Data Link and Physical Layers

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Data Link and Physical Layers
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(No Transcript)
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Data Link and Physical Layers

• Data Link
• Logical Link Control
• Medium Access
• Physical
• Still protocols, not the physical cable

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Medium Access Control

• Contention based vs. controlled access
• Contention
• Aloha, Ethernet (IEEE 802.3)
• Controlled access
• Token based protocols (token ring, token bus)

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Goals

• Minimize Wait
• Maximize use of resources
• These are often conflicting goals. If the
  customer never has to wait at all, we will
  probably have servers idle some of the time. If
  the servers are always busy, customers will have
  to wait sometimes.
• The trick is to compromise in such a way that
  service is good and resources are well used.

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Aloha

• One central site and many satellite sites
• Satellite sites cannot hear each others
  transmissions -- all go through the central site.
• If two or more of the satellite sites transmit at
  the same time, the transmission is garbled and
  completely lost

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Aloha model
Remember, any overlapping transmission means
failure.
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Can this possibly work?

• What is the probability of successful
  communication under these conditions?
• First, we need to understand what is required for
  success
• A transmission must begin while no other
  transmissions are active
• No other transmission may begin while this one is
  active.

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Probability of success

• Lets call the duration of the transmission t
• We need to be sure that, at the time we start to
  transmit
• no other station has started transmitting within
  the previous time interval, t, and
• no other station will start transmitting for a
  time interval, t, after we begin to transmit.
• How can we possibly know what the other stations
  have done or will do?

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A probability model

• This is not an uncommon problem. To predict the
  number of servers needed in a restaurant or a
  store, someone has to predict how many people
  will show up in certain periods of time. The
  problem has been studied and there are
  well-analyzed models.
• The Poisson model is good at analyzing potential
  results of some kinds of unpredictable events
  over time.

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Poisson

• The Poisson distribution predicts the number of
  events to occur in a time period.

? the arrival rate the number of frames per
unit of time t time duration k, n numbers of
frames (events, in general) The Poisson
distribution is Pn(t) (? t)n e- ? t /n!
So, if we know the rate of occurrence of some
event, we can predict the likelihood of exactly n
occurrences in a period of time t
Note that the occurrences are random, not
regularly scheduled. There will be periods of
time with no occurrences, and others with many
occurrences.
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Examples of the Poisson model

• Suppose a new piece of equipment has a rate of
  failure of 2 per 1000 hours. What is the
  probability of 2 failures in a single hour?
• ? 2/1000 n2 t1 hour
• Pn(t) (? t)n e- ? t /n!
• P2(1) (2/1000 1)2 e- 2/1000 1 /2!
• P2(1) 0.00000199600399733467

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Poisson model for Aloha

• Pn(t) (? t)n e- ? t /n!
• ? is the rate at which frames are sent
• t is the time under consideration
• n is the number of frames sent in the time
  period.
• We need 0 frames sent (by others) in a period 2t
• P0(2t) (? 2t)0 e- ? 2t /0! e- ? 2t

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Probability of success

• The probability of success will clearly depend on
  the amount of traffic
• arrival rate and time remain in the equation
• P0(2t) (? 2t)0 e- ? 2t /0! e- ? 2t
• G ? t is called the offered load and is a
  measure of how busy the resource is.
• P0(2t) (2G)0 e- 2G /0! e- 2G
• So, the probability of success for our
  transmission is e- 2G

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Throughput
Throughput (S) is a measure of how well we serve
the actual load on the system. SG e- 2G The
offered load times the probability of delivering
the frame successfully
The value of S when the slope turns down is
0.183939721 Aloha works perfectly well as long as
the throughput is not more than 18.4 of capacity
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Aloha and its descendants

• Pure alohapacket vulnerable for two times the
  packet transmission time
• Slotted aloha reduced the vulnerable period to
  one times the packet transmission time SG e- G

Throughput drops off at G 1 max value is
0.367879 usage limited to 36.8 of capacity
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Improving performance

• Slotted aloha allows use of more of the capacity
  of the data channel at the cost of more wait for
  the senders.
• Even if no one else is transmitting, the would-be
  sender must wait for the next slot boundary.
• Balance overall efficiency of the system with
  needs of the individual users
• Problem was that the stations could not hear each
  others transmission. What happens if we fix
  that?

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CSMA Carrier Sense Multiple Access

• Also called Listen Before Talk
• listen if channel clear, transmit otherwise
  wait
• vulnerability reduced to the time it takes most
  remote system to know about a transmission
• depends upon the fact that propagation delay is
  much less than transmission time
• variations
• non persistent
• p-persistent
• 1-persistent
• Still dont know about interference that occurs
  during transmission due to almost simultaneous
  transmissions by several stations.

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Add Collision Detection

• CSMA/CD or Listen While Talk
• requires the sender to listen as long as needed
  to detect any possible collision
• requires a recovery process
• binary exponential backoff
• Scalability problem throughput decreases as
  transmission rates increase.
• Simplicity in initialization, station addition
  and deletion, fault management
• This is Ethernet/IEEE 802.3

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Ethernet/IEEE 802.3
7 1 2/6 2/6 2 46-1500
4 Preamble SFD DA SA L Data FCS
Bytes field
Preamble 56 bits alternating 10 used to
synchronize the receiver SFD Start Frame
Delimiter. 10101011 to signal beginning of
the transmission DA/SA
Destination and Source Addresses L Length
of the data part Data Minimum length
required for proper operation. Padding
used if needed FCS Frame Check
Sequence Extra bits appended for error
checking the CRC
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Ethernet specification

• Maximum frame size
• 1518 bytes not counting the 8-byte preamble
• 1526 bytes counting everything
• Minimum packet size
• 72 bytes counting the preamble
• Minimum packet spacing
• 9.6 ? seconds between end of one transmission and
  beginning of next

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Scaling

• What happens if the 10Mbps Ethernet is changed to
  100Mbps?
• Bit speed does not change what does?
• How big is a bit?
• Speed of light is 299,792,458 meters per second
• transmission of data cannot quite achieve that,
  but comes fairly close.
• Assume .8 speed of light for the speed of the
  electrons representing our bits.
• Lets make life easier by using round numbers
• 300,000,000 meters per second for light .8 times
  that for the bits 100Mbps 100 Million (not the
  proper power of 2)
• So, how big is a bit? How big is a bit at 10 Mbps?

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How big is a bit?
At 100Mbps 1 second .8 300,000,000
meters --------- --------------------- 2.4
meters/bit 100,000,000 bits second At
10Mbps 1 second .8 300,000,000
meters --------- ---------------------
24 meters/bit 10,000,000 bits
second For a 72 byte minimum frame size 72
bytes 8 bits 24 meters -------- ------
-------- 13824 meters/frame frame
byte bit (Remember that 1 kilometer is
1,000 meters and is about .6 mile) A minimum
frame is measured in miles!
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Scaling, mathematically speaking
Throughput is the reciprocal of the average time
required to launch a message successfully m is
the time needed to transmit the message ? is the
time to travel the entire length of the
network. Worst case time to detect collision
2? If average number of retransmissions is J,
then 2 ? J time is required to resolve the
collision Average time required to launch a
message successfully is then m ? 2 ? J. The
only unknown is J, the average number of
retransmissions. v probability of success
after one attempt to transmit (1-v)v
probability of success after two
attempts. (1-v)k-1v probability of success
after k attempts, etc. Average number of
retransmissions required is sum over all k of
kprobability of success after k tries
J ?v(1-v) k-1 1/v
k1,?
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Scaling, continued
J ? v(1-v) k-1 1/v k1,?
Probability that a station wants to transmit in a
2? interval is p The probability that exactly one
station transmits and is successful is v n
p(1-p)n-1 Maximum v occurs when p 1/n and
thus vmax n(1/n)(1-1/n)n-1 (1-1/n) n-1 --gt
e-1, n--gt ? So, J 1/v --gt e and t
m(1a(12J)) m(1a(12e)) time to launch a
message successfully is t m(16.44a)
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Scaling the bottom line
tm(16.44a) S m/tv lt 1/(16.44a)
a ?/m This is the bottom line. Throughput is
dependent on a a is the ratio between the time
it takes to reach the farthest removed station
and the length of time it takes to transmit the
message. (Propagation time/transmission time)
This shows that if we decrease the length of
time it takes to launch a message (by increasing
the transmission speed), we increase a and
decrease throughput. To compensate, we have to
reduce the maximum propagation delay -- which can
only be done by making the longest cable distance
smaller. That is why 100 Mbps Ethernet have
shorter distance restrictions
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The Token Ring

• Controlled access no collisions to resolve
• Having a token requires a way to recover from
  token loss -- a monitor

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Token Ring Frame Format
1 1 1 2/6 2/6 lt5000 4 1 1
bytes SD AC FC DA SA LLS FCS ED
FS field
SD START DELIMITER ED END DELIMITER AC
ACCESS CONTROL FC FRAME CONTROL SA/DA
SOURCE/DESTINATION ADDRESSES FS FRAME
STATUS FCS FRAME CHECK SEQUENCE

Token SD AC ED
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The token

• start delimiter JK0KJ000
• J, K are signals that cannot be mistaken for 0 or
  1
• end delimiter JK1JK1IE
• I for intermediate (1) or last (0)
• E for error detected
• Access Control PPPTMRRR
• PPP priority
• T token (1) or data frame (0)
• M monitor has seen this (1) or not (0)
• frame control FFZZZZZZ
• identifies the frame as MAC or LLC
• allows the MAC and LLC protocols to communicate
  with peers

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Token Ring continued

• Source and destination addresses
• 16 or 48 bits, first bit indicates individual or
  group address same standard as ethernet
• FCS Frame check sequence (CRC)
• Frame status ACxxACxx
• Address recognized
• Frame Copied to host
• Scalability
• same problem as CSMA/CD

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FDDI

• Similar to Token Ring, but with differences to
  help it scale
• Five bit codes to represent four data bits
• latency each station must receive 10 bits before
  passing on
• No designated monitor
• responsibilities shared by all the interface
  units
• Early token release

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DQDB

• IEEE 802.6 (MAN)
• Two unidirectional buses
• communicate by inserting message into the first
  available slot
• slots are 53 octets to match ATM technology
• Though it is accepted as a standard, some
  fairness problems remain and are the subject of
  current study

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802.11

• IEEE Standard for wireless networks
• CSMA/CA (collision avoidance)
• Cannot use CSMA/CD because the transmitting node
  cannot hear other transmissions
• Listen. If channel clear, transmit. If channel
  busy, generate a random backoff factor. Continue
  to listen. If channel is clear, decrement the
  backoff counter otherwise do not decrement.
  When backoff counter reaches 0, transmit.
• Idea is that it is unlikely that two or more
  nodes will choose the same random number. If
  they do, then a collision occurs, the
  transmission fails, and higher layers will
  recognize the failure and cause a retry.

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Wireless transmission initiation

• Sender transmits special RTS (ready to send)
  signal. Intended receiver responds with CTS
  (clear to send).
• This is necessary to avoid problem of
  interference from a node that the sender cannot
  hear but the receiver can hear (the hidden node
  problem).

If A wants to transmit to B, it would not hear C
transmitting to B
A
B
C
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Standards variations

• 802.11 -- applies to wireless LANs and
  provides 1 or 2 Mbps transmission in the 2.4 GHz
  band using either frequency hopping spread
  spectrum (FHSS) or direct sequence spread
  spectrum (DSSS).
• 802.11a -- an extension to 802.11 that
  applies to wireless LANs and provides up to 54
  Mbps in the 5GHz band. 802.11a uses an orthogonal
  frequency division multiplexing encoding scheme
  rather than FHSS or DSSS.
• 802.11b (also referred to as 802.11 High Rate
  or Wi-Fi) -- an extension to 802.11 that applies
  to wireless LANS and provides 11 Mbps
  transmission (with a fallback to 5.5, 2 and 1
  Mbps) in the 2.4 GHz band. 802.11b uses only
  DSSS. 802.11b was a 1999 ratification to the
  original 802.11 standard, allowing wireless
  functionality comparable to Ethernet.
• 802.11g -- applies to wireless LANs and
  provides 20 Mbps in the 2.4 GHz band.

http//www.webopedia.com/TERM/8/802_11.html
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Wireless vs. Mobile

• Wireless networks connect to wired hubs which
  participate in the local network according to
  standard protocols.
• Mobile hosts move from network to network using
  one of several methods to stay connected

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Error Control

• Error detection vs. error correction
• detection is relatively easy correction comes
  with high cost
• Block parity check
• Arrange the bits in a two-dimensional array,
  perhaps one character per row and some number of
  characters giving the number of rows.
• Use a parity bit for each row and one for each
  column
• vulnerable to burst type errors, which are common
• Polynomial encoding -- the CRC, for example
• Very reliable, but not perfect

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CRC

• The basic idea choose a divisor, send a value
  that is known to be evenly divisible by the
  divisor.
• Also send the modification to the original data
  that made it evenly divisible.
• Receiver checks the division. If any remainder,
  the transmission contains an error.

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CRC polynomial

• Interpret the bits to be sent as the coefficients
  of a polynomial
• 0001010010 represents
• 0x9 0x8 0 x7 1x6 0x5 1x4 0x3 0x2
  1x1 0x0
• Agree on a common polynomial to divide by such
  that the remainder will be 0
• Shift the polynomial by the degree (n) of the
  common polynomial divisor (multiply the data
  polynomial by xn )
• Divide, then subtract the remainder from the
  shifted polynomial before transmitting.
• Receiver verifies that division yields no
  remainder and can easily see the original data by
  dropping off the extra bits

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CRC checking

• Append 0s to the data --- as many as the degree
  of the divisor polynomial
• Divide
• Subtract the remainder from the augmented
  polynomial and transmit.
• Example
• data 0010100100101010010101001001 generator
  polynomial 100101

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How dependable is the CRC?

• Andrew Tannenbaum analysis
• Suppose an error occurs in a transmission
• T(x) is the correct transmission
• T(x) E(x) is received. E(x) represents the
  error.
• Each 1 in E(x) corresponds to a bit that has been
  inverted.
• If there are k 1 bits in E(x), k single bit
  errors
• A single burst error a 1 followed by some
  combination of 0s and 1s and a final 1. All
  other bits 0.
• Receiver dividesT(x) E(x) by the generating
  polynomial, G(x)
• (T(x) E(x))/G(x) T(x)/G(x) E(x)/G(x) 0
  E(x)/G(x)
• If E(x)/G(x) 0, the error will not be detected.
  How likely is that?

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Dependability of CRC

• Undetected errors ? E(x)/G(x) 0
• Single bit error E(x) xi, i error bit.
• If G(x) has more than one term, it will not
  divide evenly. So any G(x) of more than one term
  will detect all single bit errors.
• Two isolated single bit errors E(x) xi xj,
  igtj
• E(x) xj (xi-j 1). Assume not divisible by x
  (case 1). Then sufficient to be sure that G(x)
  does not divide xk 1 for any k up to maximum
  value of i-j.
• Good examples are known. X15 x14 1 will not
  divide xk 1 for any value of k below 32,768.

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Dependability of CRC

• An odd number of error bits E(x) - odd number of
  terms
• Fact no polynomial with an odd number of terms
  has x1 as a factor in the modulo 2 system.
  Thus, if x1 is a factor of G(x), all errors
  consisting of an odd number of bits will be
  caught.
• Demonstration
• Assume E(x) has an odd number of terms and is
  divisible by x1. Factor E(x)(x1)Q(x).
• Evaluate E(1) (11)Q(1)
• Modulo 2 E(1) 0 Q(1) E(1) 0.
• If E(x) has an odd number of terms, substituting
  1 for each x will always yield 1 as a result.
  Thus, no polynomial with an odd number of terms
  is divisible by x1.

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Dependability of CRC

• A polynomial code with r check bits will always
  detect burst errors of length ltr.
• Burst error of length k xi(xk-11) where I
  determines how far from the right hand end of the
  received frame the error burst begins.
• For the error to be missed, either xi or xk-1
  must 0.
• If G(x) contains an x0 term (ends in 1), xi is
  not divisible by G(x).
• If k-1 is lt degree of G(x), the remainder will
  not 0.
• So, all burst errors of length lt r will be
  detected if G(r) has at least degree r.

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Dependability

• Burst error gt r bits
• Will go undetected iff the remainder after
  division by G(x) 0
• That will happen only if the burst is exactly the
  same as G(x). First and last bits of each are 1,
  so must look at r-1 intermediate bits.
• If all combinations equally likely, probability
  (1/2) r-1

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Dependability of CRC summary

• International Standards
• CRC-12 x12x11x3x2x1
• CRC-16 x16x15x21
• CRC-CCITT x16x12x51
• All contain x1 as a factor
• The 16 bit codes catch all single and double
  errors, all erros with an odd number of bits, all
  burst errors of length 16 or less, 99.997 of
  17-bit bursts and 98.998 of 18-bit and longer
  bursts.

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Framing

• Checksums assume that we have a fixed length
  transmission of known size.
• One job of the data link layer is to break the
  stream of bits into discrete chunks, append an
  error check sequence (often the CRC) and
  transmit. The receiver rechecks the CRC but
  needs to know where the fame boundaries are.

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Framing

• Framing is the process of marking the endpoints
  of a transmission.
• Principal methods
• Character count
• Begin and end marker characters
• Begin and end marker flags
• Physical layer coding violations

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Character count

• We saw it used in the header at higher layers,
  where bit errors have been corrected.
• Problem at the data link layer is that the
  character count itself may be corrupted. Then
  there is no clue where the frame ends

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Framing

• Special character codes
• ASCII codes available for the purpose
• DLE (data link escape, means something with
  special significance to the DL layer is coming)
• STX (start Text) Beginning of the frame.
• ETX (end Text) Ending of the frame
• Disadvantage is the character orientation. What
  to do with binary data
• Suppose the data contains DLE? Stuff in an extra
  one. Anytime DLE is followed by DLE it means DLE
  was transmitted. Suppose DLE STX is in the data?
  The character stuffing works then also.

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Framing

• Flags and bit stuffing
• flag 01111110
• bit stuffing make sure the flag does not occur
  anywhere in the actual data transmitted
• Whenever the sender sees a pattern of five
  consecutive 1's, insert a 0.
• Whenever the receiver sees a pattern of five
  consecutive 1's followed by 0, delete the 0.

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Physical Layer Coding Violations

• Physical layer coding violations
• using something other than the agreed
  representation of 1 or 0 and giving it a special
  meaning. (the J and K in the token ring start
  delimiter, for example)

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Reference

• Tannenbaum, Andrew S. Computer Networks 3rd
  edition. Prentice Hall. 1996