Transcript CSC 335 Data Communications and Networking Lecture 7: Local Area Networking Dr. Cheer-Sun Yang Fall 2000

CSC 335
Data Communications
and
Networking
Lecture 7: Local Area Networking
Dr. Cheer-Sun Yang
Fall 2000
Motivation
• Up to this point, we’ve talked about point-to-point
communication.
• We many need to connect many computers
together.
• Local Area Network(LAN): if they are located in a
relatively close geographic area.
• Metropolitan Area Network (MAN) : extends over
entire city
• Wide Area Network (WAN) : extends across
public switching network.
LAN Applications (1)
• Personal computer LANs
– Low cost
– Limited data rate
• Back end networks and storage area networks
– Interconnecting large systems (mainframes and large
storage devices)
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High data rate
High speed interface
Distributed access
Limited distance
Limited number of devices
LAN Applications (2)
• High speed office networks
– Desktop image processing
– High capacity local storage
• Backbone LANs
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Interconnect low speed local LANs
Reliability
Capacity
Cost
LAN Architecture
• Topologies
• Protocol architecture
– Physical Layer
– Media access control
– Logical Link Control
Topologies
• Bus: A single communication line, typically
a twisted pair, coaxial cable, or optical fiber,
represents the primary medium.
• Ring: packets can only be passed from one
node to it’s neighbor.
• Star: A hub or a computer is used to connect
to all other computers.
• Tree: no loop exists (logical connection).
LAN Topologies
Frame Transmission - Bus LAN
Ring Topology
• Repeaters joined by point to point links in closed
loop
– Receive data on one link and retransmit on another
– Links unidirectional
– Stations attach to repeaters
• Data in frames
– Circulate past all stations
– Destination recognizes address and copies frame
– Frame circulates back to source where it is removed
• Media access control determines when station can
insert frame
Frame
Transmission
Ring LAN
Star Topology
• Each station connected directly to central
node
– Usually via two point to point links
• Central node can broadcast
– Physical star, logical bus
– Only one station can transmit at a time
• Central node can act as frame switch
Bus
and
Tree
Multipoint medium
•
• Transmission propagates throughout medium
• Heard by all stations
– Need to identify target station
• Each station has unique address
• Full duplex connection between station and tap
– Allows for transmission and reception
• Need to regulate transmission
– To avoid collisions
– To avoid hogging
• Data in small blocks - frames
• Terminator absorbs frames at end of medium
Protocol Architecture
• Protocol layering (IEEE 802.X)
– Physical Layer
– Media access control (MAC) Sublayer
– Logical link control (LLC)
IEEE 802.X
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IEEE 802.3 : Ethernet LAN
IEEE 802.4 : Token Bus LAN
IEEE 802.5 : Token Ring LAN
Other Ring Networks: FDDI, Slotted Rings.
IEEE 802.6 : Distributed Queue Dual Bus
(DQDB) MAN standard.
IEEE 802 vs. OSI
Fig 6.4
LAN Protocols in Context
802 Physical Layer
Design Issues
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Encoding/decoding
Preamble generation/removal
Bit transmission/reception
Transmission medium and topology
802 Physical Layer
• Required hardware for connecting a PC to
Ethernet directly:
– Transceiver
– Attachment Unit Interface (AUI) cable
– Network Interface Card (NIC) also known as Network
Adapter
• Required hardware for connecting a PC to a
remote computer: modem (with the help of PPP)
Bus LANs
• Signal balancing
– Signal must be strong enough to meet receiver’s
minimum signal strength requirements
– Give adequate signal to noise ration
– Not so strong that it overloads transmitter
– Must satisfy these for all combinations of sending and
receiving station on bus
– Usual to divide network into small segments
– Link segments with amplifies or repeaters
Transmission Media
• Twisted pair
– Not practical in shared bus at higher data rates
• Baseband coaxial cable
– Used by Ethernet
• Broadband coaxial cable
– Included in 802.3 specification but no longer made
• Optical fiber
– Expensive
– Difficulty with availability
– Not used
• Few new installations
– Replaced by star based twisted pair and optical fiber
Baseband Coaxial Cable
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Uses digital signaling
Manchester or Differential Manchester encoding
Entire frequency spectrum of cable used
Single channel on cable
Bi-directional
Few kilometer range
Ethernet (basis for 802.3) at 10Mbps
50 ohm cable
10Base5
• Ethernet and 802.3 originally used 0.4 inch
diameter cable at 10Mbps
• Max cable length 500m
• Distance between taps a multiple of 2.5m
– Ensures that reflections from taps do not add in
phase
• Max 100 taps
• 10Base5
10Base2
• Cheapernet
• 0.25 inch cable
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More flexible
Easier to bring to workstation
Cheaper electronics
Greater attenuation
Lower noise resistance
Fewer taps (30)
Shorter distance (200m)
Cable Specifications for 802.3
• 10BaseT: 10 Mbps, baseband, unshield
twisted
• 10Base2: 10Mbps, Cat. 2 coaxial
• 10Base5: 10 Mbps, Cat. 5, Cat. 5e coaxial
• 100BaseTX: 100 Mbps, twisted cable (Fast
Ethernet)
• 10Broad36: maximum segment length 3600
meters
Gigabit Ethernet
• 1000Base-SX
– Short wavelength, multimode fiber
• 1000Base-LX
– Long wavelength, Multi or single mode fiber
• 1000Base-CX
– Copper jumpers <25m, shielded twisted pair
• 1000Base-T
– 4 pairs, cat 5 UTP
• Signaling - 8B/10B
Connectors
• T-connector: used to form a bus topology
• RJ-45 connectors: for connecting a PC to
another PC, Ethernet, or hub.
– Cross-over: a direct connection to another PC
– Straight-through: connection with the Ethernet
jack or hub.
Repeaters
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Transmits in both directions
Joins two segments of cable
No buffering
No logical isolation of segments
If two stations on different segments send at
the same time, packets will collide
• Only one path of segments and repeaters
between any two stations
Media Access Control Sublayer
• Assembly of data into frame with address and
error detection fields
• Disassembly of frame
– Address recognition
– Error detection
• Govern access to transmission medium
– Not found in traditional layer 2 data link control
– Also known as Contention protocols (section 3.4)
Collision vs. Contention
• When the communication link is used by one
station to transmit a frame, another station
connecting to the same link tries to send a packet–
collision
• Contention: accessing the medium with the
consideration that a collision may occur.
• Contention Protocols: the protocol is designed to
deal with collision using contention.
• Collision-free Protocols: the protocol is designed
so that collision will not occur.
Contention Protocols
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Pure ALOHA
Slotted ALOHA
Carrier Sense Multiple Access (CSMA)
Persistent and non-persistent CSMA
CSMA with Collision Detection
(CSMA/CD)
Collision-Free Protocols
• A Bit-Map Protocol: reservation protocol
• Binary Countdown
Pure Aloha
• Packet Radio
• When station has frame, it sends
• Station listens (for max round trip time)plus small
increment
• If ACK, fine. If not, retransmit
• If no ACK after repeated transmissions, give up
• Frame check sequence (as in HDLC)
Pure Aloha(cont’d)
• If frame OK and address matches receiver, send
ACK
• Frame may be damaged by noise or by another
station transmitting at the same time (collision)
• Any overlap of frames causes collision
• Max utilization 18% (WHY?)
The Efficiency of Pure Aloha
G = the traffic measured as the average number of frames
generated per slot
S = the success rate, success frame / slot
Pr[k frames are generated] = G k e –G / k !
This is called a probability distribution function(pdf) for
Poisson distribution. (e = 2.7818…)
S = Pr[no frame is generated]= e -G
= G e –2G (pure Aloha)
S = G e –G (slotted Aloha)
The Efficiency of Pure Aloha
If there is no negative acknowledgement frame received after
sending out one frame, the transmission is successful. So
P0 = Pr[no frames are generated in 2 time slots]
= e -G * e –G
= e –2G
S=G*P0
= G e –2G (pure Aloha)
The Efficiency of Pure Aloha
S = G * P 0= G e –2G (pure Aloha)
We need to find the value of G such that S is
maximized.
S’ = G (-2) e –2G + e –2G = (1 – 2G) * e –2G
Let S’ = 0 => G = ½
When G = ½, S = 1/ 2e = 0.184 = 18%
Slotted ALOHA
• A computer is not allowed to send until the
beginning of the next slot.
• Time in uniform slots equal to frame transmission
time
• When a frame is allowed to be transmitted, there is
no collision.
• Need central clock (or other sync mechanism)
• Transmission begins at slot boundary
• Max utilization 37% (WHY?)
The Efficiency of Slotted Aloha
If there is no other frame received after sending out one frame,
the transmission is successful. So
P0 = Pr[no frames are generated in one time slots]
= e -G
S=G*P0
= G e –G (slotted Aloha)
The Efficiency of Slotted Aloha
S = G * P 0= G e –G (slotted Aloha)
We need to find the value of G such that S is
maximized.
S’ = G (-1) e –G + e –G = (1 –G) * e –G
Let S’ = 0 => G = 1
When G = 1, S = 1/ e = 0.368 = 37%
Carrier Sense Multiple Access
(CSMA) Protocols
• Protocols in which stations listen for a
carrier (i.e., a transmission) and act
accordingly are called carrier sense
protocols.
– 1-persistent CSMA
– Non-persistent CSMA
– p-persistent CSMA
CSMA
• Propagation time is much less than
transmission time
• All stations know that a transmission has
started almost immediately
• First listen for clear medium (carrier sense)
If Busy?
• If medium is idle, transmit
• If busy, listen for idle then transmit
immediately
• No ACK then retransmit
• If two stations are waiting, it is called a
collision.
1-persistent CSMA
• When a station has data to send, it first
listens to the channel to see if anyone else is
transmitting at that moment.
• If the channel is busy, the station waits until
it becomes idle.
• The station retransmits with a probability of
1 when it finds that the channel is idle.
Non-persistent CSMA
• When a station has data to send, it first
listens to the channel to see if anyone else is
transmitting at that moment.
• If the channel is busy, the station waits until
it becomes idle.
• The station does not keep trying. It waits for
a random number of time and retries.
Non-persistent CSMA
• This applies to slotted channels. When a station
has data to send, it first listens to the channel to
see if anyone else is transmitting at that moment.
• If the channel is idle, it transmits with a
probability p. With a probability of 1-p, it defers
until the next slot.
• If the next slot is also idle, it transmits or defers
again with probability p and q.
CSMA
• Max utilization depends on propagation time
(medium length) and frame length. Longer frame
and shorter propagation gives better utilization.
• Collisions still can be a problem, especially with
p-persistent CSMA.
• One way to reduce the frequency of collision with
CSMA is to lower the probability that a station
will send when a previous is done. (Fig. 3.26)
• Smaller values of p => fewer collision.
Any Other Way?
• Is there another way to improve the
successful rate?
• Yes if there is a way to detect collision prior
to transmission.
• Why is this faster?
Collisions with and without
Detection
• Without collision detection, a station must
send and then wait for 2 time slots before
another attempt to send.
• With collision detection, a station can stop
transmission if collision detection requires
less time than sending a frame.
Collision Detection
• On baseband bus, collision produces much higher
signal voltage than signal
• Collision detected if cable signal greater than
single station signal
• Signal attenuated over distance
• Limit distance to 500m (10Base5) or 200m
(10Base2)
• For twisted pair (star-topology) activity on more
than one port is collision
• Special collision presence signal
CSMA/CD
• With CSMA, collision occupies medium for
duration of transmission
• Stations listen while transmitting
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If medium idle, transmit
If busy, listen for idle, then transmit
If collision detected, jam then ease transmission
After jam, wait random time then start again
– Binary exponential back off
CSMA/CD
Operation
Binary Exponential Back Off
• If a station’s frame collides for the first time, wait
0 or 1 time slot (chosen randomly) before trying
again.
• If it collides a second time, wait 0, 1, 2, or 3 slots
(again, chosen randomly).
• After a third collision, wait anywhere from 0 to 2 n
–1 slots if n <= 10, if n > 10, wait between 0 to
1024 (2 10) slots.
• After 16 collisions, give up. Further recovery is up
to the upper layer, such as a user.
IEEE 802.3 Frame Format
Frame Format (802.3)
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Start of frame delimiter: 10101011
Destination address
Source address
Data length field
Data field
Pad field: the data field must be at least 46 octets.
Frame check sequence: using 32-bit CRC.
Efficiency of 802.3(p.375)
P: the probability that a frame is sent without
a collision
Ps: the probability that a station sends
The probability of a collision = 1 – P.
The probability of a transmission requiring
exactly N attempts =
= N-1 collisions followed by a success
= N * Ps * ( 1- Ps) N - 1
Efficiency of 802.3
(p.376-377)
We would like to know under what
conditions the largest number of frames
are sent successfully.
Efficiency of 802.3(p.375)
The probability of a transmission requiring
exactly N attempts =P
= N * Ps * ( 1- Ps) N – 1
dP/dPs = N(1-Ps)N-1 + N Ps (N-1)(1-Ps)N-2
(1-Ps) N-2 [1- Ps – Ps (N –1) ]
Let dP/dPs = 0 => Ps = 1/N
=N
Efficiency of 802.3
How many time slots has passed before a
frame is sent successfully?
The Efficiency of 802.3
The contention period = the number of time
slots passed before a successful transmission
The Efficiency of 802.3
Assume that the probability of a success in
each attempt = p.
The probability of a collision = 1 – p.
The probability of a transmission requiring
exactly i+1 attempts = P(i+1)
= i collisions followed by a success
= p (1- p) i
The Efficiency of 802.3

Given i * x i  x /(1  x) 2
i 0
The contention period

=
 i * (1  p) * p  p * (1  p) / p
i
i 0
= (1-p)/p
2
The Efficiency of 802.3
The contention period (C)
= (1/p) –1
0 <= p <= 1 (p: the successful rate)
If p -> 1 C -> 0
If p -> 0, C = large
We’ve found that when Ps=1/N, p is maximized.
So, C = (1-1/N)1-N -1 when p is maximized.
If N->large, C = 2.718 – 1 = 1.718 (close to 2).
The Utilization Rate of Ethernet
The percent utilization (U) does not depend on
the number of stations in practice. A station
will try to send regardless of how many other
stations there are. The previous result often is
used as benchmarks against which measures
are made to estimate efficiency.
The Utilization Rate of Ethernet
The percent utilization (U) is defined as the
amount of time spent on transmitting a frame
as a percentage of the total time spent on
contending and transmitting. Assume:
R = transmission rate
F = number of bits in a frame
T = slot time
F
So
R
*100%
U=
F
R
 T *C
Required Reading
• Section 6.1, 3.4, 6.2