header Error Control (HEC)
method to delineate ATM cells
 PLI can be used to help finding the beginning of a frame and the beginning of
the next frame.
 cHEC (CRC-16) protects against errors in count field (single-bit error
correction + error detection)
GFP is designed to operate over octet-synchronous physical layers (e.g. SONET)
 In the frame-mapped mode, GFP can be used to carry variable-length
payloads such as Ethernet frames, PPP/IP packets, or HDLC-framed PDU
 In the transparent mode, GFP carries synchronous information stream with
low delay in fixed-length frames. Transport mode GFP is suitable for Gigabit
Ethernet, and storage devices
GFP Synchronization and Scrambling

Synchronization in three-state process
 Hunt state: the initial state where it examines 4 bytes to see if CRC
of the first two bytes equals the contents of the next two bytes



Pre-sync state: the intermediate state where the tentative PLI
indicates the boundary of the next frame



If N successful frame detections, move to the sync state
If no match, go to the hunt state
Sync state: the normal state




If no, move forward by one byte and check again
If yes, move to the pre-sync state
Validate PLI/cHEC, extract payload, go to the next frame
Use single-error correcting capability of CRC-16
Go to hunt state if non-correctable error happens
Scrambling
 Payload is scrambled to prevent malicious users from inserting
long strings of 0s which cause SONET equipment to lose bit clock
synchronization (as discussed in line code section).
 For example, 1+x43 scrambler for ATM
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
5.6 High-Level Data Link Control
(HDLC)
High-Level Data Link Control
(HDLC)



Bit-oriented data link control
Derived from Synchronous Data Link Control
(SDLC) developed by IBM
Related to Link Access Procedure Balanced
(LAPB) developed by CCITT-ITU


LAPD in ISDN
LAPM in cellular telephone signaling
NLPDU
Network
layer
DLSDU
The data link
layer can be
configured to
provide the
connectionoriented or
connectionless
services.
“Packet”
DLSAP
DLSAP
Data link
layer
Network
layer
DLSDU
DLPDU
“Frame
”
Physical
layer
Data link
layer
Physical
layer
Fig. 5.43 The data link layer
5.6.2 HDLC Configuration and
Transfer Modes

Unbalanced Configuration: Normal Response Mode (NRM) of HDLC
 Primary station uses Command to poll and Secondary replies
using response frame
Commands
Primary
Responses
Secondary

Secondary
Balanced Configuration: Asynchronous Balanced Mode (ABM)
 Two stations act as peers and information frames are
transmitted in full-duplex manner
Primary Commands
Secondary

Secondary
Responses
Responses Secondary
Commands
Mode is selected during connection establishment
Primary
5.6.3 HDLC Frame Format
Flag Address Control

Information
FCS
Flag
Codes in fields have specific meanings and uses


Flag: delineate frame boundaries
Address: identify secondary station (1 or more octets) in
unbalanced configuration; in ABM mode, a station can act as
primary or secondary so address changes accordingly



Control: gives the main functionality of HDLC (1 or 2 octets)
Information: contains user data; length not standardized, but
implementations impose maximum
Frame Check Sequence: 16- or 32-bit CRC
Control Field Format
Information Frame
1
2-4
0
N(S)
5
6-8
P/F
N(R)
P/F
N(R)
Supervisory Frame
1
0
S
S
Unnumbered Frame
1



1
M
M
S: Supervisory Function Bits
N(S): Send Sequence Number
N(R): Receive Sequence Number
P/F


M
M
M
M: Unnumbered Function Bits
P/F: Poll/final bit used in interaction
between primary and secondary
Information frame (I-frame)

Each I-frame contains the send sequence number N(S)


Positive piggybacked ACK N(R)


N(R) = Sequence number of the next frame expected, meaning to
acknowledge all frames up to and including N(R)-1
3 or 7 bit sequence number for N(S) and N(R)


N(S) = Sequence number of the sending I-frame
Maximum window sizes 7 or 127
Poll/Final Bit


NRM: Primary polls station by setting P=1; Secondary sets F=1 in last
I-frame in response
Primary and secondary always interact via paired P/F bits
Error Detection & Loss Recovery






Frames lost due to loss-of-synch or receiver buffer
overflow
Frames may undergo errors in transmission
CRCs detect errors and such frames are treated as lost
Recovery through ACKs, timeouts, and retransmission
Sequence number to identify out-of-sequence and
duplicated frames
HDLC provides for options that implement several ARQ
methods
Supervisory frames

Supervisory frame is used for error control (ACK, NAK)
and flow control (Don’t Send)

Receive Ready (RR), SS=00 with N(R)


Reject (REJ), SS=01 with N(R)


REJ frames are used by the receiver to send a negative
acknowledgement, indicating the transmitter should go back and
retransmit frames from N(R) onwards (for Go-Back-N ARQ).
Receive Not Ready (RNR), SS=10 with N(R)


RR frames are used to ACK frames when no I-frames are available to
piggyback the acknowledgement
RNR frame acknowledges all frames up to N(R)-1 and informs the
transmitter that the receiver has temporary problems, that is, no buffers,
and will not accept any more frames.
Selective Reject (SREJ), SS=11 with N(R)

SREJ frame indicates to the transmitter that it should retransmit the
frame indicated in the N(R) subfield (for selective repeat ARQ)
Unnumbered Frames

Set the data transfer mode during call setup or release





Unnumbered I-Information transfer between stations


UI: Unnumbered information frame
Recovery used when normal error/flow control fails



SABM: Set Asynchronous Balanced Mode
SNRM: Set Normal Response Mode
UA: acknowledges acceptance of mode setting command
DISC: terminates logical link connection
FRMR: frame with correct FCS but impossible semantics
RSET: indicates that the transmitter station is resetting the
sequence numbers
XID: exchange station id and characteristics
5.6.4 Typical Frame Exchange


Example: Exchange of frames for connection
establishment and release in data link connection
In HDLC



Set Asynchronous Balanced Mode (SABM)
Disconnect (DISC)
Unnumbered Acknowledgment (UA)
SABM
UA
Data
transfer
Fig. 5.47
DISC
UA
Example: HDLC using NRM to
Exchange Frames
Address of secondary
A polls B
Secondaries B, C
Primary A
B, RR, 0, P
Supervisory frame N(R)
N(S) N(R)
X
Unnumbered frame
A rejects frame 1
B, SREJ, 1
A polls C
C, RR, 0, P
B, I, 0, 0
B, I, 1, 0
B, I, 2, 0,F
Information frame
C, RR, 0, F
A polls B, requests
selective
retransmission of
frame1
A send info frame 0
to B, ACKs up to 4
B sends 3 info
frames
C nothing to
send
B, SREJ, 1,P
B, I, 1, 0 B resends frame1
B, I, 3, 0 Then frame 3 & 4
B, I, 4, 0, F
B, I, 0, 5
Time
Example: HDLC using Asynchronous
Balanced Mode to Exchange Frame
Combined Station B
Combined Station A
N(S), N(R)
B sends 5
frames
B, I, 0, 0
A, I, 0, 0
B, I, 1, 0
A, I, 1, 1
B, I, 2, 1
A, I, 2, 1
B, I, 3, 2
B, REJ, 1
B, I, 4, 3
B goes
back to 1
X
ACKs frame 0
from A
Rejects frame 1
of A
A, I, 3, 1
B, I, 1, 3
B, I, 2, 4
B, I, 3, 4
B, RR, 2
ACKs frame 1
from A
B, RR, 3
ACKs frame 2
Flow Control



Flow control is required to prevent transmitter from
overrunning receiver buffers
Receiver can control flow by delaying
acknowledgement messages
Receiver can also use supervisory frames to
explicitly control transmitter

Receive Not Ready (RNR) and Receive Ready (RR)
I3
I4
I5
RNR5
RR6
I6
Chapter 5
Peer-to-Peer Protocols
and Data Link Layer
5.7 Link Sharing Using
Statistical Multiplexing
118
Statistical Multiplexing


Multiplexing is just to concentrate bursty traffic onto a shared
line; statistical multiplexing is to multiplex on demand
Greater efficiency and lower cost
Header
Data payload
A
B
Buffer
Output line
C
119
Input lines
Tradeoff Delay for Efficiency
(a)
Dedicated lines
A2
A1
B2
B1
C1
(b)
Shared lines


A1
C2
C1
B1
A2
B2
C2
Dedicated lines involve not waiting for other users, but lines
are used inefficiently when user traffic is bursty
Shared lines concentrate packets into shared line; packets
buffered (delayed) when line is not immediately available
120
Multiplexers are inherent in
Packet Switches
1
1
2
2
.
.
.
.
.
.
N


N
Packets/frames forwarded to buffer prior to transmission from
switch
Multiplexing occurs in these buffers
121
Multiplexer Modeling
Input lines
A
Output line
B
Buffer
C





Arrivals: What is the packet interarrival pattern?
Service Time: How long are the packets?
Service Discipline: What is order of transmission?
Buffer Discipline: If buffer is full, which packet is dropped?
Performance Measures: [Delay Distribution; Packet Loss Probability;
Line Utilization] versus Traffic Load Intensity or Buffer Size
122
Delay = Waiting + Service Times
P2
P1
Packet completes
transmission
P3
P5
Service
time
Packet begins
transmission
Packet arrives
at queue
P4
Waiting
time
P1
P2
P3
P4
P5




Packets arrive and wait for service
Waiting Time: from the instant of arrival to the instant of service
beginning
Service Time: time to transmit packet
Delay: total time in system = waiting time + service time
123
Fluctuations in Packets in the
System
(a)
Dedicated lines
A1
A2
B2
B1
C2
C1
(b)
Shared line
(c)
N(t)
Number of
packets in the
system
A1
C1
B1
A2
B2
C2
124
Burst Multiplexing / Speech
Interpolation
Many
Voice
Calls
Fewer
Trunks
Part of this burst is lost





Voice active < 40% time
No buffering, on-the-fly switch bursts to available trunks
Can handle 2 to 3 times as many calls
Tradeoff: Trunk Utilization vs. Speech Loss
 Fractional Speech Loss: fraction of active speech lost
Demand Characteristics
 Talkspurt and Silence Duration Statistics
125
 Proportion of time speaker active/idle
Speech Loss vs. Trunks
speech loss
trunks
Typical
requirement
48
24
32
40
# connections
n
ænö k
(k - m)ç ÷ p (1 - p) n -k
èkø
n!
ænö
k = m +1
speech loss =
where ç ÷ =
.
np
k
!
(
n
k
)!
èkø
å
126
Effect of Scale


Larger flows lead to better performance
Multiplexing Gain = # speakers / # trunks
Trunks required for 1% speech loss
Speakers
Trunks
Multiplexing
Gain
Utilization
24
13
1.85
0.74
32
16
2.00
0.80
40
20
2.00
0.80
48
23
2.09
0.83
127
Packet Speech Multiplexing
Many voice A3
terminals
generating
B3
voice packets
A2
A1
B2
B1
C3
C2
C1
D3
D2
D1
Buffer
B3 C3 A2 D2 C2 B1 C1 D1 A1
Buffer overflow
B2





Digital speech carried by fixed-length packets
No packets when speaker silent
Synchronous packets when speaker active
Buffer packets & transmit over shared high-speed line
Tradeoffs: Utilization vs. Delay/Jitter & Loss
128
Packet Switching of Voice
Sent
Received



1
2
1
3
t
2
3
t
Packetization delay: time for speech samples to fill a packet
Jitter: variable inter-packet arrivals at destination
Playback strategies required to compensate for jitter/loss
 Flexible delay inserted to produce fixed end-to-end delay
 Need buffer overflow/underflow countermeasures
 Need clock recovery algorithm
129
Home Work
Problem 1: 5.15
 Problem 2: 5.21
 Problem 3: 5.33 Please hand in on December 12
 Problem 4: 5.36
 Problem 5: 5.52
=========================================
 Problem 6: 5.56
 Problem 7: 5.67
 Problem 8: 5.69 Please hand in on December 19
 Problem 9: 5.72
 Problem 10: 5.73

130