btech 2012
TRANSCRIPT
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K.Balachandran
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To introduce the concepts, terminologies andtechnologies used in modern days data
communication and computer networking.
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To understand the concepts of datacommunications.
To study the functions of different layers. To introduce IEEE standards employed in
computer networking. To make the students to get familiarized with
different protocols and network components.
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Text book: Behrouz A. Forouzan, Data communication and
Networking, Tata McGraw-Hill, 2004. Reference books James F. Kurose and Keith W. Ross, Computer
Networking: A Top-Down Approach Featuring theInternet, Pearson Education, 2003.
Larry L.Peterson and Peter S. Davie, ComputerNetworks, Harcourt Asia Pvt. Ltd., Second Edition.
Andrew S. Tanenbaum, Computer Networks, PHI, FourthEdition, 2003.
William Stallings, Data and Computer Communication,Sixth Edition, Pearson Education, 2000.
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DATA COMMUNICATIONS DATA LINK LAYER
NETWORK LAYER TRANSPORT LAYER APPLICATION LAYER
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Components Direction of Data flow networks Components and Categories
types of Connections Topologies Protocolsand Standards ISO / OSI model Transmission Media Coaxial Cable FiberOptics Line Coding Modems RS232Interfacing sequences.
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Why study data communications? Data communication exchange of data
between two devices via a transmissionmedium
Effectiveness depends on:
Delivery, Accuracy, Timeliness, Jitter
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Sender Receiver
Message Medium Protocol
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Text represented as a bit pattern; codesoften used:
ASCII; Extended ASCII; Unicode; ISO Numbers represented by binary equivalent Images bit patterns representing pixels Audio Video
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Simplex unidirectional; one transmits, otherreceives
Half-duplex each can transmit/receive;communication must alternate
Full-duplex both can transmit/receivesimultaneously
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Set of devices (nodes) connected by media Distributed processing
Advantages
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Performance affected by # users, type ofmedium, HW/SW
Reliability measured by freq of failure,recovery time, catastrophe vulnerability Security protection from unauthorized
access, viruses/worms
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Point-to-point dedicated
Multipoint shared
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Physical or logical arrangement 4 basic types: mesh, star, bus, ring
May often see hybrid
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LAN MAN
WAN
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Protocols:
Syntax: structure and format of the data
Semantics: meaning of each section of bits Timing
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De facto: standards not approved by standardagencies
De jure: legislated and officially recognized
Standards Organizations: ISO
ITU- International Telecommunication Union
CCITT: Consultative committee for InternationalTelegraphy and Telephony
ANSI: American National Standards Institute
IEEE: institute of Electrical and Electronics Engineers
EIA: Electronics Industries Association
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OSI (Open Systems Interconnect) Network architecture based on a proposal developed by
ISO (International Standards Organization) to standardize the protocols used in various layers
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Dedicated point-to-pointlinks to every other device
Advantages Disadvantages
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Dedicated point-to-pointlinks to central controller(hub)
Controller acts as exchange Advantages Disadvantages
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Transmission Medium and Physical Layer
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Categories of unshielded twisted-pair cables
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(a)Category 3 UTP
(b)Category 5 UTP
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UTPconnector
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Physical description:
Each wire with copper conductor
Separately insulated wires
Twisted together to reduce cross talk Often bundled into cables of two or four twisted pairs
If enclosed in a sheath then is shielded twisted pair (STP)
otherwise often for home usage unshielded twisted pair (UTP).
Must be shield from voltage lines
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Application:
Common in building for digital signaling used at
speed of 10s Mb/s (CAT3) and 100Mb/s (CAT5)over 100s meters.
Common for telephone interconnection at home
and office buildings
Less expensive medium; limited in distance,bandwidth, and data rate.
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Category Maximum data
rate
Usual application
CAT 1 Less than 1 Mbps analog voice (plain old telephone
service) Integrated Services Digital
Network Basic Rate Interface in ISDN
Doorbell wiringCAT 2 4 Mbps Mainly used in the IBM Cabling System
for token ring networks
CAT 3 16 Mbps Voice and data on 10BASE-T Ethernet
(certify 16Mhz signal)
CAT 4 20 Mbps Used in 16Mbps Token Ring
Otherwise not used much
CAT 5 100 Mbps 100 Mbps TPDDI
155 Mbps asynchronous transfer
mode (certify 100 Mhz signal)
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Widely installed for use in business and corporationEthernet and other types of LANs.
Consists of inter copper insulator covered by claddingmaterial, and then covered by an outer jacket
Physical Descriptions
Covered by sheath material
Outer conductor is braided shielded (ground)
Separated by insulating material
Inner conductor is solid copper metal
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Applications:
TV distribution (cable tv); long distance telephone transmission;short run computer system links
Local area networks Transmission characteristics: Can transmit analog and digital signals
Usable spectrum for analog signaling is about 400 Mhz
Amplifier needed for analog signals for less than 1 Km and less
distance for higher frequency Repeater needed for digital signals every Km or less distance for
higher data rates
Operation of 100s Mb/s over 1 Km.
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Physical Description: Glass or plastic core of optical fiber = 2 to125 m
Cladding is an insulating material
Jacket is a protective cover
Laser or light emitting diode providestransmission light source
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Applications:
Long distance telecommunication
Greater capacity; 2 Gb/s over 10s of Km
Smaller size and lighter weight
Lower attenuation (reduction in strength of signal)
Electromagnetic isolation not effected by externalelectromagnetic environment. more privacy
Greater repeater spacing fewer repeaters, reduces lineregeneration cost
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multimode fiber is optical fiber that is designedto carry multiple light rays or modesconcurrently, each at a slightly different
reflection angle within the optical fiber core.used for relatively short distances because themodes tend to disperse over longer lengths (thisis called modal dispersion) .
For longer distances, single mode fiber(sometimes called monomode) fiber is used. Insingle mode fiber a single ray or mode of lightact as a carrier
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0.85, 1.30, 1.55 bands Bands are 25000 to 30000 GHz wide
Dispersion: Spreading out in length as theypropagate- Hyperbolic cosine Solitons: pulses which can travel 1000s of
KMs without any appreciable shape
dispersion
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Band Range Propagation Application
VLF 330 KHz Ground Long-range radio navigation
LF 30300 KHz GroundRadio beacons and
navigational locators
MF 300 KHz3 MHz Sky AM radio
HF 330 MHz SkyCitizens band (CB),
ship/aircraft communication
VHF 30300 MHzSky and
line-of-sight
VHF TV,
FM radio
UHF 300 MHz3 GHz Line-of-sightUHF TV, cellular phones,
paging, satellite
SHF 330 GHz Line-of-sight Satellite communication
EHF 30300 GHz Line-of-sight Long-range radio navigation
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Process of converting binary data to a digital signal
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Residual direct-current(dc) components or zerofrequencies areundesirable
Some systems do not allow
passage of a dc component;
may distort the signal and
create output errors DC component is extra
energy and is useless
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Includes timinginformation in the databeing transmitted toprevent misinterpretation
Lack of synchronization
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Unipolar Polar
Bipolar
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Simplest method; inexpensive Uses only one voltage level Polarity is usually assigned to binary 1; a 0 is
represented by zero voltage
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Potential problems:
DC component
Lack of synchronization
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Uses two voltage levels, one positive and onenegative
Alleviates DC component Variations
Nonreturn to zero (NRZ)
Return to zero (RZ)
Manchester
Differential Manchester
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Value of signal is always positive or negative NRZ-L Signal level depends on bit represented; positive
usually means 0, negative usually means 1 Problem: synchronization of long streams of 0s or
1s NRZ-I (NRZ-Invert)
Inversion of voltage represents a 1 bit 0 bit represented by no change Allows for synchronization
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Uses an inversion at the middle of each bit interval forboth synchronization and bit representation
Negative-to-positive represents binary 1 Positive-to-negative represents binary 0
Achieves same level of synchronization with only 2levels of amplitude
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Inversion at middle of bit interval is used for synchronization Presence or absence of additional transition at beginning ofinterval identifies the bit
Transition means binary 0; no transition means 1 Requires two signal changes to represent binary 0; only one to
represent 1
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RS-232 is the Serial interface on the PC
Three major wires for the Serial interface:
Transmit -Pin 2
Receive -Pin 3
Ground -Pin 7 (25 pin connector)
- Pin 5 (9 pin connector)
Tx Tx
RxRx
GndGnd
Computer
Device
Transmit connects to Receive
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TD: transmitted data RD: received data DSR: data set ready
indicate whether DCE is powered on DTR: data terminal ready
indicate whether DTR is powered on
turning off DTR causes modem to hang up the line
RI: ring indicator ON when modem detects phone call
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DCD: data carrier detect ON when two modems have negotiated successfully and
the carrier signal is established on the phone line RTS: request to send
ON when DTE wants to send data Used to turn on and off modems carrier signal in multi-
point (i.e. multi-drop) lines
Normally constantly ON in point-to-point lines CTS: clear to send
ON when DCE is ready to receive data SG: signal ground
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Means to ask the transmitter to stop/resume sending indata
Required when: DTE to DCE speed > DCE to DCE speed
(e.g. terminal speed = 115.2kbps and line speed = 33.6kbps, inorder to benefit from modems data compression protocol)
without flow control, the buffer within modem will overflow sooner or later
the receiving end takes time to process the data and thus
cannot be always ready to receive
Position of the data-linklayer
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Data link layer duties
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LLC and MAC sublayers
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IEEE standards for LANs
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Error Detection
and
Correction
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Data can be corrupted during
transmission. For reliablecommunication, errors must be detected
and corrected.
Note:
T
f E
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Types of Error
Single-Bit Error
Burst Error
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In a single-bit error, only one bit in the
data unit has changed.
Note:
Single-bit error
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A burst error means that 2 or more bits
in the data unit have changed.
Note:
Burst error of length 5
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D t ti
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Detection
Redundancy
Parity Check
Cyclic Redundancy Check (CRC)
Checksum
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Error detection uses the concept of
redundancy, which means adding extrabits for detecting errors at the
destination.
Note:
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Detection methods
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Even-pari ty concept
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In parity check, a parity bit is added to
every data unit so that the total numberof 1s is even
(or odd for odd-parity).
Note:
Example 1
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Example 1
Suppose the sender wants to send the word world. In
ASCII the five characters are coded as
1110111 1101111 1110010 1101100 1100100
The following shows the actual bits sent
11101110 11011110 11100100 11011000 11001001
Example 2
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Example 2
Now suppose the word world in Example 1 is received by
the receiver without being corrupted in transmission.
11101110 11011110 11100100 11011000 11001001
The receiver counts the 1s in each character and comes up
with even numbers (6, 6, 4, 4, 4). The data are accepted.
Example 3
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Example 3
Now suppose the word world in Example 1 is corrupted
during transmission.
11111110 11011110 11101100 11011000 11001001
The receiver counts the 1s in each character and comes up
with even and odd numbers (7, 6, 5, 4, 4). The receiver
knows that the data are corrupted, discards them, and asks
for retransmission.
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Simple parity check can detect all
single-bit errors. It can detect bursterrors only if the total number of errors
in each data unit is odd.
Note:
Two-dimensional pari ty
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Example 4
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Example 4
Suppose the following block is sent:
10101001 00111001 11011101 11100111 10101010
However, it is hit by a burst noise of length 8, and some
bits are corrupted.
10100011 10001001 11011101 11100111 10101010
When the receiver checks the parity bits, some of the bits
do not follow the even-parity rule and the whole block is
discarded.
10100011 10001001 11011101 11100111 10101010
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In two-dimensional parity check, a block
of bits is divided into rows and aredundant row of bits is added to the
whole block.
Note:
CRC generator and checker
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Binary division in a CRC generator
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Binary division in CRC checker
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A polynomial representi ng a divisor
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Standardpolynomials
Name Polynomial Application
CRC-8 x8 +x2 +x+1 ATM header
CRC-10 x10+x9 +x5 +x4 +x2+1 ATM AAL
ITU-16 x16+x12+x5 + 1 HDLC
ITU-32x32+x26 +x23+x22+x16+x12+x11+x10
+x8+x7+x5+x4+x2+x +1LANs
Example 5
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p 5
It is obvious that we cannot choose x(binary 10) or x2+ x
(binary 110) as the polynomial because both are divisibleby x. However, we can choose x + 1(binary 11) because
it is not divisible by x, but is divisible by x + 1. We can
also choose x2+ 1(binary 101) because it is divisible by
x + 1 (binary division).
Example 6
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p
The CRC-12
x12+ x11+ x3+ x + 1
which has a degree of 12, will detect all burst errors
affecting an odd number of bits, will detect all bursterrors with a length less than or equal to 12, and will
detect, 99.97 percent of the time, burst errors with a
length of 12 or more.
Checksum
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Data uni t and checksum
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The sender follows these steps:
The unit is divided into k sections, each of n bits.
All sections are added using ones complement to get the sum.
The sum is complemented and becomes the checksum.
The checksum is sent with the data.
Note:
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The receiver follows these steps:
The unit is divided into k sections, each of n bits.
All sections are added using ones complement to get the sum.
The sum is complemented.
If the result is zero, the data are accepted: otherwise, rejected.
Note:
Example 7
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Suppose the following block of 16 bits is to be sent using a
checksum of 8 bits.10101001 00111001
The numbers are added using ones complement
10101001
00111001
------------
Sum 11100010
Checksum 00011101
The pattern sent is 10101001 00111001 00011101
Example 8
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Now suppose the receiver receives the pattern sent in Example 7
and there is no error.10101001 00111001 00011101
When the receiver adds the three sections, it will get all 1s, which,
after complementing, is all 0s and shows that there is no error.
10101001
00111001
00011101
Sum 11111111
Complement 00000000 means that the pattern is OK.
Example 9
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Now suppose there is a burst error of length 5 that affects 4 bits.
10101111 11111001 00011101
When the receiver adds the three sections, it gets
10101111
11111001
00011101
Partial Sum 111000101
Carry 1
Sum 11000110
Complement 00111001 the pattern is corrupted.
Correction
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Correction
Retransmission
Forward Error Correction
Burst Error Correction
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Data and redundancy bits
Number of
data bits
m
Number of
redundancy bits
r
Total
bits
m + r
1 2 3
2 3 5
3 3 6
4 3 7
5 4 9
6 4 10
7 4 11
Positions of redundancy bits in Hamming code
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Redundancy bits calculation
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Example of redundancy bit calculation
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Er ror detection using Hamming code
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Burst error correction example
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How to find an error? Introducing redundancy -- using 2 bits message to send 1 bit
information in the previous example.
Message = information bits + redundant bits (checksum).
How to design codes that have errorcorrection/detection capability? Hamming distancebetween two code words: the
number of different bits between the two code words.
E.g 010101 and 111000? Hamming distance = ?
Hamming distanceof a complete code: the minimum
Hamming distance any of the two codewords in the code.
E.g 010101, 111000, 000111, 111111 Hamming_distance= ?
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Redundancy bits calculation
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10.16 Example of redundancy bit calculation
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Data transmission in one direction only Both Transmitting and receiving host are always
ready for operation Processing time can be ignored
Infinite buffer size available Communication channels between IMPs never
damages or loses frame No sequence Number, No acknowledgements Only event is FrameArrival arrival of undamaged
frame
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typedef enum {frme_arraival} event_type;#include protocol.hvoid sender1(void){ frame s; /*buffer for an outbound frame*/
packet buffer; /*buffer for an outbound packet*/while (true)
{from_network_layer(&buffer);s.info = buffer;
to_physical_layer(&s);}
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void receiver1(void){
frame r;/*buffers for the frames*/
event_type event;while (true){
wait_for_event(&event);from_physical_layer(&r);to_network_layer(&r.info);
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typedef enum {frme_arraival} event_type;#include protocol.hvoid sender2(void){ frame s; /*buffer for an outbound frame*/
packet buffer; /*buffer for an outbound packet*/
while (true){
from_network_layer(&buffer);s.info = buffer;
to_physical_layer(&s);wait_for_event(&event);}
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void receiver1(void){frame r,s;/*buffers for the frames*/event_type event;while (true)
{wait_for_event(&event);from_physical_layer(&r);to_network_layer(&r.info);
to_physical_layer(&s); /*sending dummy frame to awaken sender */}
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ACKs introduce a new issue how long doesreceiver wait before sending ONLY an ACKframe.
We need anACKTimer!!sender timeout periodneeds to set longer.
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Each outbound frame must contain a sequence number. With nbits for the sequence number field, the numbers range from 0to maxseq.
Sliding window :: sender has a windowof frames and maintainsa list of consecutive sequence numbers for frames that it ispermitted to send without waiting for ACKs.
receiver has a window that is a list of frame sequence numbersit is permitted to accept.
Notesending and receiving windows do NOT have to be the
same size.Windows can be fixed size or dynamically growing and
shrinking.
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Host is oblivious, message order at transportlevel is maintained.
senders window ::frames sent but not yet ACKed.
new packets from the Host cause the upper
edge inside sender window to be incremented.
ACKed frames from the receiver cause the
lower edge inside window to be incremented.
All frames in the senders window must be savedfor possible retransmission and we need onetimer per frame in the window.
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If the maximum sender window size is B,the sender needs B buffers.
If the sender windowgets full (i.e., reaches
its maximum window size, the protocolmust shut off the Host (the network layer)until buffers become available.
receiver window
Frames received with sequence numbers outsidethe receiver windoware not accepted.
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receiver window Frames received with sequence numbers
outside the receiver windoware not accepted.
The receiver window size is normally static.The set of acceptable sequence numbers isrotated as acceptable frames arrive.
a receiver window size = 1the protocol
only accepts frames in order.There is referred to as Go Back N.
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A sliding window of size 1, with a 3-bit sequence number.(a)Initially.(b)After the first frame has been sent.(c)After the first frame has been received.
(d)After the first acknowledgement hasbeen received.
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(a) Normal case. (b) Abnormal case.The notation is (seq, ack, packet number). Anasterisk indicates where a network layer accepts apacket.
Go Back N
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A
B
fr0
timefr1
fr2
fr3
fr4
fr5
fr6
fr3
ACK
1 error
Out-of-sequence frames
Go-Back-4: 4 frames are outstanding; so go back 4
fr5
fr6
fr4
fr7
fr8
fr9
ACK
2
ACK
3
ACK
4
ACK
5
ACK
6
ACK7
ACK
8
ACK
9
ACKing next frame expected
Go Back Nwith NAK error recovery
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A
B
fr0
timefr1
fr2
fr3
fr4
fr5
fr1
fr2
ACK
1
error
Out-of-sequenceframes
Go-Back-7:
fr4
fr5
fr3
fr6
fr7
fr0
NAK
1
ACK
3
ACK
4
ACK
5
ACK
6
ACK
7
ACK
2
Transmitter goes back to frame 1
Figure 5.17Leon-Garcia & Widjaja: Communication NetworksCopyright 2000 The McGraw Hill Companies
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A
B
fr0
timefr1
fr2
fr3
fr4
fr5
fr6
fr2
AC
K1 error
fr8
fr9
fr7
fr10
fr11
fr12
AC
K2
NA
K2
AC
K7
AC
K8
AC
K9
AC
K10
AC
K11
AC
K12
AC
K2
AC
K2
AC
K2
Figure 5.21
Selective Repeatwith NAK error recovery
Leon-Garcia & Widjaja: Communication NetworksCopyright 2000 The McGraw Hill Companies
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HDLC
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Con f igu rat ions and Trans fer Modes
Frames
Frame Format
Examples
Data Transparency
HDLC frame
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HDLC frame types
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I-frame
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S-frame control field in HDLC
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Table 11.1 U
-
frame control command and response
Command/response Meaning
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SNRM Set normal response mode
SNRME Set normal response mode (extended)
SABM Set asynchronous balanced mode
SABME Set asynchronous balanced mode (extended)
UP Unnumbered poll
UI Unnumbered information
UA Unnumbered acknowledgmentRD Request disconnect
DISC Disconnect
DM Disconnect mode
RIM Request information mode
SIM Set initialization mode
RSET Reset
XID Exchange ID
FRMR Frame reject
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Address Control Upper-layer data FCS
DSAP
SSAP
Control
Upperlayerdata
MACheader
MACpayload
FCS
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CS 640 137
Most popular packet-switched LAN technology Bandwidths: 10Mbps, 100Mbps, 1Gbps Max bus length: 2500m
500m segments with 4 repeaters
Bus and Star topologies are used to connect hosts
Hosts attach to network via Ethernet transceiver or hub or switch
Detects line state and sends/receives signals
Hubs are used to facilitate shared connections
All hosts on an Ethernet are competing for access to the medium
Switches break this model
Problem: Distributed algorithm that provides fair access
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CS 640 138
Ethernet by definition is a broadcastprotocol Any signal can be received by all hosts
Switching enables individual hosts tocommunicate Network layer packets are transmitted
over an Ethernet by encapsulating
Frame Format
Destaddr
64 48 32
CRCPreambleSrcaddr
Type Body
1648
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CS 640 139
Physical layer configurations are specified in three parts Data rate (10, 100, 1,000)
10, 100, 1,000Mbps Signaling method (base, broad)
Baseband Digital signaling
Broadband Analog signaling
Cabling (2, 5, T, F, S, L) 5 - Thick coax (original Ethernet cabling)
F Optical fiber
S Short wave laser over multimode fiber
L Long wave laser over single mode fiber
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CS 640 140
Developed in late 60s by Norm Abramson at Univ. of Hawaii(!!) for use with packet radio systems Any station can send data at any time
Receiver sends an ACK for data
Timeout for ACK signals that there was a collision What happens if timeout is poorly timed?
If there is a collision, sender will resend data after a random backoff
Utilization (fraction of transmitted frames avoiding collisionfor N nodes) was pretty bad
Max utilization = 18% Slotted Aloha (dividing transmit time into windows) helped Max utilization increased to 36%
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CS 640 141
In Aloha, decisions to transmit are made without paying
attention to what other nodes might be doing Ethernet uses CSMA/CD listens to line before/during sending If line is idle (no carrier sensed) send packet immediately upper bound message size of 1500 bytes
must wait 9.6us between back-to-back frames If line is busy (carrier sensed) wait until idle and transmit packet immediately
called 1-persistent sending If collision detected Stop sending and jam signal Try again later
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CS 640 142
Packet?
Sense
Carrier
Discard
Packet
Send Detect
Collision
Jam channel
b=CalcBackoff();wait(b);
attempts++;
No
Yes
attempts < 16
attempts == 16
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CS 640 143
A B
A B
Collisions are caused when two adaptors transmit at the samtime (adaptors sense collision based on voltage differences)
Both found line to be idleBoth had been waiting to for a busy line to become idle
A starts attime 0
Message almostthere at time T when
B starts collision!
How can we be sure A knows about the collision?
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CS 640 144
How can A know that a collision has taken place? There must be a mechanism to insure retransmission on collision As message reaches B at time T Bs message reaches A at time 2T So, A must still be transmitting at 2T
IEEE 802.3 specifies max value of 2T to be 51.2us This relates to maximum distance of 2500m between hosts At 10Mbps it takes 0.1us to transmit one bit so 512 bits (64B) take 51.2us to send So, Ethernet frames must be at least 64B long
14B header, 46B data, 4B CRC
Padding is used if data is less than 46B Send jamming signal after collision is detected to insure all hosts see
collision 48 bit signal
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CS 640 145
A B
A B
A B
time = 0
time = T
time = 2T
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CS 640 146
If a collision is detected, delay and try again Delay time is selected using binary exponential backoff 1st time: choose K from {0,1} then delay = K * 51.2us 2nd time: choose K from {0,1,2,3} then delay = K * 51.2us nthtime: delay = K x51.2us, for K=0..2n1
Note max value for k = 1023
give up after several tries (usually 16) Report transmit error to host
If delay were not random, then there is a chance that sourceswould retransmit in lock step
Why not just choose from small set for K
This works fine for a small number of hosts
Large number of nodes would result in more collisions
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CS 640 147
Senders handle all access control Receivers simply read frames with acceptable
address
Address to host
Address to broadcast
Address to multicast to which host belongs
All frames if host is in promiscuous mode
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CS 640 148
Fast Ethernet (100Mbps) has technology very similar to10Mbps Ethernet Uses different physical layer encoding (4B5B)
Many NICs are 10/100 capable Can be used at either speed
Gigabit Ethernet (1,000Mbps) Compatible with lower speeds
Uses standard framing and CSMA/CD algorithm
Distances are severely limited
Typically used for backbones and inter-router connectivity
Becoming cost competitive
How much of this bandwidth is realizable?
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a carrier-sense, multiple access with collisiondetection (CSMA/CD) protocol for a bustopology
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Token Bus protocol for a token-passingaccess method on a bus topology
The topology of the computer network can
include groups of workstations connectedby long trunk cables
token busis used in some manufacturing
environments, Ethernet and token ringstandards have become more prominent inthe office environment.
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Token Ring protocol which, like Token Bus, isanother token-passing access method, butfor a ring topology
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TOKEN RING NETWORK LIKE IEEE 802.5
TOKEN: A SPECIAL SEQUENCE OF BITS TOKEN CIRCULATES AROUND THE RING
A STATION REMOVES THE TOKEN FROM RINGBEFORE TRANSMISSION
AFTER TRANSMISSION, THE STATION
RETURNS THE TOKEN TO THE RING
COLLISIONS ARE PREVENTED AS THERE ISONLY ONE TOKEN IN THE RING
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DUAL-COUNTER-ROTATING TOKENRING ARCHITECTURE
ONE RING IS PRIMARY AND THEOTHER SECONDARY
UP TO 500 STATIONS WITH A
MAXIMUM DISTANCE OF 2 KMBETWEEN ANY PAIR OF STATIONS
FOR MULTIMODE FIBER
WITH SINGLE-MODE FIBER THE
DISTANCE CAN BE UP TO 40 KM
MAXIMUM RING LENGTH IS 100 KM(TOTAL FIBER LENGTH IS 200 KM
FOR TWO RINGS)
USES 4B/5B ENCODING
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154
FEATURES FDDI ETHERNET TOKEN RING
TRANSMISSION
RATE
125 MBAUD 20 MBAUD 8 & 32 MBAUD
DATA RATE 100 MBPS 10 MBPS 4 & 16 MBPS
SIGNALENCODING
4B/5B (80%EFFICIENT)
MANCHESTER(50%
EFFICIENT)
DIFFERENTIALMANCHESTER
(50% EFFICIENT)
MAXIMUM
COVERAGE
100 KM 2.5 KM CONFIGURATION
DEPENDENT
MAXIMUMNODES
500 1024 250
MAXIMUM
DISTANCE
BETWEEN
NODES
2 KM (MULTIMODE
FIBER)
40 KM (SINGLE-
MODE FIBER)
2.5 KM 300 M
(RECOMMENDED
100 M)
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802.11 is primarily concerned withthe lower layers of the OSI model.
Data Link Layer
Logical Link Control (LLC).
Medium Access Control (MAC).
Physical Layer
Physical Layer ConvergenceProcedure (PLCP).
Physical Medium Dependent (PMD).
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Well-supported, stable, and cost effective,but runs in the 2.4 GHz range that makes itprone to interference from other devices(microwave ovens, cordless phones, etc) andalso has security disadvantages.
Limits the number of access points in rangeof each other to three.
Has 11 channels, with 3 non-overlapping, and
supports rates from 1 to 11 Mbps, butrealistically about 4-5 Mbps max.
Uses direct-sequence spread-spectrumtechnology.
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Completely different from 11b and 11g. Flexible because multiple channels can be combined for
faster throughput and more access points can be co-located. Shorter range than 11b and 11g. Runs in the 5 GHz range, so less interference from other
devices. Has 12 channels, 8 non-overlapping, and supports rates from
6 to 54 Mbps, but realistically about 27 Mbps max Uses frequency division multiplexing
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Speed Slower than cable. Range Affected by various medium. Travels best through open space.
Reduced by walls, glass, water, etc Security Greater exposure to risks. Unauthorized access.
Compromising data.
Denial of service.
N t
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A bridge has a table used in filtering
decisions.
Note:
Bridge
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Networks: SONET 163
Digital carrier systems
The hierarchy of digital signals that the telephone networkuses.
Trunks and access links organized in DS (digital signal)hierarchy
Problem: rates are not multiples of each other.
In the 1980s Bellcore developed the Synchronous
Optical Network (SONET) standard. Previous efforts include: ISDNand BISDN.
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SONET:: encodes bit streams into optical signals propagatedover optical fiber. SONET defines a technology for carryingmany signals of different capacities through a synchronous,flexible, optical hierarchy.
A bit-way implementation providing end-to-end transportof bit streams.
All clocks in the network are locked to a common masterclock so that simple TDM can be used.
Multiplexing done by byte interleaving.
SONETis backward compatible to DS-1 and E-1 and forwardcompatible to ATM cells.
Demultiplexing is easy.
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Networks: SONET 165
SONET topology can be a mesh, but mostoften it is a dual ring.
Standard component of SONET ring is an
ADM (Add/Drop Multiplexer) Drop one incoming multiplexed stream and
replace it with another stream.
Used to make up bi-directional line switchingrings.
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Networks: SONET
a
b
c
d
a
b
c
d
(a) Dual ring (b) Loop-around in response to fault
Figure 4.12
ADM
ADM
ADM
ADM
Leon-Garcia & Widjaja: Communication NetworksCopyright 2000 The McGraw Hill Companies
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Networks: SONET
SONET Ring
(a)
STS
PTE
STS
PTE
SONET Architecture
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STE: Section Terminating Equipment, e.g. a repeaterLTE: Line Terminating Equipment, e.g. a STS-1 to STS-3 multiplexer
PTE: Path Terminating Equipment, e.g. an STS-1 multiplexer
optical
section
optical
section
optical
section
optical
section
line
optical
section
line
optical
section
line
path
optical
section
line
path
(b)
PTELTE
STE
STS-1 Path
STS Line
Section Section
Mux Muxreg reg reg
SONETTerminal
STE STELTE
SONETTerminal
Learning bridge
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Loop problem
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Prior to spanning tree application
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Position of network layer
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Network layer duties
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Internetwork
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Links in an internetwork
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Network layer in an internetwork
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Network layer at the source
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Note
:
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An IP address is a 32-bit address.
Note:
Note
:
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The IP addresses are unique
and universal.
Note:
Dotted-decimal notation
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Note
:
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The binary, decimal, and hexadecimal
number systems are reviewed in
Appendix B.
Note:
Example 1
Change the following IP addresses from binary notation to
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dotted-decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11111001 10011011 11111011 00001111
Solution
We replace each group of 8 bits with its equivalent decimalnumber (see Appendix B) and add dots for separation:
a. 129.11.11.239b. 249.155.251.15
Example 2
Change the following IP addresses from dotted-decimal notation
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to binary notation.
a. 111.56.45.78
b. 75.45.34.78
SolutionWe replace each decimal number with its binary equivalent (seeAppendix B):
a. 01101111 00111000 00101101 01001110b. 01001011 00101101 00100010 01001110
Finding the address class
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Example 3
Find the class of each address:
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a. 00000001 00001011 00001011 11101111b. 11110011 10011011 11111011 00001111
Solution
See the procedure in Figure 19.11.
a. The first bit is 0; this is a class A address.b. The first 4 bits are 1s; this is a class E address.
Finding the class in decimal notation
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Example 4
Find the class of each address:
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a. 227.12.14.87b. 252.5.15.111
c. 134.11.78.56
Solutiona. The first byte is 227 (between 224 and 239); the class is D.
b. The first byte is 252 (between 240 and 255); the class is E.c. The first byte is 134 (between 128 and 191); the class is B.
Netid and hostid
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Blocks in class A
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Blocks in class B
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Blocks in class C
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Network address
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First Octet IP Address Characteristics
MostSignificant
BITS
ValueRanges
Addr.Class
Networkvs.
Host
#NETWORK
S# HOSTS
0000 0-126 A N.h.h.h 256 16,777,214
-- 127 - - Special - Local Loopback1000 128-191 B N.N.h.h 65,536 65,534
1100 192-223 C N.N.N.h 16,777,216 254
1110 224 - 239 D Special N/A N/A
1111 240 + E Special N/A N/A
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5-4
The Optimality Principle
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The Optimality Principle
Shortest Path Routing Flooding Distance Vector Routing
Link State Routing
Hierarchical Routing
Broadcast Routing
Multicast Routing
Routing for Mobile Hosts Routing in Ad Hoc Networks
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The first 5 steps used in computing the shortest path from A to D.The arrows indicate the working node.
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(a)A subnet. (b)Input from A, I, H, K, and the newrouting table for J.
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The count-to-infinity problem.
Each router must do the following:
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Each router must do the following:
1. Discover its neighbors, learn their network address.2. Measure the delay or cost to each of its neighbors.3. Construct a packet telling all it has just learned.
4. Send this packet to all other routers.
5. Compute the shortest path to every other router.
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(a)Nine routers and a LAN. (b)A graph modelof (a).
A subnet in which the East and West parts are
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A subnet in which the East and West parts areconnected by two lines.
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(a) A subnet. (b) The link state packets for thissubnet.
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The packet buffer for router B in the previousslide (Fig. 5-13).