Unit - 4 — — Transport Layer

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CS 1302
Computer Networks
— Unit - 4 —
— Transport Layer —

Text Book
Behrouz .A. Forouzan, “Data communication and
Networking”, Tata McGrawHill, 2004
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Unit-4 : Transport Layer
1
Transport Layer
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Unit-4 : Transport Layer
2
Position of transport layer
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Unit-4 : Transport Layer
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Transport layer duties
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Chapters
Chapter 22 Process-to-Process Delivery
Chapter 23 Congestion Control and QoS
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Chapter 22
Process-to-Process
Delivery:
UDP and TCP
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22.1 Process-to-Process Delivery
Client-Server Paradigm
Addressing
Multiplexing and Demultiplexing
Connectionless/Connection-Oriented
Reliable/Unreliable
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Note:
The transport layer is responsible for
process-to-process delivery.
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Figure 22.1 Types of data deliveries
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Figure 22.2
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Port numbers
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Figure 22.3
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IP addresses versus port numbers
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Figure 22.4
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IANA ranges
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Figure 22.5
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Socket address
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Figure 22.6
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Multiplexing and demultiplexing
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Figure 22.7
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Connection establishment
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Figure 22.8
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Connection termination
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Figure 22.9
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Error control
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22.2 UDP
Port Numbers
User Datagram
Applications
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Note:
UDP is a connectionless, unreliable
protocol that has no flow and error
control. It uses port numbers to
multiplex data from the application
layer.
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Table 22.1 Well-known ports used by UDP
Port
Protocol
Description
7
Echo
9
Discard
11
Users
13
Daytime
17
Quote
19
Chargen
53
Nameserver
67
Bootps
Server port to download bootstrap information
68
Bootpc
Client port to download bootstrap information
69
TFTP
Trivial File Transfer Protocol
111
RPC
Remote Procedure Call
123
NTP
Network Time Protocol
161
SNMP
Simple Network Management Protocol
162
SNMP
Simple Network Management Protocol (trap)
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Echoes a received datagram back to the sender
Discards any datagram that is received
Active users
Returns the date and the time
Returns a quote of the day
Returns a string of characters
Domain Name Service
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Figure 22.10
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User datagram format
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Note:
The calculation of checksum and its
inclusion in the user datagram are
optional.
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Note:
UDP is a convenient transport-layer
protocol for applications that provide
flow and error control. It is also used
by multimedia applications.
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22.3 TCP
Port Numbers
Services
Sequence Numbers
Segments
Connection
Transition Diagram
Flow and Error Control
Silly Window Syndrome
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Table 22.2 Well-known ports used by TCP
Port
Protocol
Description
7
Echo
9
Discard
11
Users
13
Daytime
17
Quote
19
Chargen
20
FTP, Data
21
FTP, Control
23
TELNET
25
SMTP
53
DNS
67
BOOTP
79
Finger
Finger
80
HTTP
Hypertext Transfer Protocol
111
RPC
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Echoes a received datagram back to the sender
Discards any datagram that is received
Active users
Returns the date and the time
Returns a quote of the day
Returns a string of characters
File Transfer Protocol (data connection)
File Transfer Protocol (control connection)
Terminal Network
Simple Mail Transfer Protocol
Domain Name Server
Bootstrap Protocol
Remote Procedure Call
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Figure 22.11 Stream delivery
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Figure 22.12
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Sending and receiving buffers
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Figure 22.13 TCP segments
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Example 1
Imagine a TCP connection is transferring a file of 6000 bytes. The
first byte is numbered 10010. What are the sequence numbers for
each segment if data are sent in five segments with the first four
segments carrying 1000 bytes and the last segment carrying 2000
bytes?
Solution
The following shows
Segment 1 ==>
Segment 2 ==>
Segment 3 ==>
Segment 4 ==>
Segment 5 ==>
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the sequence number for each segment:
sequence number: 10,010 (range: 10,010
sequence number: 11,010 (range: 11,010
sequence number: 12,010 (range: 12,010
sequence number: 13,010 (range: 13,010
sequence number: 14,010 (range: 14,010
Unit-4 : Transport Layer
to 11,009)
to 12,009)
to 13,009)
to 14,009)
to 16,009)
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Note:
The bytes of data being transferred in
each connection are numbered by
TCP. The numbering starts with a
randomly generated number.
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Note:
The value of the sequence number
field in a segment defines the number
of the first data byte contained in that
segment.
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Note:
The value of the acknowledgment field
in a segment defines the number of the
next byte a party expects to receive.
The acknowledgment number is
cumulative.
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Figure 22.14 TCP segment format
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Figure 22.15
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Control field
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Table 22.3 Description of flags in the control field
Flag
Description
URG
The value of the urgent pointer field is valid.
ACK
The value of the acknowledgment field is valid.
PSH
Push the data.
RST
The connection must be reset.
SYN
Synchronize sequence numbers during connection.
FIN
Terminate the connection.
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Figure 22.16 Three-step connection establishment
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Figure 22.17
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Four-step connection termination
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Table 22.4 States for TCP
State
Description
CLOSED
There is no connection.
LISTEN
The server is waiting for calls from the client.
SYN-SENT
A connection request is sent; waiting for acknowledgment.
SYN-RCVD
A connection request is received.
ESTABLISHED Connection is established.
FIN-WAIT-1
The application has requested the closing of the
connection.
FIN-WAIT-2
The other side has accepted the closing of the connection.
TIME-WAIT
Waiting for retransmitted segments to die.
CLOSE-WAIT
The server is waiting for the application to close.
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LAST-ACK
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The server is waiting for the last acknowledgment.
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Figure 22.18
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State transition diagram
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Note:
A sliding window is used to make
transmission more efficient as well as
to control the flow of data so that the
destination does not become
overwhelmed with data. TCP’s sliding
windows are byte-oriented.
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Figure 22.19
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Sender buffer
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Figure 22.20
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Receiver window
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Figure 22.21
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Sender buffer and sender window
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Figure 22.22
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Sliding the sender window
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Figure 22.23
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Expanding the sender window
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Figure 22.24
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Shrinking the sender window
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Note:
In TCP, the sender window size is
totally controlled by the receiver
window value (the number of empty
locations in the receiver buffer).
However, the actual window size can
be smaller if there is congestion in the
network.
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Note:
Some points about TCP’s sliding windows:
The source does not have to send a full
window’s worth of data.
The size of the window can be increased or
decreased by the destination.
The destination can send an acknowledgment
at any time.
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Figure 22.25
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Lost segment
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Figure 22.26
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Lost acknowledgment
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Figure 22.27 TCP timers
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Multiplexing
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Figure 6.1
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Dividing a link into channels
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Figure 6.2
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Categories of multiplexing
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6.1 FDM
Multiplexing Process
Demultiplexing Process
The Analog Hierarchy
Other Applications of FDM
Implementation
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Figure 6.3
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FDM
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Note:
FDM is an analog multiplexing
technique that combines signals.
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Figure 6.4
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FDM process
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Figure 6.5
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FDM demultiplexing example
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Example 1
Assume that a voice channel occupies a bandwidth of 4
KHz. We need to combine three voice channels into a link
with a bandwidth of 12 KHz, from 20 to 32 KHz. Show
the configuration using the frequency domain without the
use of guard bands.
Solution
Shift (modulate) each of the three voice channels to a
different bandwidth, as shown in Figure 6.6.
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Figure 6.6
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Example 1
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Example 2
Five channels, each with a 100-KHz bandwidth, are to be
multiplexed together. What is the minimum bandwidth of
the link if there is a need for a guard band of 10 KHz
between the channels to prevent interference?
Solution
For five channels, we need at least four guard bands.
This means that the required bandwidth is at least
5 x 100 + 4 x 10 = 540 KHz,
as shown in Figure 6.7.
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Figure 6.7
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Example 2
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Example 3
Four data channels (digital), each transmitting at 1 Mbps,
use a satellite channel of 1 MHz. Design an appropriate
configuration using FDM
Solution
The satellite channel is analog. We divide it into four
channels, each channel having a 250-KHz bandwidth.
Each digital channel of 1 Mbps is modulated such that
each 4 bits are modulated to 1 Hz. One solution is 16QAM modulation. Figure 6.8 shows one possible
configuration.
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Figure 6.8
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Example 3
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Figure 6.9
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Analog hierarchy
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Example 4
The Advanced Mobile Phone System (AMPS) uses two
bands. The first band, 824 to 849 MHz, is used for
sending; and 869 to 894 MHz is used for receiving. Each
user has a bandwidth of 30 KHz in each direction. The 3KHz voice is modulated using FM, creating 30 KHz of
modulated signal. How many people can use their cellular
phones simultaneously?
Solution
Each band is 25 MHz. If we divide 25 MHz into 30 KHz,
we get 833.33. In reality, the band is divided into 832
channels.
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6.2 WDM
Wave Division Multiplexing
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Figure 6.10
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WDM
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Note:
WDM is an analog multiplexing
technique to combine
optical signals.
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Figure 6.11 Prisms in WDM multiplexing and demultiplexing
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6.3 TDM
Time Slots and Frames
Interleaving
Synchronizing
Bit Padding
Digital Signal (DS) Service
T Lines
Inverse TDM
More TDM Applications
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Figure 6.12
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TDM
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Note:
TDM is a digital multiplexing
technique to combine data.
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Figure 6.13
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TDM frames
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Example 5
Four 1-Kbps connections are multiplexed together. A unit
is 1 bit. Find (1) the duration of 1 bit before multiplexing,
(2) the transmission rate of the link, (3) the duration of a
time slot, and (4) the duration of a frame?
Solution
We can answer the questions as follows:
1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms).
2. The rate of the link is 4 Kbps.
3. The duration of each time slot 1/4 ms or 250 ms.
4. The duration of a frame 1 ms.
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Note:
In a TDM, the data rate of the link is n
times faster, and the unit duration is n
times shorter.
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Figure 6.14
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Interleaving
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Example 6
Four channels are multiplexed using TDM. If each
channel sends 100 bytes/s and we multiplex 1 byte per
channel, show the frame traveling on the link, the size of
the frame, the duration of a frame, the frame rate, and the
bit rate for the link.
Solution
The multiplexer is shown in Figure 6.15.
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Figure 6.15
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Example 6
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Example 7
A multiplexer combines four 100-Kbps channels using a
time slot of 2 bits. Show the output with four arbitrary
inputs. What is the frame rate? What is the frame
duration? What is the bit rate? What is the bit duration?
Solution
Figure 6.16 shows the output for four arbitrary inputs.
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Figure 6.16
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Example 7
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Figure 6.17
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Framing bits
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Example 8
We have four sources, each creating 250 characters per
second. If the interleaved unit is a character and 1
synchronizing bit is added to each frame, find (1) the data
rate of each source, (2) the duration of each character in
each source, (3) the frame rate, (4) the duration of each
frame, (5) the number of bits in each frame, and (6) the
data rate of the link.
Solution
See next slide.
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Solution (continued)
We can answer the questions as follows:
1. The data rate of each source is 2000 bps = 2 Kbps.
2. The duration of a character is 1/250 s, or 4 ms.
3. The link needs to send 250 frames per second.
4. The duration of each frame is 1/250 s, or 4 ms.
5. Each frame is 4 x 8 + 1 = 33 bits.
6. The data rate of the link is 250 x 33, or 8250 bps.
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Example 9
Two channels, one with a bit rate of 100 Kbps and
another with a bit rate of 200 Kbps, are to be multiplexed.
How this can be achieved? What is the frame rate? What
is the frame duration? What is the bit rate of the link?
Solution
We can allocate one slot to the first channel and two slots
to the second channel. Each frame carries 3 bits. The
frame rate is 100,000 frames per second because it
carries 1 bit from the first channel. The frame duration is
1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x
3 bits/frame, or 300 Kbps.
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Figure 6.18
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DS hierarchy
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Table 6.1 DS and T lines rates
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Service
Line
Rate
(Mbps)
Voice
Channels
DS-1
T-1
1.544
24
DS-2
T-2
6.312
96
DS-3
T-3
44.736
672
DS-4
T-4
274.176
4032
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Figure 6.19
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T-1 line for multiplexing telephone lines
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Figure 6.20
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T-1 frame structure
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Table 6.2 E line rates
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E Line
Rate
(Mbps)
Voice
Channels
E-1
2.048
30
E-2
8.448
120
E-3
34.368
480
E-4
139.264
1920
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Figure 6.21
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Multiplexing and inverse multiplexing
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Client-Server
Model:
Socket Interface
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24.1 Client-Server Model
Relationship
Concurrency
Processes
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Figure 24.1
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Client-server model
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Figure 24.2
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Client-server relationship
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Figure 24.3
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Connectionless iterative server
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Figure 24.4
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Connection-oriented concurrent server
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24.2 Socket Interface
Sockets
Connectionless Iterative Server
Connection-Oriented Server
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Figure 24.5
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Socket structure
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Figure 24.6
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Socket types
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Figure 24.7
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Socket interface for connectionless iterative server
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Figure 24.8
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Socket interface for connection-oriented concurrent server
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Congestion Control
and
Quality of Service
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23.1 Data Traffic
Traffic Descriptor
Traffic Profiles
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Figure 23.1 Traffic descriptors
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Figure 23.2
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Constant-bit-rate traffic
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Figure 23.3 Variable-bit-rate traffic
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Figure 23.4
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Bursty traffic
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23.2 Congestion
Network Performance
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Figure 23.5
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Incoming packet
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Figure 23.6
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Packet delay and network load
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Figure 23.7 Throughput versus network load
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23.3 Congestion Control
Open Loop
Closed Loop
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23.4 Two Examples
Congestion Control in TCP
Congestion Control in Frame Relay
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Note:
TCP assumes that the cause of a lost
segment is due to congestion
in the network.
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Note:
If the cause of the lost segment is
congestion, retransmission of the
segment does not remove
the cause—it aggravates it.
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Figure 23.8
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Multiplicative decrease
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Figure 23.9
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BECN
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Figure 23.10
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FECN
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Figure 23.11 Four cases of congestion
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23.5 Quality of Service
Flow Characteristics
Flow Classes
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23.6 Techniques to Improve QoS
Scheduling
Traffic Shaping
Resource Reservation
Admission Control
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Figure 23.12
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Flow characteristics
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Figure 23.13
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FIFO queue
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Figure 23.14
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Priority queuing
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Figure 23.15 Weighted fair queuing
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Figure 23.16
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Leaky bucket
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Figure 23.17
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Leaky bucket implementation
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Note:
A leaky bucket algorithm shapes
bursty traffic into fixed-rate traffic by
averaging the data rate. It may drop
the packets if the bucket is full.
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Figure 23.18 Token bucket
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Note:
The token bucket allows bursty traffic
at a regulated maximum rate.
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23.7 Integrated Services
Signaling
Flow Specification
Admission
Service Classes
RSVP
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Note:
Integrated Services is a flow-based
QoS model designed for IP.
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Figure 23.19
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Path messages
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Figure 23.20
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Resv messages
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Figure 23.21
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Reservation merging
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Figure 23.22
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Reservation styles
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