Transmission Media-Taxonomy
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
1
Design Factors
for Transmission Media
• Bandwidth: All other factors remaining constant, the greater
the band-width of a signal, the higher the data rate that can
be achieved.
• Transmission impairments. Limit the distance a signal can
travel.
• Interference: Competing signals in overlapping frequency
bands can distort or wipe out a signal.
• Number of receivers: Each attachment introduces some
attenuation and distortion, limiting distance and/or data rate.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
2
Twisted Pair (TP) Wires
• Commonly used for telephones and LANs
• Reduced electromagnetic interference
– Via twisting two wires together
(Usually several twists per inch)
• TP cables have a number of pairs of wires
– Telephone lines: two pairs (4 wires, usually only one pair is used by the
telephone)
– LAN cables: 4 pairs (8 wires)
• Also used in telephone trunk lines (up to several thousand
pairs)
• Shielded twisted pair also exists, but is more expensive
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
3
Types-UTP and STP and connectors
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
4
UTP Categories
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
5
Coaxial Cables
• Used for cable television, LANs, telephony
• Has an inner conductor surrounded by a braided mesh
• Both conductors share a common center axial, hence the
term “co-axial”.
• Advantages– Higher bandwidth
• 400 to 600Mhz
• up to 10,800 voice conversations
– Can be tapped easily (pros and cons)
– Much less susceptible to interference than twisted pair
• Disadvantages– High attenuation rate makes it expensive over long distance
– Bulky
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
6
Coax Layers
outer jacket
(polyethylene)
shield
(braided wire)
insulating material
copper or aluminum
conductor
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
7
Coaxial Cable-Categories and Connectors
RG: RadioGuide
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
8
Optical Fiber
Relatively new transmission medium used by telephone
companies in place of long-distance trunk lines.
Also used by private companies in implementing local data
communications networks
Require a light source with injection laser diode (ILD) or lightemitting diodes (LED)
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
9
Optical Fiber-Benefits and applications
• Greater capacity
• Data rates of hundreds of Gbps
• Smaller size & weight
• Lower attenuation
• Electromagnetic isolation
• Long-haul trunks
• Greater repeater spacing
• 1500km, 20 – 60k voice channels
• 10s of km at least
• Metropolitan trunks
• 12 km, 100k channels
• Rural exchange trunks
• 40 – 160Km, 5k voice channels
• Subscriber loops
• Voice data cables leased by corporate clients
• LANs
• 100Mbps – 1 Ghz
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
10
Unguided Media
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
11
A Realtime Implementation
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
12
Wireless Media
• Radio
– Wireless transmission of electrical waves over air
– Each device has a radio transceiver with a specific frequency
• Low power transmitters (few miles range)
• Often attached to portables (Laptops, PDAs, cell phones)
– Includes
• AM and FM radios, Cellular phones
• Wireless LANs – WiFi (IEEE 802.11) and Bluetooth
• Microwaves and Satellites
• Infrared
– “invisible” light waves (frequency is below red light)
– Requires line of sight; generally subject to interference from heavy rain,
smog, and fog
– Used in remote control units (e.g., TV)
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
13
Microwave Radio
• High frequency form of radio communications
– Extremely short (micro) wavelength (1 cm to 1 m)
– Requires line-of-sight
• Perform same functions as cables
– Often used for long distance, terrestrial transmissions
(over 50 miles without repeaters)
– No wiring and digging required
– Requires large antennas (about 10 ft) and high towers
• Possesses properties similar to light
– Reflection, Refraction, and focusing
– Can be focused into narrow powerful beams for long distance
Microwave is the general term used to describe RF waves that starts from UHF
(Ultra High Frequency) to EHF (Extremely High Frequency) which covers all
frequencies between 300Mhz to 300GHz, lower frequencies are refered to
as radio waves while higher frequencies are called millimeter waves.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
14
Satellite Communications
in a geosynchronous orbit
A special form of
microwave
communications
• Long propagation delay
– Due to great distance
between ground station and
satellite (Even with signals
traveling at light speed)
8/18/2019
Computer Networks
Signals sent
from the
ground to a
satellite; Then
relayed to its
destination
ground station
- PRONAYA BHATTACHARYA, CSE, ITNU
15
Factors Used in Media Selection
• Type of network
– LAN, WAN, or Backbone
• Cost
– Always changing; depends on the distance
• Transmission distance
– Short: up to 300 m; medium: up to 500 m
• Security
– Wireless media is less secure
• Error rates
– Wireless media has the highest error rate (interference)
• Transmission speeds
– Constantly improving; Fiber has the highest
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
16
Noise
• There are different types of noise
– Thermal - random noise of electrons in the
wire creates an extra signal
– Induced - from motors and appliances, devices
act are transmitter antenna and medium as
receiving antenna.
– Crosstalk - same as above but between two
wires.
– Impulse - Spikes that result from power lines,
lightning, etc.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
17
Network Devices
Repeater
Hub
 Switch
8/18/2019
Computer Networks
 Router
Bridge
Gateways
- PRONAYA BHATTACHARYA, CSE, ITNU
18
REPEATER-LAYER 1 DEVICE
A repeater operates at the physical
layer. Its job is to regenerate the
signal over the same network before
the signal becomes too weak or
corrupted so as to extend the length
to which the signal can be
transmitted over the same network.
An important point to be noted about
repeaters is that they do no amplify
the signal. When the signal becomes
weak, they copy the signal bit by bit
and regenerate it at the original
strength. It is a 2 port device.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
19
HUB-LAYER 2 DEVICE
A hub is basically a multiport
repeater. A hub connects multiple
wires
coming
from
different
branches. For example, the connector
in star topology which connects
different stations. Hubs cannot filter
data, so data packets are sent to all
connected devices.
Also, they do not have intelligence to
find out best path for data packets
which leads to inefficiencies and
wastage.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
20
SWITCH-LAYER 2 DEVICE
A switch is a multi port bridge
with a buffer and a design that
can boost its efficiency(large
number of ports imply less traffic)
and performance.
Switch is a data link layer device. Switch
can perform error checking before
forwarding data, that makes it very
efficient as it does not forward packets
that have errors and forward good
packets selectively to correct port only. In
other words, switch divides collision
domain of hosts, but broadcast
domain remains same.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
21
ROUTERS-LAYER 3 DEVICE
A router is a device like a switch
that routes data packets based on
their IP addresses. Router is
mainly a Network Layer device.
Routers normally connect LANs
and WANs together and have a
dynamically updating routing table
based on which they make
decisions on routing the data
packets. Router divide broadcast
domains of hosts connected
through it.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
22
NETWORK BRIDGE-LAYER 2
DEVICE
A network bridge is a product that
connects a local area network
(LAN) to another local area
network that uses the same
protocol.
Bridges are similar to—but more
intelligent than—simple repeaters,
which also extend signal range. It
has a single input and single output
port, thus making it a 2 port
device.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
23
NETWORK GATEWAY-ALL LAYERS
A gateway, as the name suggests,
is a passage to connect two
networks together that may work
upon
different
networking
models.
They basically works as the messenger
agents that take data from one system,
interpret it, and transfer it to another
system. Gateways are also called protocol
converters and can operate at any
network layer. Gateways are generally
more complex than switch or router. A
gateway can be implemented completely
in software, hardware, or in a
combination of both.
8/18/2019
Computer Networks
- PRONAYA BHATTACHARYA, CSE, ITNU
24
3-4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not
perfect. The imperfection causes signal impairment. This
means that the signal at the beginning of the medium is
not the same as the signal at the end of the medium.
What is sent is not what is received. Three causes of
impairment are attenuation, distortion, and noise.
Topics discussed in this section:
Attenuation
Distortion
Noise
3.25
Figure 3.25 Causes of impairment
3.26
Figure 3.26 Attenuation
3.27
Example 3.26
Suppose a signal travels through a transmission medium
and its power is reduced to one-half. This means that P2
is (1/2)P1. In this case, the attenuation (loss of power)
can be calculated as
A loss of 3 dB (–3 dB) is equivalent to losing one-half
the power.
3.28
Example 3.27
A signal travels through an amplifier, and its power is
increased 10 times. This means that P2 = 10P1 . In this
case, the amplification (gain of power) can be calculated
as
3.29
Example 3.28
One reason that engineers use the decibel to measure the
changes in the strength of a signal is that decibel
numbers can be added (or subtracted) when we are
measuring several points (cascading) instead of just two.
In Figure 3.27 a signal travels from point 1 to point 4. In
this case, the decibel value can be calculated as
3.30
Figure 3.27 Decibels for Example 3.28
3.31
Example 3.29
Sometimes the decibel is used to measure signal power
in milliwatts. In this case, it is referred to as dBm and is
calculated as dBm = 10 log10 Pm , where Pm is the power
in milliwatts. Calculate the power of a signal with dBm =
−30.
Solution
We can calculate the power in the signal as
3.32
Example 3.30
The loss in a cable is usually defined in decibels per
kilometer (dB/km). If the signal at the beginning of a
cable with −0.3 dB/km has a power of 2 mW, what is the
power of the signal at 5 km?
Solution
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.
We can calculate the power as
3.33
Figure 3.29 Noise
3.34
Example 3.31
The power of a signal is 10 mW and the power of the
noise is 1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as
follows:
3.35
Example 3.32
The values of SNR and SNRdB for a noiseless channel
are i.e whatever we transmit is exactly received.
We can never achieve this ratio in real life; it is an ideal.
For a perfectly noisy channel, Noise Factor will be
infinite, hence SNR = 0. i.e whatever we transmit is lost
in the channel.
3.36
DELAY ANALYSIS
Causes of end-to-end delay:




Processing delays
Buffer delays
Transmission delays
Propagation delays
TOTAL END TO END DELAY=Processing delays+ Buffer
delays+ Transmission delays+ Propagation delays
3.37
PROCESSING AND QUEUING DELAY
Processing Delay
• Required time to analyze a packet header and decide
where to send the packet (e.g. a routing decision)
• Inside a router this depends on the number of entries in the
routing table, the implementation of data structures, hardware
in use, etc.
• This can include error verification, such as IPv4, IPv6
header checksum calculations.
Queueing or Buffer Delays
• The time a packet is enqueued until it is transmitted
• The number of packets waiting in the queue will
depend on traffic intensity and of the type of traffic
(bursty or sustained)
• Router queue algorithms try to adapt delays to specific
preferences, or impose equal delay On all traffic.
3.38
TRANSMISSION DELAY AND PROPOGATION
DELAY
Transmission Delay
The time required to push all the bits in a MESSAGE on
the transmission medium in use.
B= bandwidth of channel/link, S=Size of message, Td= transmission
delay
Td = S/B
3.39
Propogation Delay
Once a bit is 'pushed' on to the transmission medium, the time
required for the bit to propagate to the end of its physical
trajectory
• The velocity of propagation of the circuit depends mainly on
the actual distance of the physical circuit
For d = distance, s = propagation velocity, Pd= propogation
delay
Pd = d/s
Some important formulae





3.40
For M hops and N packets –
Total delay = M*(Transmission delay + propagation delay)+ (M1)*(Processing delay + Queuing delay) + (N-1)*(Transmission delay)
For N connecting link in the circuit –
Transmission delay = N*L/R
Propagation delay = N*(d/s)
DELAY ANALYSIS
Ans is: 0.01secs propagation time+ 1 sec
transmission time+ 0.00002secs queuing
time+0.00001 secs propagation delay
3.41
Example 3.46
What are the propagation time and the transmission
time for a 2.5-kbyte message (an e-mail) if the
bandwidth of the network is 1 Gbps? Assume that the
distance between the sender and the receiver is 12,000
km and that light travels at 2.4 × 108 m/s.
Solution
We can calculate the propagation and transmission time
as shown on the next slide:
3.42
Example 3.46 (continued)
Transmission Time = Message Size/Bandwidth
First part = 2 X 106 / 56 X 103
= 35.71 secs.
Second Part = 2 X 106 / 1 X 106 = 2 secs.
3.43
Example 3.47
What are the propagation time and the transmission
time for a 5-Mbyte message (an image) if the bandwidth
of the network is 1 Mbps? Assume that the distance
between the sender and the receiver is 12,000 km and
that light travels at 2.4 × 108 m/s.
Solution
We can calculate the propagation and transmission
times as shown on the next slide.
3.44
Example 3.47 (continued)
Note that in this case, because the message is very long
and the bandwidth is not very high, the dominant factor
is the transmission time, not the propagation time. The
propagation time can be ignored.
3.45
QUES.1





3.46
Question : How much time will it take to send a packet
of size L bits from A to B (A---R1---R2---B) if Bandwidth
is R bps, propagation speed is t meter/sec and distance
b/w any two points is d meters (ignore processing and
queuing delay) ?
Ans: N = no. of links = no. of hops = no. of routers +1 = 3 ,File size = L bits ,Bandwidth = R bps,
Propagation speed = t meter/sec, Distance = d meters
Transmission delay = (N*L)/R = (3*L)/R sec
Propagation delay = N*(d/t) = (3*d)/t sec
Total time = 3*(L/R + d/t) sec
QUES.2
3.47
Answer
3.48
Average Queuing delay = (N-1)L/(2*R)
where N = no. of packets L=size of packet R=bandwidth
3.49
Queueing delay
Ques. Compute the average queueing delay in a ADSL
network if 1000 packets are processed with an average
packet length of 400bytes and bandwidth of the link is
2kbps.
 QT= (999*400*8)/(2*103)=1598.4 s.
Will the processing delay at router increase if pkt size
increases?.

3.50
Switching-Motivation



A network consists of many switching devices. In order to
connect multiple devices, one solution could be to have a point to
point connection in between pair of devices. But this increases
the number of connection.
The other solution could be to have a central device and connect
every device to each other via the central device which is
generally known as Star Topology.
Both these methods are wasteful and impractical for very large
network. The other topology also can not be used at this stage.
Hence a better solution for this situation is SWITCHING. A
switched network is made up of a series of interconnected nodes
called switches.
Taxonomy of Switched Networks
Circuit Switching
A circuit-switched network is made of a
set of switches connected by physical
links, in which each link is
divided into n channels by using FDM
or TDM
Explain Scenario 1
Explain Scenario 2
Circuit Switching
•
Circuit switching is a technique that directly connects the sender and the receiver in an unbroken path.
•
Phase 1-Connection Establishment
– With this type of switching technique, once a connection is established, a dedicated path exists between both ends until the
connection is terminated.
– Routing decisions must be made when the circuit is first established, but there are no decisions made after that time.
Phase 2- Data Transfer
– Once the connection has been initiated and completed to the destination device, the destination device must acknowledge that it
is ready and willing to carry on a transfer.
Phase 3- Connection Teardown
– After Data Transfer is complete, the Resources reserved during setup are released.
•
•
Delay in a Circuit-Switched Network
What is the delay in CSN in the above scenario?
Message Switching(Store-and-Forward
N/W)
With message switching there is no need to establish a
dedicated path between two stations.
• When a station sends a message, the destination
address is appended to the message.
• The message is then transmitted through the
network, in its entirety, from node to node.
• Each node receives the entire message, stores it in
its entirety on disk, and then transmits the message to
the next node.
• A message-switching node is typically a generalpurpose computer. The device needs sufficient
secondary-storage capacity to store the incoming
messages, which could be long.
• A variable time delay is introduced using this type of
scheme due to store- and-forward time, plus the time
required to find the next node in the transmission path.
Packet Switching
•Packet switching can be seen as a solution that tries to combine the advantages of
message and circuit switching and to minimize the disadvantages of both.
• There are two methods of packet switching: Datagram and virtual
circuit.
•In both packet switching methods, a message is broken into small parts, called
packets. Each packet is tagged with appropriate source and destination addresses.
• Since packets have a strictly defined maximum length, they can be stored in main
memory instead of disk, therefore access delay and cost are minimized.
• Also the transmission speeds, between nodes, are optimized.
• With current technology, packets are generally accepted onto the network on a
first-come, first-served basis. If the network becomes overloaded, packets are delayed
or discarded (``dropped''). The size of the packet can vary from depending on the
Network to which it is forwarded.
Packet Switching: Datagram
 Datagram packet switching is similar to message switching in
that each packet is a self-contained unit with complete
addressing information attached. This fact allows packets to
take a variety of possible paths through the network.
 So the packets, each with the same destination address, do
not follow the same route, and they may arrive out of
sequence at the exit point node (or the destination).
 Reordering is done at the destination point based on the
sequence number of the packets.
Delay Analysis in Datagram Networks
Packet Switching:Virtual Circuit
In virtual circuit, the route between stations does not mean that this is a
dedicated path, as in circuit switching.
• A packet is still buffered at each node and queued for output over a line.
• The difference between virtual circuit and datagram approaches:
•
 With virtual circuit, the node does not need to make a routing decision for each
packet.
 It is made only once for all packets using that virtual circuit.
VC's offer guarantees that the packets sent
 arrive in the order sent with no duplicates or omissions
 with no errors (with high probability)
regardless of how they are implemented internally.
Virtual Circuit and VCI
Figure 8.13 Source-to-destination data transfer in a virtual-circuit network
8.64
Figure 8.14 Setup request in a virtual-circuit network
8.65
Figure 8.15 Setup acknowledgment in a virtual-circuit network
8.66
Delay Analysis in Virtual Circuit
Comparison chart
Exercise:
•
•
•
•
There is a network having bandwidth of 1 MBps.
A message of size 1000 bytes has to be sent.
Packet switching technique is used.
Each packet contains a header of 100 bytes.
• Out of the following, in how many packets the message must be divided so that total
time taken is minimum• 1 packet
• 5 packets
• 10 packets
• 20 packets[GATE-2014]
Important Points to NOTE
• NOTE
• While calculating the total time, we often ignore the
propagation delay.
• The reason is in packet switching, transmission delay
dominates over propagation delay.
• This is because each packet is transmitted over the link at
each hop.
Case-01: Sending Message in 1 PacketIn this case, the entire message is sent in a single packet.
Size Of PacketPacket size
= 1000 bytes of data + 100 bytes of header
= 1100 bytes
Transmission DelayTransmission delay
= Packet size / Bandwidth
= 1100 bytes / 1 MBps
= 1100 x 10-6 sec
= 1100 μsec
= 1.1 msec
Total Time TakenTotal time taken to send the complete message from sender to receiver
= 3 x Transmission delay
= 3 x 1.1 msec
= 3.3 msec
Exercise:
In a packet switching network, packets are routed from source to
destination along a single path having two intermediate nodes. If the
message size is 24 bytes and each packet contains a header of 3 bytes,
then the optimum packet size is-[GATE-2017]
• 4 bytes
• 6 bytes
• 7 bytes
• 9 bytes
Option 1
• Let bandwidth of the network = X Bps and 1 / X = a
• Option-A: Packet Size = 4 Byte
• In this case The entire message is divided into packets of size 4 bytes. These packets are then sent
one after the other.
• Data Sent in One Packet• Data size = Packet size – Header size = 4 bytes – 3 bytes = 1 byte
• Thus, only 1 byte of data can be sent in each packet.
• Number of packets required = Total data to be sent / Data contained in one packet = 24 bytes / 1 byte
= 24 packets
• Transmission delay = Packet size / Bandwidth = 4 bytes / X Bps = 4a sec
• Time Taken By First Packet- Time taken by the first packet to reach from sender to receiver = 3 x
Transmission delay = 3 x 4a sec = 12a sec
• Time Taken By Remaining Packets• Time taken by the remaining packets to reach from sender to receiver = Number of remaining
packets x Transmission delay = 23 x 4a sec = 92a sec
• Total Time Taken- Total time taken to send the complete message from sender to receiver = 12a sec
+ 92a sec = 104a sec
Similarly calculate for other options
•
•
•
•
•
•
•
•
•
Option-B: Packet Size = 6 bytes
Option-C: Packet Size = 7 bytes
Option-D: Packet size = 9 Bytes
ObservationsFrom here,
Total time taken when packet size is 4 bytes = 104a sec
Total time taken when packet size is 6 bytes = 60a sec
Total time taken when packet size is 7 bytes = 56a sec
Total time taken when packet size is 9 bytes = 54a sec
Problem with Crossbar Switch
M * N crosspoints i.e O(MN) or if M=N then O(N2), suppose m=1000 and n=10000, then
1000*10000 crosspoints, normally only 25% of the crosspoints are active at any time.
Figure 8.18 Multistage switch
8.76
Note
Stage 1: N/n crossbars of size n x k.
Stage 2: k crossbars of size N/n x N/n.
Stage 3: N/n crossbars of size k x n.
In a three-stage switch, the total
number of cross points is thus
2kN + k(N/n)2
which is much smaller than the number of
crosspoints in a single-stage switch (N2).
8.77
Example 8.3
Design a three-stage, 200 × 200 switch (N = 200) with
k = 4 and n = 20.
Solution
In the first stage we have N/n or 10 crossbars, each of size
20 × 4. In the second stage, we have 4 crossbars, each of
size 10 × 10. In the third stage, we have 10 crossbars,
each of size 4 × 20. The total number of crosspoints is
2kN + k(N/n)2, or 2000 crosspoints. This is 5 percent of
the number of crosspoints in a single-stage switch (200 ×
200 = 40,000).
8.78
PROBLEM OF BLOCKING IN MULTISTAGE SWITCH
"In multistage switching, blocking refers to times when one
input cannot be connected to an output because there is
no path available between them—all the possible
intermediate switches are occupied. One solution to
blocking is to increase the number of intermediate
switches based on the Clos criteria."
According to the Clos criterion for NonBlocking Switch:
n = (N/2)1/2
k > 2n – 1
1/2 – 1]
Cross
points
≥
4N
[(2N)
8.79
Example 8.4
Redesign the previous three-stage, 200 × 200 switch,
using the Clos criteria with a minimum number of
crosspoints.
Solution
We let n = (200/2)1/2, or n = 10. We calculate k = 2n − 1 =
19. In the first stage, we have 200/10, or 20, crossbars,
each with 10 × 19 crosspoints. In the second stage, we
have 19 crossbars, each with 10 × 10 crosspoints. In the
third stage, we have 20 crossbars each with 19 × 10
crosspoints. The total number of crosspoints is 20(10 ×
19) + 19(10 × 10) + 20(19 ×10) = 9500.
8.80