Cs8591 Computer Networks

CS8591 COMPUTER NETWORKS
Anna University Approved 6 Days Faculty Development Program on
Computer Networks 24.04.2019 – 30.04.2019
JEPPIAAR ENGINEERING COLLEGE, CHENNAI
25.04.2019
Dr.A.Kathirvel, Professor & Head, Computer Science and Engg.
M N M Jain Engineering College, Chennai

Topics Covered
Data-Link Layer Protocols
HDLC
PPP
Media Access Control

Perspectives on Connecting
An end-user’s view of the Internet
4

Link Capacity & Shannon-Hartley
Theorem
Gives the upper bound to the capacity of a link in
terms of bits per second (bps) as a function of
signal-to-noise ratio of the link measured in
decibels (dB).
C = Blog2(1+S/N)
Where B = 3300 – 300 = 3000Hz, S is the signal power,
N the average noise.
The signal to noise ratio (S/N) is measured in decibels is
related to dB = 10 x log10(S/N). If there is 30dB of
noise then S/N = 1000.
Now C = 3000 x log2(1001) = 30kbps.
How can we get 56kbps?
5

Links
 All practical links rely on some sort of electromagnetic
radiation propagating through a medium or, in some cases,
through free space
 One way to characterize links, then, is by the medium they use
Typically copper wire in some form (as in Digital Subscriber
Line (DSL) and coaxial cable),
Optical fiber (as in both commercial fiber-to-the home
services and many long-distance links in the Internet’s
backbone), or
Air/free space (for wireless links)
 Another important link characteristic is the frequency
Measured in hertz, with which the electromagnetic waves
oscillate
6

Links
 Distance between the adjacent pair of maxima or minima of a wave
measured in meters is called wavelength
 Speed of light divided by frequency gives the wavelength.
 Frequency on a copper cable range from 300Hz to 3300Hz; Wavelength for
300Hz wave through copper is speed of light on a copper / frequency
 2/3 x 3 x 108 /300 = 667 x 103 meters.
Electromagnetic
spectrum
7

Links
Common services available to connect your home
Placing binary data on a signal is called encoding.
Modulation involves modifying the signals in terms of
their frequency, amplitude, and phase.
8

Encoding
Signals travel between signaling components; bits flow between adaptors
NRZ encoding of a bit stream
9

Encoding
Problem with NRZ
Baseline wander
The receiver keeps an average of the signals it has seen so far
Uses the average to distinguish between low and high signal
When a signal is significantly low than the average, it is 0, else it is 1
Too many consecutive 0’s and 1’s cause this average to change, making it
difficult to detect
Clock recovery
Frequent transition from high to low or vice versa are necessary to enable
clock recovery
Both the sending and decoding process is driven by a clock
Every clock cycle, the sender transmits a bit and the receiver recovers a bit
The sender and receiver have to be precisely synchronized
10

Encoding
NRZI
Non Return to Zero Inverted
Sender makes a transition from the current signal to encode 1
and stay at the current signal to encode 0
Solves for consecutive 1’s
Manchester encoding
Merging the clock with signal by transmitting Ex-OR of the
NRZ encoded data and the clock
Clock is an internal signal that alternates from low to high, a
low/high pair is considered as one clock cycle
In Manchester encoding
0: low high transition
1: high low transition 11

Encoding
Different encoding strategies
12

Encoding
 Problem with Manchester encoding
 Doubles the rate at which the signal transitions are made on the link
Which means the receiver has half of the time to detect each pulse of the
signal
 The rate at which the signal changes is called the link’s baud rate
 In Manchester the bit rate is half the baud rate
 4B/5B encoding
 Insert extra bits into bit stream so as to break up the long sequence of 0’s and
1’s
 Every 4-bits of actual data are encoded in a 5- bit code that is transmitted to
the receiver
 5-bit codes are selected in such a way that each one has no more than one
leading 0(zero) and no more than two trailing 0’s.
 No pair of 5-bit codes results in more than three consecutive 0’s
13

Encoding
4B/5B encoding
0000  11110 16 left
0001  01001 11111 – when the line is idle
0010  10100 00000 – when the line is dead
.. 00100 – to mean halt
..
1111  11101 13 left : 7 invalid, 6 for various
control signals
14

Framing
 We are focusing on packet-switched networks, which means that
blocks of data (called frames at this level), not bit streams, are
exchanged between nodes.
 It is the network adaptor that enables the nodes to exchange
frames.
 When node A wishes to transmit a frame to node B, it tells its
adaptor to transmit a frame from the node’s memory. This results in
a sequence of bits being sent over the link.
Bits flow between adaptors,
frames between hosts
15

Framing
The adaptor on node B then collects together the
sequence of bits arriving on the link and deposits the
corresponding frame in B’s memory.
Recognizing exactly what set of bits constitute a frame—
that is, determining where the frame begins and ends—is
the central challenge faced by the adaptor
Byte-oriented Protocols
To view each frame as a collection of bytes (characters) rather
than bits
BISYNC (Binary Synchronous Communication) Protocol
Developed by IBM (late 1960)
DDCMP (Digital Data Communication Protocol)
Used in DECNet
16

Framing
 BISYNC – sentinel approach
 Frames transmitted beginning with leftmost field
 Beginning of a frame is denoted by sending a special SYN (synchronize)
character
 Data portion of the frame is contained between special sentinel character STX
(start of text) and ETX (end of text)
 SOH : Start of Header DLE : Data Link Escape CRC: Cyclic Redund.Check
 Recent PPP which is commonly run over Internet links uses sentinel approach
 Special start of text character denoted as Flag
0 1 1 1 1 1 1 0
 Address, control : default numbers
 Protocol for demux : IP / IPX
 Payload : negotiated (1500 bytes)
 Checksum : for error detection
17

Framing
 Byte-counting approach
DDCMP
count : how many bytes are contained in the frame body
If count is corrupted
Framing error
 Bit-oriented Protocol
HDLC : High Level Data Link Control
Beginning and Ending Sequences
0 1 1 1 1 1 1 0
On the sending side, any time five consecutive 1’s have been
transmitted from the body of the message (i.e. excluding when the
sender is trying to send the distinguished 01111110 sequence)
The sender inserts 0 before transmitting the next bit
18

Framing
HDLC Protocol
On the receiving side
5 consecutive 1’s
Next bit 0 : Stuffed, so discard it
 1 : Either End of the frame marker Or Error has been
introduced in the bitstream
 Look at the next bit
 If 0 ( 01111110 )  End of the frame marker
 If 1 ( 01111111 )  Error, discard the whole frame
The receiver needs to wait for next 01111110 before it can start
receiving again
19

Error Detection
 Bit errors are introduced into frames
Because of electrical interference and thermal noises
 Detecting Error and Correction Error
 Two approaches when the recipient detects an error
Notify the sender that the message was corrupted, so the sender can send
again.
If the error is rare, then the retransmitted message will be error-free
Using some error correct detection and correction algorithm, the receiver
reconstructs the message
 Common technique for detecting transmission error
CRC (Cyclic Redundancy Check)
Used in HDLC, DDCMP, CSMA/CD, Token Ring
Other approaches
Two Dimensional Parity (BISYNC)
Checksum (IP) 20

Error Detection
 Basic Idea of Error Detection
To add redundant information to a frame that can be used to
determine if errors have been introduced
Imagine (Extreme Case)
Transmitting two complete copies of data
Identical  No error
Differ  Error
Poor Scheme ???
n bit message, n bit redundant information
Error can go undetected
In general, we can provide strong error detection technique
k redundant bits, n bits message, k << n
In Ethernet, a frame carrying up to 12,000 bits of data
requires only 32-bit CRC 21

Error Detection
 Extra bits are redundant
 They add no new information to the message
 Derived from the original message using some algorithm
 Both the sender and receiver know the algorithm
Sender Receiver
 Receiver computes r using m
 If they match, no error
m rm r
22

Two-dimensional parity
Two-dimensional parity is exactly what the name suggests
It is based on “simple” (one-dimensional) parity, which
usually involves adding one extra bit to a 7-bit code to
balance the number of 1s in the byte. For example,
Odd parity sets the eighth bit to 1 if needed to give an odd
number of 1s in the byte, and
Even parity sets the eighth bit to 1 if needed to give an even
number of 1s in the byte
Two-dimensional parity does a similar calculation
for each bit position across each of the bytes
contained in the frame
23

Two-dimensional parity
This results in an extra parity
byte for the entire frame, in
addition to a parity bit for each
byte
Two-dimensional parity catches
all 1-, 2-, and 3-bit errors and
most 4-bit errors
24

Internet Checksum Algorithm
Not used at the link level
Add up all the words that are transmitted and then transmit
the result of that sum
The result is called the checksum
The receiver performs the same calculation on the received
data and compares the result with the received checksum
If any transmitted data, including the checksum itself, is
corrupted, then the results will not match, so the receiver
knows that an error occurred
Consider the data being checksummed as a sequence of 16-
bit integers.
25

Add them together using 16-bit ones complement
arithmetic (explained next slide) and then take the ones
complement of the result.
That 16-bit number is the checksum
In ones complement arithmetic, a negative integer −x is
represented as the complement of x;
Each bit of x is inverted.
When adding numbers in ones complement arithmetic, a
carryout from the most significant bit needs to be added
to the result.
26

Consider, for example, the addition of −5 and −3 in
ones complement arithmetic on 4-bit integers
+5 is 0101, so −5 is 1010; +3 is 0011, so −3 is 1100
If we add 1010 and 1100 ignoring the carry, we get
0110
In ones complement arithmetic, the fact that this
operation caused a carry from the most significant bit
causes us to increment the result, giving 0111, which is
the ones complement representation of −8 (obtained by
inverting the bits in 1000), as we would expect
27

Cyclic Redundancy Check (CRC)
 Reduce the number of extra bits and maximize protection
 Given a bit string 110001 we can associate a polynomial on a single
variable x for it.
1.x5+1.x4+0.x3+0.x2+0.x1+1.x0 = x5+x4+1 and the degree is 5.
A k-bit frame has a maximum degree of k-1
 Let M(x) be a message polynomial and C(x) be a generator
polynomial.
 Let M(x)/C(x) leave a remainder of 0.
 When M(x) is sent and M’(x) is received we have M’(x) =
M(x)+E(x)
 The receiver computes M’(x)/C(x) and if the remainder is nonzero,
then an error has occurred.
 The only thing the sender and the receiver should know is C(x).
28

Polynomial Arithmetic Modulo 2
Any polynomial B(x) can be divided by a divisor polynomial
C(x) if B(x) is of higher degree than C(x).
Any polynomial B(x) can be divided once by a divisor
polynomial C(x) if B(x) is of the same degree as C(x).
The remainder obtained when B(x) is divided by C(x) is
obtained by subtracting C(x) from B(x).
To subtract C(x) from B(x), we simply perform the exclusive-
OR (XOR) operation on each pair of matching coefficients.
Let M(x) be a frame with m bits and let the generator
polynomial have less than m bits say equal to r.
Let r be the degree of C(x). Append r zero bits to the
low-order end of the frame, so it now contains m+r bits
and corresponds to the polynomial xrM(x).
29

 Divide the bit string
corresponding to xrM(x) by
the bit string corresponding to
C(x) using modulo 2 division.
 Subtract the remainder (which
is always r or fewer bits) from
the string corresponding to
xrM(x) using modulo 2
subtraction (addition and
subtraction are the same in
modulo 2).
 The result is the
checksummed frame to be
transmitted. Call it
polynomial M’(x).
CRC Calculation using
Polynomial Long Division
30

 Properties of Generator Polynomial
 Let P(x) represent what the sender sent and P(x) + E(x) is the received string.
A 1 in E(x) represents that in the corresponding position in P(x) the message
the bit is flipped.
 We know that P(x)/C(x) leaves a remainder of 0, but if E(x)/C(x) leaves a
remainder of 0, then either E(x) = 0 or C(x) is factor of E(x).
 When C(x) is a factor of E(x) we have problem; errors go unnoticed.
 If there is a single bit error then E(x) = xi, where i determines the bit in error.
If C(x) contains two or more terms it will never divide E(x), so all single bit
errors will be detected.
 In general, it is possible to prove that the following types of errors can be detected
by a C(x) with the stated properties
 All single-bit errors, as long as the xk and x0 terms have nonzero coefficients.
 All double-bit errors, as long as C(x) has a factor with at least three terms.
 Any odd number of errors, as long as C(x) contains the factor (x+1).
 Any “burst” error (i.e., sequence of consecutive error bits) for which the length
of the burst is less than k bits. (Most burst errors of larger than k bits can also
be detected.) 31

Six generator polynomials that have become
international standards are:
CRC-8 = x8+x2+x+1
CRC-10 = x10+x9+x5+x4+x+1
CRC-12 = x12+x11+x3+x2+x+1
CRC-16 = x16+x15+x2+1
CRC-CCITT = x16+x12+x5+1
CRC-32 =
x32+x26+x23+x22+x16+x12+x11+x10+x8+x7+x5+x4+x2+x+1
32

Reliable Transmission
 CRC is used to detect errors.
 Some error codes are strong enough to correct errors.
 The overhead is typically too high.
 Corrupt frames must be discarded.
 A link-level protocol that wants to deliver frames reliably must
recover from these discarded frames.
 This is accomplished using a combination of two fundamental
mechanisms
Acknowledgements and Timeouts
 An acknowledgement (ACK for short) is a small control frame that
a protocol sends back to its peer saying that it has received the
earlier frame.
A control frame is a frame with header only (no data).
33

Reliable Transmission
The receipt of an acknowledgement indicates to the
sender of the original frame that its frame was
successfully delivered.
If the sender does not receive an acknowledgment after a
reasonable amount of time, then it retransmits the original
frame.
The action of waiting a reasonable amount of time is
called a timeout.
The general strategy of using acknowledgements and
timeouts to implement reliable delivery is sometimes
called Automatic Repeat reQuest (ARQ).
34

Stop and Wait Protocol
Idea of stop-and-wait
protocol is
straightforward
After transmitting one
frame, the sender waits
for an acknowledgement
before transmitting the
next frame.
If the acknowledgement
does not arrive after a
certain period of time,
the sender times out and
retransmits the original
frame
Timeline showing four different scenarios for the stop-
and-wait algorithm. (a) The ACK is received before the
timer expires; (b) the original frame is lost; (c) the
ACK is lost; (d) the timeout fires too soon
35

 If the acknowledgment is lost or delayed
in arriving
 The sender times out and retransmits the
original frame, but the receiver will think
that it is the next frame since it has correctly
received and acknowledged the first frame
 As a result, duplicate copies of frames will
be delivered
 How to solve this
 Use 1 bit sequence number (0 or 1)
 When the sender retransmits frame 0, the
receiver can determine that it is seeing a
second copy of frame 0 rather than the first
copy of frame 1 and therefore can ignore it
(the receiver still acknowledges it, in case
the first acknowledgement was lost)
Timeline for stop-and-wait
with 1-bit sequence
number
36

 The sender has only one outstanding frame on the link at a
time
 This may be far below the link’s capacity
 Consider a 1.5 Mbps link with a 45 ms RTT
 The link has a delay  bandwidth product of 67.5 Kb or approximately
8 KB
 Since the sender can send only one frame per RTT and assuming a
frame size of 1 KB
 Maximum Sending rate
 Bits per frame  Time per frame = 1024  8  0.045 = 182 Kbps
 Or about one-eighth of the link’s capacity
 To use the link fully, then sender should transmit up to eight frames
before having to wait for an acknowledgement
37

Sliding Window Protocol
Timeline for Sliding Window Protocol
 Sender assigns a sequence
number denoted as SeqNum to
each frame.
 Assume it can grow infinitely large
 Sender maintains three variables
 Sending Window Size (SWS)
Upper bound on the number of
outstanding (unacknowledged)
frames that the sender can transmit
 Last Acknowledgement Received
(LAR)
Sequence number of the last
acknowledgement received
 Last Frame Sent (LFS)
Sequence number of the last frame
sent
38

 Sender also maintains the following invariant
 LFS – LAR ≤ SWS
 When an acknowledgement arrives
 the sender moves LAR to right, thereby allowing the sender to transmit
another frame
 Also the sender associates a timer with each frame it transmits
 It retransmits the frame if the timer expires before the ACK is received
 Note that the sender has to be willing to buffer up to SWS
frames
 WHY?
Sliding Window
on Sender
39

 Receiver maintains three variables
 Receiving Window Size (RWS)
Upper bound on the number of out-of-order frames that the receiver is willing to
accept
 Largest Acceptable Frame (LAF)
Sequence number of the largest acceptable frame
 Last Frame Received (LFR)
Sequence number of the last frame received
 Receiver also maintains the following invariant
LAF – LFR ≤ RWS
Sliding Window
on Receiver
40

 When a frame with sequence number SeqNum arrives, what does the
receiver do?
 If SeqNum ≤ LFR or SeqNum > LAF
Discard it (the frame is outside the receiver window)
 If LFR < SeqNum ≤ LAF
Accept it
Now the receiver needs to decide whether or not to send an ACK
 Let SeqNumToAck
Denote the largest sequence number not yet acknowledged, such that all frames
with sequence number less than or equal to SeqNumToAck have been received
 The receiver acknowledges the receipt of SeqNumToAck even if high-
numbered packets have been received
This acknowledgement is said to be cumulative.
 The receiver then sets
LFR = SeqNumToAck and adjusts
LAF = LFR + RWS 41

 For example, suppose LFR = 5 and RWS = 4
 (i.e. the last ACK that the receiver sent was for seq. no. 5)
 LAF = 9
 If frames 7 and 8 arrive, they will be buffered because they are
within the receiver window
 But no ACK will be sent since frame 6 is yet to arrive
 Frames 7 and 8 are out of order
 Frame 6 arrives (it is late because it was lost first time and had to
be retransmitted)
 Now Receiver Acknowledges Frame 8
 and bumps LFR to 8
 and LAF to 12
42

Issues with Sliding Window Protocol
 When timeout occurs, the amount of data in transit decreases
 Since the sender is unable to advance its window
 When the packet loss occurs, this scheme is no longer keeping the
pipe full
 The longer it takes to notice that a packet loss has occurred, the more severe
the problem becomes
 How to improve this
 Negative Acknowledgement (NAK)
 Additional Acknowledgement
 Selective Acknowledgement
 Negative Acknowledgement (NAK)
 Receiver sends NAK for frame 6 when frame 7 arrive (in the previous example)
 However this is unnecessary since sender’s timeout mechanism will be sufficient to catch
the situation
 Additional Acknowledgement
 Receiver sends additional ACK for frame 5 when frame 7 arrives
 Sender uses duplicate ACK as a clue for frame loss
43

 Selective Acknowledgement
 Receiver will acknowledge exactly those frames it has received, rather than the
highest number frames
Receiver will acknowledge frames 7 and 8
Sender knows frame 6 is lost
Sender can keep the pipe full (additional complexity)
 How to select the window size
 SWS is easy to compute
Delay  Bandwidth
 RWS can be anything
Two common setting
RWS = 1
No buffer at the receiver for frames that arrive out of order RWS = SWS
The receiver can buffer frames that the sender transmits
It does not make any sense to keep RWS > SWS
WHY?
44

Finite Sequence Number
Frame sequence number is specified in the header field
Finite size
3 bit: eight possible sequence number: 0, 1, 2, 3, 4, 5, 6, 7
It is necessary to wrap around
How to distinguish between different incarnations of the
same sequence number?
Number of possible sequence number must be larger than the
number of outstanding frames allowed
Stop and Wait: One outstanding frame
2 distinct sequence number (0 and 1)
Let MaxSeqNum be the number of available sequence numbers
SWS + 1 ≤ MaxSeqNum
Is this sufficient?
45

 SWS + 1 ≤ MaxSeqNum
Is this sufficient?
Depends on RWS
If RWS = 1, then sufficient
If RWS = SWS, then not good enough
 For example, we have eight sequence numbers 0, 1, 2, 3, 4, 5, 6, 7
 RWS = SWS = 7
 Sender sends 0, 1, …, 6
 Receiver receives 0, 1, … ,6
 Receiver acknowledges 0, 1, …, 6
 ACK (0, 1, …, 6) are lost
 Sender retransmits 0, 1, …, 6
 Receiver is expecting 7, 0, …., 5
46

To avoid this,
 If RWS = SWS
 SWS < (MaxSeqNum + 1)/2
Serves three different roles
 Reliable
 Preserve the order
Each frame has a sequence number
The receiver makes sure that it does not pass a frame up to the next
higher-level protocol until it has already passed up all frames with a
smaller sequence number
 Frame control
Receiver is able to throttle the sender
Keeps the sender from overrunning the receiver
From transmitting more data than the receiver is able to process
47

The Data Link Layer in the Internet
A home personal computer acting as an internet
host.
Technology like Ethernet cannot provide “high-level” functionality
like connection management and parameter negotiation
49

Point to Point Data Link Control
 One sender, one receiver, one link: easier than broadcast link:
No Media Access Control
No need for explicit MAC addressing
E.g., dialup link, ISDN line
 Popular point-to-point and high-level DLC protocols:
PPP (point-to-point protocol)
HDLC: High level data link control (Data link used to be considered
“high layer” in protocol stack). HDLC is also used in multi-point links
(one station many receivers)
 These protocols can often be run over other data link technologies
providing best of both worlds
E.g., PPPoE, HDLC encapsulation by Ethernet
50

PPP Design Requirements [RFC 1557]
Functionality : (similar to link layer services +
extra management functions)
Packet framing - encapsulation of network-layer datagram in
data link frame
Multi-protocol - carry network layer data of any network
layer protocol (not just IP) at same time ability to demultiplex
upwards
Bit transparency - must carry any bit pattern in the data field
(even if underlying channel can't)
Error detection - not correction
51

PPP Design Requirements (cont.)
The extra stuff:
 Connection liveness: detect, signal link failure to
network layer
 Network layer address negotiation: endpoint can
learn/configure each other’s network address and other
characteristics.
 Authentication: who are you (or at least whose account
do I bill for your dial-in time?)
 This information is used by traffic management software to control
bandwidth to individual subscribers
 Management features: loopback detection
52

PPP non-requirements
No error correction/recovery
(modems do one layer FEC, one layer packetization +
retransmission “under the covers” anyway; other
technologies are pretty reliable)
No flow control
Out of order delivery OK
Error recovery, flow control, data re-ordering
all relegated to higher layers!
53

PPP Data Frame
Flag: delimiter (framing)
Address: ignored. (historical)
Control: ignored. (historical)
Protocol: upper layer protocol to which frame
delivered (e.g., PPP-LCP, IP, IPCP, etc)
54

PPP Data Frame
info: upper layer data being carried
check: cyclic redundancy check for error detection
55

Byte Stuffing
flag byte
pattern
in data
to send
flag byte pattern plus
stuffed byte in
transmitted data
56

PPP Data Control Protocol
Before exchanging network-layer
data, data link peers must
Configure PPP link (max.
frame length, authentication)
Learn/configure network
 layer information
for IP: carry IP Control
Protocol (IPCP) msgs
(protocol field: 8021) to
configure/learn IP address
57

Where does PPP get used?
Dial-up – PPP over async serial, over modem
ADSL – PPP over Ethernet
Backbone – Packet over SONET (POS)
Why?
Framing (dialup, POS)
Efficiency (POS)
Authentication, address negotiation (PPPoE)
58

High-Level Data Link Control (HDLC)
HDLC was defined by ISO for use on both point-to-point
and multipoint data links.
It supports full-duplex communication
Other similar protocols are
Synchronous Data Link Control (SDLC) by IBM
Advanced Data Communication Control Procedure (ADCCP) by
ANSI
Link Access Procedure, Balanced (LAP-B) by CCITT, as part of
its X.25 packet-switched network standard
59

HDLC Overview
Broadly HDLC features are as follows:
Reliable protocol
selective repeat or go-back-N
Full-duplex communication
receive and transmit at the same time
Bit-oriented protocol
use bits to stuff flags occurring in data
Flow control
adjust window size based on receiver capability
Uses physical layer clocking and synchronization to send
and receive frames
60

HDLC Overview
Defines three types of stations
Primary
Secondary
Combined
Defines three types of data transfer mode
Normal Response mode
Asynchronous Response mode
Asynchronous Balanced mode
Three types of frames
Unnumbered
information
Supervisory
61

HDLC
The three stations are :
Primary station
Has the responsibility of controlling the operation of data flow
the link.
Handles error recovery
Frames issued by the primary station are called commands.
Secondary station,
Operates under the control of the primary station.
Frames issued by a secondary station are called responses.
The primary station maintains a separate logical link with each
secondary station.
Combined station,
Acts as both as primary and secondary station.
Does not rely on other for sending data
62

HDLC
Primary
Secondary Secondary
Commands
Responses
Combined Combined
commands/Responses
Unbalanced Mode
Balanced mode
63

HDLC
 The three modes of data transfer operations are
Normal Response Mode (NRM)
Mainly used in terminal-mainframe networks. In this case,
Secondaries (terminals) can only transmit when specifically
instructed by the primary station in response to a polling
Unbalanced configuration, good for multi-point links
Asynchronous Response Mode (ARM)
Same as NRM except that the secondaries can initiate
transmissions without direct polling from the primary station
Reduces overhead as no frames need to be sent to allow
secondary nodes to transmit
Transmission proceeds when channel is detected idle , used
mostly in point-to-point-links
Asynchronous Balanced Mode (ABM)
Mainly used in point-to-point links, for communication
between combined stations
64

Non-operational Modes
Normal Disconnected Mode
Asynchronous Disconnected Mode
Both the above modes mean that the secondary node is
logically disconnected from the primary node
Initialization Mode
A node negotiates transmission parameters with the
other node E.g., flow control information
Parameters negotiated in this mode are used during
any of the data transfer modes
65

Data Link Control HDLC frame structure
(a) Frame Format
(b) Control field format
66

Data Link Control
HDLC frame structure
(c) Extended address field (d) Extended control field
67

HDLC
 Flag: 01111110- start and ending delimiter. Bits are stuffed for flags in data frames
 FCS: 16-bit CRC using generating polynomial
 G(x) = x16 + x12 + x5 + 1
 Address field:
 mainly used in multidrop link configuration, and not used in point-to-point
 In unbalanced configuration, every secondary is assigned a unique address.
Contains address of secondary station in both command and response frames
 In balanced mode, command frame has destination address and response frame
has sending node’s address
 Group addresses are also possible. E.g., One command sent to all the
secondaries
 In I-frames, N(s) is the sequence number of the frame being sent, and R(s) is the
sequence number of the frame being expected.
 The P/F bit, known as the poll/final bit, is used with different meaning in different
contexts.
 It is used to indicate polling, to indicate the final I-frame, etc
68

HDLC
There are three different classes of frames used in
HDLC
Unnumbered frames, used in link setup and
disconnection, and hence do not contain ACK.
Information frames, which carry actual information.
Such frames can piggyback ACK in case of ABM
Supervisory frames, which are used for error and flow
control purposes and hence contain send and receive
sequence numbers
69

HDLC
There are four different supervisory frames
SS=00, Receiver Ready (RR), and N(R) ACKs all frames
received up to and including the one with sequence number N(R)
- 1
SS=10, Receiver Not Ready (RNR), and N(R) has the same
meaning as above
SS=01, Reject; all frames with sequence number N(R) or higher
are rejected, which in turns ACKs frames with sequence number
N(R) -1 or lower.
SS=11, Selective Reject; the receive rejects the frame with
sequence number N(R)
70

HDLC
The unnumbered frames can be grouped into the
following categories:
Mode-setting commands and responses
Recovery commends and responses
Miscellaneous commands and responses
71

Ethernet
 Most successful local area networking technology of last 20 years.
 Developed in the mid-1970s by researchers at the Xerox Palo Alto
Research Centers (PARC).
 Uses CSMA/CD technology
 Carrier Sense Multiple Access with Collision Detection.
 A set of nodes send and receive frames over a shared link.
 Carrier sense means that all nodes can distinguish between an idle and a busy
link.
 Collision detection means that a node listens as it transmits and can therefore
detect when a frame it is transmitting has collided with a frame transmitted by
another node.
 Uses ALOHA (packet radio network) as the root protocol
 Developed at the University of Hawaii to support communication across the
Hawaiian Islands.
 For ALOHA the medium was atmosphere, for Ethernet the medium is a coax
cable.
73

Ethernet
 DEC and Intel joined Xerox to define a 10-Mbps Ethernet
standard in 1978.
 This standard formed the basis for IEEE standard 802.3
 More recently 802.3 has been extended to include a 100-Mbps
version called Fast Ethernet and a 1000-Mbps version called
Gigabit Ethernet.
 An Ethernet segment is implemented on a coaxial cable of up to 500 m.
 This cable is similar to the type used for cable TV except that it typically
has an impedance of 50 ohms instead of cable TV’s 75 ohms.
 Hosts connect to an Ethernet segment by tapping into it.
 A transceiver (a small device directly attached to the tap) detects when
the line is idle and drives signal when the host is transmitting.
 The transceiver also receives incoming signal.
74

Ethernet
 The transceiver is connected to an
Ethernet adaptor which is plugged
into the host.
 The protocol is implemented on the
adaptor.
 Multiple Ethernet segments can
be joined together by repeaters.
 A repeater is a device that
forwards digital signals.
 No more than four repeaters may
be positioned between any pair of
hosts.
 An Ethernet has a total reach of only
2500 m.
75
Ethernet transceiver and adaptor

Ethernet
Ethernet repeater
76
Any signal placed on the
Ethernet by a host is
broadcast over the entire
network
Signal is propagated in both
directions.
Repeaters forward the signal
on all outgoing segments.
Terminators attached to the
end of each segment absorb
the signal.
Ethernet uses Manchester
encoding scheme.

Ethernet
New Technologies in Ethernet
 Instead of using coax cable, an Ethernet can be constructed from a thinner
cable known as 10Base2 (the original was 10Base5)
10 means the network operates at 10 Mbps
Base means the cable is used in a baseband system
2 means that a given segment can be no longer than 200 m
 Another cable technology is 10BaseT
T stands for twisted pair
Limited to 100 m in length
 With 10BaseT, the common configuration is to have several point to point
segments coming out of a multiway repeater, called Hub
77
Ethernet Hub

Access Protocol for Ethernet
 The algorithm is commonly called Ethernet’s Media Access Control
(MAC).
 It is implemented in Hardware on the network adaptor.
 Frame format
 Preamble (64bit): allows the receiver to synchronize with the signal
(sequence of alternating 0s and 1s).
 Host and Destination Address (48bit each).
 Packet type (16bit): acts as demux key to identify the higher level protocol.
 Data (up to 1500 bytes)
Minimally a frame must contain at least 46 bytes of data.
Frame must be long enough to detect collision.
 CRC (32bit)
78

Ethernet Addresses
 Each host on an Ethernet (in fact, every Ethernet host in the world)
has a unique Ethernet Address.
 The address belongs to the adaptor, not the host.
 It is usually burnt into ROM.
 Ethernet addresses are typically printed in a human readable format
 As a sequence of six numbers separated by colons.
 Each number corresponds to 1 byte of the 6 byte address and is given by a pair
of hexadecimal digits, one for each of the 4-bit nibbles in the byte
 Leading 0s are dropped.
 For example, 8:0:2b:e4:b1:2 is
00001000 00000000 00101011 11100100 10110001 00000010
 To ensure that every adaptor gets a unique address, each
manufacturer of Ethernet devices is allocated a different prefix that
must be prepended to the address on every adaptor they build
AMD has been assigned the 24bit prefix 8:0:20
79

Ethernet Addresses
 Each frame transmitted on an Ethernet is received by every adaptor
connected to that Ethernet.
 Each adaptor recognizes those frames addressed to its address and
passes only those frames on to the host.
 In addition, to unicast address, an Ethernet address consisting of all
1s is treated as a broadcast address.
 All adaptors pass frames addressed to the broadcast address up to the host.
 Similarly, an address that has the first bit set to 1 but is not the
broadcast address is called a multicast address.
 A given host can program its adaptor to accept some set of
multicast addresses. To summarize, an Ethernet adaptor receives all
frames and accepts
 Frames addressed to its own address
 Frames addressed to the broadcast address
 Frames addressed to a multicast addressed if it has been instructed
80

Ethernet Transmitter Algorithm
 When the adaptor has a frame to send and the line is idle, it transmits
the frame immediately.
 The upper bound of 1500 bytes in the message means that the adaptor can
occupy the line for a fixed length of time.
 When the adaptor has a frame to send and the line is busy, it waits
for the line to go idle and then transmits immediately.
 The Ethernet is said to be 1-persistent protocol because an adaptor
with a frame to send transmits with probability 1 whenever a busy
line goes idle.
 Since there is no centralized control it is possible for two (or more)
adaptors to begin transmitting at the same time,
 Either because both found the line to be idle,
 Or, both had been waiting for a busy line to become idle.
 When this happens, the two (or more) frames are said to be collide
on the network.
81

 Since Ethernet supports collision detection, each sender is able
to determine that a collision is in progress.
 At the moment an adaptor detects that its frame is colliding
with another, it first makes sure to transmit a 32-bit jamming
sequence and then stops transmission.
 Thus, a transmitter will minimally send 96 bits in the case of collision
64-bit preamble + 32-bit jamming sequence
 One way that an adaptor will send only 96 bit (called a runt
frame) is if the two hosts are close to each other.
 Had they been farther apart,
 They would have had to transmit longer, and thus send more bits,
before detecting the collision.
82

The worst case scenario happens when the two hosts are at
opposite ends of the Ethernet.
To know for sure that the frame its just sent did not collide
with another frame, the transmitter may need to send as
many as 512 bits.
Every Ethernet frame must be at least 512 bits (64 bytes) long.
14 bytes of header + 46 bytes of data + 4 bytes of CRC
Why 512 bits?
Why is its length limited to 2500 m?
The farther apart two nodes are, the longer it takes for a
frame sent by one to reach the other, and the network is
vulnerable to collision during this time
83

 A begins transmitting a frame at time t
 d denotes the one link latency
 The first bit of A’s frame arrives at B at
time t + d
 Suppose an instant before host A’s frame
arrives, host B begins to transmit its own
frame
 B’s frame will immediately collide with A’s
frame and this collision will be detected by
host B
 Host B will send the 32-bit jamming
sequence
 Host A will not know that the collision
occurred until B’s frame reaches it, which
will happen at t + 2 * d
 Host A must continue to transmit until this
time in order to detect the collision
Host A must transmit for 2 * d to be
sure that it detects all possible
collisions
Worst-case scenario: (a) A sends a frame
at time t; (b) A’s frame arrives at B at time
t + d; (c) B begins transmitting at time t +
d and collides with A’s frame; (d) B’s runt
(32-bit) frame arrives at A at time t + 2d.
84

Consider that a maximally configured Ethernet is 2500 m
long, and there may be up to four repeaters between any two
hosts, the round trip delay has been determined to be 51.2
s
Which on 10 Mbps Ethernet corresponds to 512 bits
The other way to look at this situation,
We need to limit the Ethernet’s maximum latency to a fairly small
value (51.2 s) for the access algorithm to work
Hence the maximum length for the Ethernet is on the order of 2500 m.
 Once an adaptor has detected a collision, and stopped its transmission,
it waits a certain amount of time and tries again.
 Each time the adaptor tries to transmit but fails, it doubles the amount
of time it waits before trying again.
85

 This strategy of doubling the delay interval between each
retransmission attempt is known as Exponential Backoff.
 The adaptor first delays either 0 or 51.2 s, selected at random.
 If this effort fails, it then waits 0, 51.2, 102.4, 153.6 s (selected
randomly) before trying again;
 This is k * 51.2 for k = 0, 1, 2, 3
 After the third collision, it waits k * 51.2 for k = 0…23 – 1 (again
selected at random).
 In general, the algorithm randomly selects a k between 0 and 2n –
1 and waits for k * 51.2 s, where n is the number of collisions
experienced so far.
86

Experience with Ethernet
 Ethernets work best under lightly loaded conditions.
 Under heavy loads, too much of the network’s capacity is wasted by
collisions.
 Most Ethernets are used in a conservative way.
 Have fewer than 200 hosts connected to them which is far fewer than the
maximum of 1024.
 Most Ethernets are far shorter than 2500m with a round-trip
delay of closer to 5 s than 51.2 s.
 Ethernets are easy to administer and maintain.
 There are no switches that can fail and no routing and configuration tables
that have to be kept up-to-date.
 It is easy to add a new host to the network.
 It is inexpensive.
 Cable is cheap, and only other cost is the network adaptor on each host.
87

Wireless Links
 Wireless links transmit electromagnetic signals
 Radio, microwave, infrared
 Wireless links all share the same “wire” (so to speak)
 The challenge is to share it efficiently without unduly interfering with each other
 Most of this sharing is accomplished by dividing the “wire” along the
dimensions of frequency and space
 Exclusive use of a particular frequency in a particular geographic area
may be allocated to an individual entity such as a corporation
 Devices that use license-exempt frequencies are still subject to certain
restrictions
 The first is a limit on transmission power
 This limits the range of signal, making it less likely to interfere with another
signal
 For example, a cordless phone might have a range of about 100 feet.
88

Wireless Links
These allocations are determined by government agencies
such as FCC (Federal Communications Commission) in
USA
Specific bands (frequency) ranges are allocated to certain
uses.
Some bands are reserved for government use
Other bands are reserved for uses such as AM radio, FM radio,
televisions, satellite communications, and cell phones
Specific frequencies within these bands are then allocated to
individual organizations for use within certain geographical areas.
Finally, there are several frequency bands set aside for “license
exempt” usage
Bands in which a license is not needed
89

Wireless Links
The second restriction requires the use of Spread
Spectrum technique
Idea is to spread the signal over a wider frequency band
So as to minimize the impact of interference from other devices
Originally designed for military use
Frequency hopping
Transmitting signal over a random sequence of frequencies
First transmitting at one frequency, then a second, then a third…
The sequence of frequencies is not truly random, instead computed
algorithmically by a pseudorandom number generator
The receiver uses the same algorithm as the sender, initializes it with
the same seed, and is
Able to hop frequencies in sync with the transmitter to correctly receive
the frame
90

Wireless Links
 A second spread spectrum technique called Direct sequence
 Represents each bit in the frame by multiple bits in the transmitted signal.
 For each bit the sender wants to transmit
It actually sends the exclusive OR of that bit and n random bits
 The sequence of random bits is generated by a pseudorandom number
generator known to both the sender and the receiver.
 The transmitted values, known as an n-bit chipping code, spread the
signal across a frequency band that is n times wider
Example 4-bit chipping sequence
91

Wireless Links
 Wireless technologies differ in a variety of dimensions
 How much bandwidth they provide
 How far apart the communication nodes can be
 Four prominent wireless technologies: Bluetooth, Wi-Fi (more formally
known as 802.11), WiMAX (802.16), 3G cellular wireless
Overview of leading wireless technologies
92

Wireless Links
 Mostly widely used
wireless links today
are usually
asymmetric
 Two end-points are
usually different kinds
of nodes
 One end-point usually
has no mobility, but
has wired connection
to the Internet
(known as base
station)
 The node at the other
end of the link is often
mobile
A wireless network using a base station
93

Wireless Links
 Wireless communication supports point-to-multipoint communication
 Communication between non-base (client) nodes is routed via the base
station
 Three levels of mobility for clients
 No mobility: the receiver must be in a fix location to receive a directional
transmission from the base station (initial version of WiMAX)
 Mobility is within the range of a base (Bluetooth)
 Mobility between bases (Cell phones and Wi-Fi)
 Mesh or Ad-hoc network
 Messages may be forwarded via a chain of peer nodes
 Nodes are peers
94

IEEE 802.11
 Also known as Wi-Fi
 Like its Ethernet and token ring siblings, 802.11 is designed for use
in a limited geographical area (homes, office buildings, campuses)
 Primary challenge is to mediate access to a shared communication medium –
in this case, signals propagating through space
 802.11 supports additional features
 power management and
 security mechanisms
 Original 802.11 standard defined two radio-based physical layer standard
 One using the frequency hopping
 Over 79 1-MHz-wide frequency bandwidths
 Second using direct sequence
 Using 11-bit chipping sequence
 Both standards run in the 2.4-GHz and provide up to 2 Mbps
95

IEEE 802.11
Then physical layer standard 802.11b was added
Using a variant of direct sequence 802.11b provides up to
11 Mbps
Uses license-exempt 2.4-GHz band
Then came 802.11a which delivers up to 54 Mbps
using OFDM
802.11a runs on license-exempt 5-GHz band
Most recent standard is 802.11g which is backward
compatible with 802.11b
Uses 2.4 GHz band, OFDM and delivers up to 54 Mbps
96

IEEE 802.11 – Collision Avoidance
 Consider the situation in the following figure where each of four
nodes is able to send and receive signals that reach just the nodes to
its immediate left and right
 For example, B can exchange frames with A and C, but it cannot reach D
 C can reach B and D but not A
Example of a wireless network
97

 Suppose both A and C want
to communicate with B and
so they each send it a frame.
 A and C are unaware of each
other since their signals do not
carry that far
 These two frames collide with
each other at B
But unlike an Ethernet,
neither A nor C is aware of
this collision
 A and C are said to hidden
nodes with respect to each
other
The “Hidden Node” Problem. Although
A and C are hidden from each
other, their signals can collide at B. (B’s
reach is not shown.)
98

 Another problem called exposed
node problem occurs
 Suppose B is sending to A. Node
C is aware of this communication
because it hears B’s transmission.
 It would be a mistake for C to
conclude that it cannot transmit to
anyone just because it can hear B’s
transmission.
 Suppose C wants to transmit to
node D.
 This is not a problem since C’s
transmission to D will not interfere
with A’s ability to receive from B.
Exposed Node Problem. Although B and
C are exposed to each other’s signals,
there is no interference if B transmits to A
while C transmits to D. (A and D’s reaches
are not shown.)
99

802.11 addresses these two problems with an algorithm
called Multiple Access with Collision Avoidance (MACA).
Key Idea
Sender and receiver exchange control frames with each other
before the sender actually transmits any data.
This exchange informs all nearby nodes that a transmission is
about to begin
Sender transmits a Request to Send (RTS) frame to the receiver.
The RTS frame includes a field that indicates how long the sender wants to
hold the medium
- Length of the data frame to be transmitted
Receiver replies with a Clear to Send (CTS) frame
This frame echoes this length field back to the sender
100

Any node that sees the CTS frame knows that
it is close to the receiver, therefore
cannot transmit for the period of time it takes to
send a frame of the specified length
Any node that sees the RTS frame but not the
CTS frame
is not close enough to the receiver to interfere with
it, and
so is free to transmit
101

 Using ACK in MACA
 Proposed in MACAW: MACA for Wireless LANs
 Receiver sends an ACK to the sender after successfully receiving a
frame
 All nodes must wait for this ACK before trying to transmit
 If two or more nodes detect an idle link and try to transmit an RTS
frame at the same time
 Their RTS frame will collide with each other
 802.11 does not support collision detection
 So the senders realize the collision has happened when they do not receive the
CTS frame after a period of time
 In this case, they each wait a random amount of time before trying again.
 The amount of time a given node delays is defined by the same exponential
backoff algorithm used on the Ethernet.
102

IEEE 802.11 – Distribution System
802.11 is suitable for an ad-hoc configuration of nodes that
may or may not be able to communicate with all other
nodes.
Nodes are free to move around
The set of directly reachable nodes may change over time
To deal with this mobility and partial connectivity,
802.11 defines additional structures on a set of nodes
Instead of all nodes being created equal,
some nodes are allowed to roam
some are connected to a wired network infrastructure
 they are called Access Points (AP) and they are connected to each other by a
so-called distribution system
103

 Following figure illustrates a distribution system that connects three access points,
each of which services the nodes in the same region
 Each of these regions is analogous to a cell in a cellular phone system with the
APIs playing the same role as a base station
 The distribution network runs at layer 2 of the ISO architecture
 Although two nodes can communicate directly with each other if they are within
reach of each other, the idea behind this configuration is
 Each nodes associates itself with one access point
 For node A to communicate with node E, A first sends a frame to its AP-1 which
forwards the frame across the distribution system to AP-3, which finally transmits the
frame to E
Access points connected
to a distribution network
104

 How do the nodes select their access points
 How does it work when nodes move from one cell to another
 The technique for selecting an AP is called scanning
 The node sends a Probe frame
 All APs within reach reply with a Probe Response frame
 The node selects one of the access points and sends that AP an Association
Request frame
 The AP replies with an Association Response frame
 A node engages this protocol whenever
 it joins the network, as well as
 when it becomes unhappy with its current AP
This might happen, for example, because the signal from its current AP has
weakened due to the node moving away from it
Whenever a node acquires a new AP, the new AP notifies the old AP of the
change via the distribution system
105

 Consider the situation shown in the following figure when node C moves from the
cell serviced by AP-1 to the cell serviced by AP-2.
 As it moves, it sends Probe frames, which eventually result in Probe Responses
from AP-2.
 At some point, C prefers AP-2 over AP-1 , and so it associates itself with that
access point.
 This is called active scanning since the node is actively searching for an access point
 APs also periodically send a Beacon frame that advertises the capabilities of the
access point; these include the transmission rate supported by the AP
 This is called passive scanning
 A node can change to this AP based on the Beacon frame simply by sending it an
Association Request frame back to the access point.
Node Mobility
106

IEEE 802.11 – Frame Format
Source and Destinations addresses: each 48 bits
Data: up to 2312 bytes
CRC: 32 bit
Control field: 16 bits
Contains three subfields (of interest)
6 bit Type field: indicates whether the frame is an RTS or CTS frame or
being used by the scanning algorithm
A pair of 1 bit fields : called ToDS and FromDS
Frame Format
107

 Frame contains four addresses
 How these addresses are interpreted depends on the settings of
the ToDS and FromDS bits in the frame’s Control field
 This is to account for the possibility that the frame had to be
forwarded across the distribution system which would mean
that,
 the original sender is not necessarily the same as the most recent
transmitting node
 Same is true for the destination address
 Simplest case
 When one node is sending directly to another, both the DS bits are 0,
Addr1 identifies the target node, and Addr2 identifies the source node
108

Most complex case
Both DS bits are set to 1
Indicates that the message went from a wireless node onto the
distribution system, and then from the distribution system to
another wireless node
With both bits set,
Addr1 identifies the ultimate destination,
Addr2 identifies the immediate sender (the one that forwarded the
frame from the distribution system to the ultimate destination)
Addr3 identifies the intermediate destination (the one that accepted
the frame from a wireless node and forwarded across the
distribution system)
Addr4 identifies the original source
Addr1: E, Addr2: AP-3, Addr3: AP-1, Addr4: A
109

Bluetooth
 Used for very short range
communication between mobile
phones, PDAs, notebook
computers and other personal or
peripheral devices
 Operates in the license-exempt
band at 2.45 GHz
 Has a range of only 10 m
 Communication devices
typically belong to one
individual or group
 Sometimes categorized as Personal
Area Network (PAN) A Bluetooth Piconet
110

Bluetooth
 Version 2.0 provides speeds up to 2.1 Mbps
 Power consumption is low
 Bluetooth is specified by an industry consortium called the
Bluetooth Special Interest Group
 It specifies an entire suite of protocols, going beyond the link layer
to define application protocols, which it calls profiles, for a range of
applications
 There is a profile for synchronizing a PDA with personal computer
 Another profile gives a mobile computer access to a wired LAN
 The basic Bluetooth network configuration is called a piconet
 Consists of a master device and up to seven slave devices
 Any communication is between the master and a slave
 The slaves do not communicate directly with each other
 A slave can be parked: set to an inactive, low-power state
111

ZigBee
ZigBee is a new technology that competes with
Bluetooth
Devised by the ZigBee alliance and standardized as IEEE
802.15.4
It is designed for situations where the bandwidth
requirements are low and power consumption must be
very low to give very long battery life
It is also intended to be simpler and cheaper than
Bluetooth, making it financially feasible to incorporate in
cheaper devices such as a wall switch that wirelessly
communicates with a ceiling-mounted fan
112

Cs8591 Computer Networks

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