Transcript Unit - 4 — — Transport Layer

CS 1302
Computer Networks
— Unit - 4 —
— Transport Layer —

Text Book
Behrouz .A. Forouzan, “Data communication and
Networking”, Tata McGrawHill, 2004
4/9/2015
Unit-4 : Transport Layer
1
Transport Layer
4/9/2015
Unit-4 : Transport Layer
2
Position of transport layer
4/9/2015
Unit-4 : Transport Layer
3
Transport layer duties
4/9/2015
Unit-4 : Transport Layer
4
Chapters
Chapter 22 Process-to-Process Delivery
Chapter 23 Congestion Control and QoS
4/9/2015
Unit-4 : Transport Layer
5
Chapter 22
Process-to-Process
Delivery:
UDP and TCP
4/9/2015
Unit-4 : Transport Layer
6
22.1 Process-to-Process Delivery
Client-Server Paradigm
Addressing
Multiplexing and Demultiplexing
Connectionless/Connection-Oriented
Reliable/Unreliable
4/9/2015
Unit-4 : Transport Layer
7
Note:
The transport layer is responsible for
process-to-process delivery.
4/9/2015
Unit-4 : Transport Layer
8
Figure 22.1 Types of data deliveries
4/9/2015
Unit-4 : Transport Layer
9
Figure 22.2
4/9/2015
Port numbers
Unit-4 : Transport Layer
10
Figure 22.3
4/9/2015
IP addresses versus port numbers
Unit-4 : Transport Layer
11
Figure 22.4
4/9/2015
IANA ranges
Unit-4 : Transport Layer
12
Figure 22.5
4/9/2015
Socket address
Unit-4 : Transport Layer
13
Figure 22.6
4/9/2015
Multiplexing and demultiplexing
Unit-4 : Transport Layer
14
Figure 22.7
4/9/2015
Connection establishment
Unit-4 : Transport Layer
15
Figure 22.8
4/9/2015
Connection termination
Unit-4 : Transport Layer
16
Figure 22.9
4/9/2015
Error control
Unit-4 : Transport Layer
17
22.2 UDP
Port Numbers
User Datagram
Applications
4/9/2015
Unit-4 : Transport Layer
18
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.
4/9/2015
Unit-4 : Transport Layer
19
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)
4/9/2015
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
Unit-4 : Transport Layer
20
Figure 22.10
4/9/2015
User datagram format
Unit-4 : Transport Layer
21
Note:
The calculation of checksum and its
inclusion in the user datagram are
optional.
4/9/2015
Unit-4 : Transport Layer
22
Note:
UDP is a convenient transport-layer
protocol for applications that provide
flow and error control. It is also used
by multimedia applications.
4/9/2015
Unit-4 : Transport Layer
23
22.3 TCP
Port Numbers
Services
Sequence Numbers
Segments
Connection
Transition Diagram
Flow and Error Control
Silly Window Syndrome
4/9/2015
Unit-4 : Transport Layer
24
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
4/9/2015
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
Unit-4 : Transport Layer
25
Figure 22.11 Stream delivery
4/9/2015
Unit-4 : Transport Layer
26
Figure 22.12
4/9/2015
Sending and receiving buffers
Unit-4 : Transport Layer
27
Figure 22.13 TCP segments
4/9/2015
Unit-4 : Transport Layer
28
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 ==>
4/9/2015
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)
29
Note:
The bytes of data being transferred in
each connection are numbered by
TCP. The numbering starts with a
randomly generated number.
4/9/2015
Unit-4 : Transport Layer
30
Note:
The value of the sequence number
field in a segment defines the number
of the first data byte contained in that
segment.
4/9/2015
Unit-4 : Transport Layer
31
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.
4/9/2015
Unit-4 : Transport Layer
32
Figure 22.14 TCP segment format
4/9/2015
Unit-4 : Transport Layer
33
Figure 22.15
4/9/2015
Control field
Unit-4 : Transport Layer
34
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.
4/9/2015
Unit-4 : Transport Layer
35
Figure 22.16 Three-step connection establishment
4/9/2015
Unit-4 : Transport Layer
36
Figure 22.17
4/9/2015
Four-step connection termination
Unit-4 : Transport Layer
37
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.
4/9/2015
LAST-ACK
Unit-4 : Transport Layer
The server is waiting for the last acknowledgment.
38
Figure 22.18
4/9/2015
State transition diagram
Unit-4 : Transport Layer
39
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.
4/9/2015
Unit-4 : Transport Layer
40
Figure 22.19
4/9/2015
Sender buffer
Unit-4 : Transport Layer
41
Figure 22.20
4/9/2015
Receiver window
Unit-4 : Transport Layer
42
Figure 22.21
4/9/2015
Sender buffer and sender window
Unit-4 : Transport Layer
43
Figure 22.22
4/9/2015
Sliding the sender window
Unit-4 : Transport Layer
44
Figure 22.23
4/9/2015
Expanding the sender window
Unit-4 : Transport Layer
45
Figure 22.24
4/9/2015
Shrinking the sender window
Unit-4 : Transport Layer
46
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.
4/9/2015
Unit-4 : Transport Layer
47
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.
4/9/2015
Unit-4 : Transport Layer
48
Figure 22.25
4/9/2015
Lost segment
Unit-4 : Transport Layer
49
Figure 22.26
4/9/2015
Lost acknowledgment
Unit-4 : Transport Layer
50
Figure 22.27 TCP timers
4/9/2015
Unit-4 : Transport Layer
51
Multiplexing
4/9/2015
Unit-4 : Transport Layer
52
Figure 6.1
4/9/2015
Dividing a link into channels
Unit-4 : Transport Layer
53
Figure 6.2
4/9/2015
Categories of multiplexing
Unit-4 : Transport Layer
54
6.1 FDM
Multiplexing Process
Demultiplexing Process
The Analog Hierarchy
Other Applications of FDM
Implementation
4/9/2015
Unit-4 : Transport Layer
55
Figure 6.3
4/9/2015
FDM
Unit-4 : Transport Layer
56
Note:
FDM is an analog multiplexing
technique that combines signals.
4/9/2015
Unit-4 : Transport Layer
57
Figure 6.4
4/9/2015
FDM process
Unit-4 : Transport Layer
58
Figure 6.5
4/9/2015
FDM demultiplexing example
Unit-4 : Transport Layer
59
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.
4/9/2015
Unit-4 : Transport Layer
60
Figure 6.6
4/9/2015
Example 1
Unit-4 : Transport Layer
61
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.
4/9/2015
Unit-4 : Transport Layer
62
Figure 6.7
4/9/2015
Example 2
Unit-4 : Transport Layer
63
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.
4/9/2015
Unit-4 : Transport Layer
64
Figure 6.8
4/9/2015
Example 3
Unit-4 : Transport Layer
65
Figure 6.9
4/9/2015
Analog hierarchy
Unit-4 : Transport Layer
66
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.
4/9/2015
Unit-4 : Transport Layer
67
6.2 WDM
Wave Division Multiplexing
4/9/2015
Unit-4 : Transport Layer
68
Figure 6.10
4/9/2015
WDM
Unit-4 : Transport Layer
69
Note:
WDM is an analog multiplexing
technique to combine
optical signals.
4/9/2015
Unit-4 : Transport Layer
70
Figure 6.11 Prisms in WDM multiplexing and demultiplexing
4/9/2015
Unit-4 : Transport Layer
71
6.3 TDM
Time Slots and Frames
Interleaving
Synchronizing
Bit Padding
Digital Signal (DS) Service
T Lines
Inverse TDM
More TDM Applications
4/9/2015
Unit-4 : Transport Layer
72
Figure 6.12
4/9/2015
TDM
Unit-4 : Transport Layer
73
Note:
TDM is a digital multiplexing
technique to combine data.
4/9/2015
Unit-4 : Transport Layer
74
Figure 6.13
4/9/2015
TDM frames
Unit-4 : Transport Layer
75
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.
4/9/2015
Unit-4 : Transport Layer
76
Note:
In a TDM, the data rate of the link is n
times faster, and the unit duration is n
times shorter.
4/9/2015
Unit-4 : Transport Layer
77
Figure 6.14
4/9/2015
Interleaving
Unit-4 : Transport Layer
78
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.
4/9/2015
Unit-4 : Transport Layer
79
Figure 6.15
4/9/2015
Example 6
Unit-4 : Transport Layer
80
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.
4/9/2015
Unit-4 : Transport Layer
81
Figure 6.16
4/9/2015
Example 7
Unit-4 : Transport Layer
82
Figure 6.17
4/9/2015
Framing bits
Unit-4 : Transport Layer
83
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.
4/9/2015
Unit-4 : Transport Layer
84
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.
4/9/2015
Unit-4 : Transport Layer
85
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.
4/9/2015
Unit-4 : Transport Layer
86
Figure 6.18
4/9/2015
DS hierarchy
Unit-4 : Transport Layer
87
Table 6.1 DS and T lines rates
4/9/2015
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
Unit-4 : Transport Layer
88
Figure 6.19
4/9/2015
T-1 line for multiplexing telephone lines
Unit-4 : Transport Layer
89
Figure 6.20
4/9/2015
T-1 frame structure
Unit-4 : Transport Layer
90
Table 6.2 E line rates
4/9/2015
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
Unit-4 : Transport Layer
91
Figure 6.21
4/9/2015
Multiplexing and inverse multiplexing
Unit-4 : Transport Layer
92
Client-Server
Model:
Socket Interface
4/9/2015
Unit-4 : Transport Layer
93
24.1 Client-Server Model
Relationship
Concurrency
Processes
4/9/2015
Unit-4 : Transport Layer
94
Figure 24.1
4/9/2015
Client-server model
Unit-4 : Transport Layer
95
Figure 24.2
4/9/2015
Client-server relationship
Unit-4 : Transport Layer
96
Figure 24.3
4/9/2015
Connectionless iterative server
Unit-4 : Transport Layer
97
Figure 24.4
4/9/2015
Connection-oriented concurrent server
Unit-4 : Transport Layer
98
24.2 Socket Interface
Sockets
Connectionless Iterative Server
Connection-Oriented Server
4/9/2015
Unit-4 : Transport Layer
99
Figure 24.5
4/9/2015
Socket structure
Unit-4 : Transport Layer
100
Figure 24.6
4/9/2015
Socket types
Unit-4 : Transport Layer
101
Figure 24.7
4/9/2015
Socket interface for connectionless iterative server
Unit-4 : Transport Layer
102
Figure 24.8
4/9/2015
Socket interface for connection-oriented concurrent server
Unit-4 : Transport Layer
103
Congestion Control
and
Quality of Service
4/9/2015
Unit-4 : Transport Layer
104
23.1 Data Traffic
Traffic Descriptor
Traffic Profiles
4/9/2015
Unit-4 : Transport Layer
105
Figure 23.1 Traffic descriptors
4/9/2015
Unit-4 : Transport Layer
106
Figure 23.2
4/9/2015
Constant-bit-rate traffic
Unit-4 : Transport Layer
107
Figure 23.3 Variable-bit-rate traffic
4/9/2015
Unit-4 : Transport Layer
108
Figure 23.4
4/9/2015
Bursty traffic
Unit-4 : Transport Layer
109
23.2 Congestion
Network Performance
4/9/2015
Unit-4 : Transport Layer
110
Figure 23.5
4/9/2015
Incoming packet
Unit-4 : Transport Layer
111
Figure 23.6
4/9/2015
Packet delay and network load
Unit-4 : Transport Layer
112
Figure 23.7 Throughput versus network load
4/9/2015
Unit-4 : Transport Layer
113
23.3 Congestion Control
Open Loop
Closed Loop
4/9/2015
Unit-4 : Transport Layer
114
23.4 Two Examples
Congestion Control in TCP
Congestion Control in Frame Relay
4/9/2015
Unit-4 : Transport Layer
115
Note:
TCP assumes that the cause of a lost
segment is due to congestion
in the network.
4/9/2015
Unit-4 : Transport Layer
116
Note:
If the cause of the lost segment is
congestion, retransmission of the
segment does not remove
the cause—it aggravates it.
4/9/2015
Unit-4 : Transport Layer
117
Figure 23.8
4/9/2015
Multiplicative decrease
Unit-4 : Transport Layer
118
Figure 23.9
4/9/2015
BECN
Unit-4 : Transport Layer
119
Figure 23.10
4/9/2015
FECN
Unit-4 : Transport Layer
120
Figure 23.11 Four cases of congestion
4/9/2015
Unit-4 : Transport Layer
121
23.5 Quality of Service
Flow Characteristics
Flow Classes
4/9/2015
Unit-4 : Transport Layer
122
23.6 Techniques to Improve QoS
Scheduling
Traffic Shaping
Resource Reservation
Admission Control
4/9/2015
Unit-4 : Transport Layer
123
Figure 23.12
4/9/2015
Flow characteristics
Unit-4 : Transport Layer
124
Figure 23.13
4/9/2015
FIFO queue
Unit-4 : Transport Layer
125
Figure 23.14
4/9/2015
Priority queuing
Unit-4 : Transport Layer
126
Figure 23.15 Weighted fair queuing
4/9/2015
Unit-4 : Transport Layer
127
Figure 23.16
4/9/2015
Leaky bucket
Unit-4 : Transport Layer
128
Figure 23.17
4/9/2015
Leaky bucket implementation
Unit-4 : Transport Layer
129
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.
4/9/2015
Unit-4 : Transport Layer
130
Figure 23.18 Token bucket
4/9/2015
Unit-4 : Transport Layer
131
Note:
The token bucket allows bursty traffic
at a regulated maximum rate.
4/9/2015
Unit-4 : Transport Layer
132
23.7 Integrated Services
Signaling
Flow Specification
Admission
Service Classes
RSVP
4/9/2015
Unit-4 : Transport Layer
133
Note:
Integrated Services is a flow-based
QoS model designed for IP.
4/9/2015
Unit-4 : Transport Layer
134
Figure 23.19
4/9/2015
Path messages
Unit-4 : Transport Layer
135
Figure 23.20
4/9/2015
Resv messages
Unit-4 : Transport Layer
136
Figure 23.21
4/9/2015
Reservation merging
Unit-4 : Transport Layer
137
Figure 23.22
4/9/2015
Reservation styles
Unit-4 : Transport Layer
138