Transcript Document 7921853

IEEE 802.3 Ethernet
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Ethernet has been adapted over the
years to run at a variety of speeds and
over a range of physical media.
Transmission media: coaxial cable,
twisted pairs, and fiber optics
Topology: star, linear bus, spine, tree,
and segmented bus
Three states of the Ethernet: contention,
transmission, or idle.
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CSMA/CD (Ethernet)
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Local-area network technology
Developed in the mid-1970's by
researchers at the Xerox Palo Alto
Research Center (PARC)
The Ethernet is a working example of
the more general CSMA/CD
CSMA/CD: Carrier Sense, Multiple
Access with Collision Detection
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IEEE 802.3 Baseband
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Coaxial cable/twisted pairs
Running digital signal
Entire bandwidth used by the signal
Digital Manchester encoding
10Base5, 10Base2, 10BaseT,
100BaseT
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IEEE 802.3 Broadband
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Coaxial cable
Analog encoding
Require RF modem
Multiple channel using FDM possible for
data, video, audio
Analog PSK, 10Broad36
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10base5 Ethernet (thick net)
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Cable diameter: 0.4-in diameter, 50ohms impedance
Data rate of 10Mbps
Maximum cable segment length: 500
meters
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10base5 Ethernet (thick net)
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Network span: 2500 meters
Node spacing: 2.5 meters; stations
attach to the cable by means of a tap,
with the distance between any two taps
being a multiple of 2.5 meters.
Maximum number of nodes per
segment: 100 taps
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10Base2 Ethernet (thin net)
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A lower cost network system compared
with 10Base5
Cable diameter: 0.25-in diameter, 50ohms impedance
Data rate of 10Mbps
Maximum cable segment length: 185
meters
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10Base2 Ethernet (continue)
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Network span 1000 meters
Node spacing: 0.5 meters; stations
attach to the cable by means of a tap,
with the distance between any two taps
being a multiple of 0.5 meters.
Maximum number of nodes per
segment: 30 taps
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10BaseT Ethernet
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A lower cost network system using an
intelligent hub with a port for each
station
Data rate of 10Mbps
Stations are connected to a hub by fourpair RJ-45 cable (eight-wire unshielded
twisted-pair cable)
Cable diameter: 0.25-in diameter, 50ohms impedance
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10BaseT Ethernet (continue)
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Maximum cable segment length: 185
meters
Network span 1000 meters
Node spacing: 0.5 meters; stations
attach to the cable by means of a tap,
with the distance between any two taps
being a multiple of 0.5 meters.
Maximum number of nodes per
segment: 30 taps
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10Broad36
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Data rate: 10 Mbps
Coaxial cable (75 ohms)
Broadband
Max segment length: 1800 m
Network span 3600 m
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Other Devices
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AUI (Attachment Unit Interface) Cables
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A 15-wire cable with plugs that perform the
physical interface functions between the
station and the transceiver
Each end of an AUI terminates in a DB-15
(15-pin) connector
Maximum length of 50 meters
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Other Devices
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Transceiver Tap
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A connecting mechanism that allows the
transceiver to tap into the cable line at any
point
BNC-T connector
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A T-shaped device with three ports: use for
the NIC and one each for the input and
output ends of cables.
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Access Method: CSMA/CD
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A multiple access network: a set of
nodes or stations send and receive
frames over a shared link
Time-division multiplexed bus
Carrier Sense - means that all the
nodes can distinguish between an idle
and a busy link
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Access Method: CSMA/CD
(continue)
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Collision Detection - means that a node
listens as it transmits, and can therefore
detect a frame when it is transmitting
has interfered (collided) with a frame
transmitted by another node
When a station is in transmission,
according to the CSMA/CD scheme no
other station may transmit.
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Collision Detection
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If two stations transmit at once, their
signals interfere with each other
Since each sender listens (sense the
carrier) before the transmission, they
know that there has been a collision
Both of the stations stop and wait a
random amount time before next
attempt to transmit
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10BaseT Ethernet Hub
Server
Hub
Ethernet Card
Station A
10BASE-T w all plate
Print Server
Printer
10BASE-T w all plate
Transmit
Ethernet Card
Station B
10BASE-T w all plate
Ethernet Card
Station C
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CSMA/CD Access Rules
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All stations listen to the transmission
medium. If the medium is idle, transmit;
otherwise, go to step 2
If the medium is busy, continue to listen
until the channel is idle, then transmit
immediately
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CSMA/CD Access Rules
(continue)
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If a collision is detected during
transmission, transmit a brief jamming
signal to ensure that all stations know
that there has been a collision and then
cease transmission
After transmitting the jamming signal,
wait a random amount of time, then
attempt to transmit again
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Meidum
Idle?
Try
again
Wait for a
random
amount of
time
Try
again
Yes
No
Transmit
Channel is
Busy
Transmit a
brief
jamming
Signal
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MAC Frame Format
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Preamble - 7 bytes
Start Frame Delimiter (SFD) - 1 byte
Destination Address (DA) - 2 or 6 bytes
Source Address (SA) - 2 or 6 bytes
Length of PDU (Protocol Data Unit) - 2
bytes
LLC or 802.2 Data frame - 46 to 1500
bytes
CRC or FCS - 4 bytes
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MAC Frame Format (continue)
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Preamble
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7-bytes (56 bits) pattern of alternating 0s
and 1’s (1010101..)
Alert the receiving station to the incoming
frame, and enable it to synchronize its
input timing.
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MAC Frame Format (continue)
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Start Frame Delimiter (SFD)
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The second field (one byte 10101011) of
the frame indicates the beginning of the
frame.
It tells the receiver that the next thing
coming is the destination address of the
receiver.
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MAC Frame Format (continue)
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Destination Address (DA)
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When the packet reaches the target
network, the DA field contains the physical
address of the destination station
DA filed is either 2- or 6-byte for holding a
unique physical address, a group address,
or a global address of the packet’s next
destination
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MAC Frame Format (continue)
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Destination Address (DA)
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DA address may hold the router connection
address at the beginning if the packet must
cross from one LAN to another
Source Address (SA)
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SA field is either 2- or 6-bytes for holding
the frame sender’s address
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MAC Frame Format (continue)
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Length of PDU (Protocol Data Unit)
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two bytes length field for indicating the
number of bytes of the data field.
LLC or 802.2 Data frame
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Entire data frame from 46 to 1500 bytes
long
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MAC Frame Format (continue)
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CRC or FCS
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4-byte field that holds a CRC-32 (Cyclic
redundancy check) for checking the frame
integrity.
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Example 1: Time needed for
collision detection
Question: Two stations begin transmitting
at exactly the same time. How long will
it take to realize that there has been a
collision? Contention period, delay, and
throughput.
Ans. The minimum time for detecting the
collision is the time that it takes to
propagate from one station to the other.
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Example 2: Elapses time and
throughput
Consider the transfer of a file containing one
million characters from one station to
another. What is the total elapsed time and
effective throughput for the 10BaseT star
topology with a data rate of 10 Mbps?
Assume that the network setup time is
negligible.
Solution:
8 bit/char x 1 M char  10 Mbits/sec = 0.8
seconds.
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Example 3: What is a Carrier
Detection?
To understand the collision detection
process, we imagine that the station’s
hardware must listen to the cable while
it is transmitting. If what it reads back is
different from it is putting out, it knows a
collision is occurring. A station wants to
acquire the medium it must first output a
1 bit, if the channel is busy the return bit
is 0, otherwise the bit is 1.
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Ethernet
Busy = 1
Idle = 0
XOR
Transmit
=1
Ethernet Card
Channel Idle = 1
Channel Busy = 0
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Example 4: Slot Time
What is the most amount of time needed
for a station to know that if it be granted
the access to the channel? This time is
normally called slot time.
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Station A
Station B
Ethernet Card
Ethernet Card
Ethernet
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Example 4: Slot Time
(continue)
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Let Tau be the time for a signal to propagate
between the two farthest stations.
At time t0, station A begins transmitting.
At time t0 - t, before the signal arrive at
station B which is the farthest station, the
station B also begins transmission.
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Example 4: Slot Time
(continue)
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The station B detects the collision
instantaneously.
But the jamming signal did not get back to the
station A until (2Tau - t).
So that the most amount of the time needed
to be sure that it has seized the channel is it
has transmitted for 2Tau time without hearing
a collision.
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Example 5: Effective data rate
calculation
What is the effective data rate, excluding
overhead? Assuming that there are no
collisions for a 10 Mbps Ethernet with
the following specification:
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Example 5: Effective data rate
calculation
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1 km long cable that has a propagation
speed of 200 m/sec. Data packets are
1024 bits long, including 32-bit header,
CRC check sum and other overhead.
The first bit slot after a successful
transmission is reserved for the receiver
to capture the channel to send a 32-bit
acknowledgement packet.
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Example 5 (continue)
Solution:
 The round trip propagation time of the
cable is
(200 m/sec  1 km)  2 = 10 s
 And we know that a complete
transmission has four phases:
 Sender seizes cable (10 sec)
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Example 5 (continue)
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Transmit data ( 1024 bits  10 Mbps =
102.4 sec)
Receiver seizes cable (10 sec)
Ack. Sent (32 bits  10 Mbps = 3.2
sec)
The sum of these time is 125.6 sec.
In this period (1024 - 32 = 992) data bits
are sent, for a rate of 7.9 Mbps
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Example 5 (continue)
Repeat the same calculation for a data packet
of 256 bit long.
Solution:
 Now, the packet size is 256 bit and 32-bit is
the overhead. The transmit time is
 256 bits  10 Mbps = 25.6 sec
 The sum of these time is then 48.8 sec =
25.6 sec + Sender seize time 10 sec +
Receiver seize time 10 sec + Ack
Send 3.2 sec
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Example 5 (continue)
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In this period (256 - 32 = 224 ) data bits are
sent.
The data rate is 224 bits  48.8 sec = 4.7
Mbps.
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Example 6: Network topology
and wiring design
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A 7-floor office building has 15 adjacent
offices per floor.
Each office contains a wall socket for a
network drop in the front wall, so the
sockets from a rectangular grid in the
vertical plane, with a separation of 4m
between sockets, both horizontally and
vertically.
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Example 6: (continue)
Assuming that it is feasible to run a straight
cable between any pair of sockets,
horizontally, vertically, or diagonally.
 How many meters of cable are needed to
connect all sockets using a star configuration
with a single hub in middle?
An Ethernet?
A ring net?
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T
1
T
2
T
T
6
7
T
T
T
3
4
5
T
T
9
10
T
T
T
13
14
15
T
T
T
3
4
5
HUB
8
T
T
11
12
4Th Floor
T
1
T
2
T
T
T
T
T
6
7
8
9
10
T
T
T
T
T
11
12
13
14
15
1, 2, 3, 5, 6, and
7Th Floor
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Example 6: (continue)
Solution: Number the floors 1 through 7.
In the star configuration, the hub is in the
middle of 4th floor. Cable are needed for
each of the 7 x 15 - 1 = 104 rooms. The
separation between sockets is 4 meters.
Wire length for the 4th floor can be
calculated through the following
equations:
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Total length = 1382 meters
15
4 ( j  8) 2
j 1
7 15
4  (i  4)  ( j  8)  1832 meters
2
2
i 1 j 1
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Example 6: (continue)
From 8th cell to 1, and 15 are of the same
length = 4 x 7
From 8th cell to 5, and 11 are of the same
length = 4 x 3
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Example 6: (continue)
len = 0;
for j =1: 15
for i = 1: 7
a = (i-4)^2;
b = (j-8)^2;
len = len + sqrt(a+b);
end
end
len = 4 * len
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Example 6: (continue)
With an Ethernet bus:
 7 floors of horizontal cables
 Each floor has 4 m x 15 - 4m = 56 m
 Vertical cable 4 m x 6 = 24 m
 Total length = 7 x 56 m + 24m = 416 m
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Example 6: (continue)
With a ring net:
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(1,1), (15,1), (15,7), (1,7), (1,2),
(14,2), and then back to (1,1)
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The total length is about 462 m.
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