CS 372 – introduction to computer networks*
Announcements:
 Assign 5 and Lab 5 will be posted today, and due
next Thursday.
* Based in part on slides by Bechir Hamdaoui and Paul D. Paulson.
Acknowledgement: slides drawn heavily from Kurose & Ross
Chapter 5, slide: 1
Chapter 5: Data Link Layer
Our goals:
 understand principles behind data link layer services:
 error detection, correction
 sharing a broadcast channel: multiple access
 link layer addressing
 instantiation and implementation of link layer
technologies

Ethernet
Chapter 5, slide: 2
Link Layer
 1 Introduction and services
 2 Error detection and correction
 3 Multiple access protocols
 4 Link-layer Addressing
 5 Ethernet
Chapter 5, slide: 3
Link Layer (also known as layer 2)
Some terminology:
 nodes = hosts or routers
 links = communication
channels that connect
adjacent nodes
• wired links
• wireless links
 frame = layer-2 packet
data-link layer has responsibility of
transferring datagram from one node
to adjacent node over a link
Chapter 5, slide: 4
Link layer: context
 links may have different
link protocols
frames may be delivered by
different link protocols over
different links: e.g.,
 Ethernet on 1st link, frame
relay on intermediate links,
802.11 on last link
transportation analogy
 Trip: Princeton to Berlin



 tourist = frame
 travel agent = routing
algorithm

 link protocols may provide
different services

limo: Princeton to JFK
plane: JFK to Munich
train: Munich to Berlin
Doesn’t know/care of mode
 transportation mode = link
layer protocol
e.g., may or may not provide
rdt over link
Chapter 5, slide: 5
Link Layer Services
 framing, link access:



encapsulate datagram into frame, adding header, trailer
channel access if shared medium
“MAC” addresses used in frame headers to identify
source, dest
• different from IP address!
 reliable delivery between adjacent nodes
 similar to what is done at transport layer
 rarely used on low bit-error link (fiber, some twisted
pair)
 often used on wireless links: high error rates
Chapter 5, slide: 6
Link Layer Services (more)
 flow control:

pacing between adjacent sending and receiving nodes
 error detection:


errors caused by signal attenuation, noise.
receiver detects presence of errors:
• signals sender for retransmission or drops frame
 error correction:

receiver identifies and corrects bit error(s) without
resorting to retransmission
Chapter 5, slide: 7
Where is the link layer implemented?
 in all hosts
 implemented in network adapter
(NIC network interface card)


Ethernet card, PCMCI card,
802.11 card
implements link to physical layer
host schematic
application
transport
network
link
cpu
memory
 attaches into host’s system
buses
 combination of hardware and
controller
link
physical
host
bus
(e.g., PCI)
physical
transmission
software
network adapter
card
Chapter 5, slide: 8
NICs Communicating
datagram
datagram
controller
controller
receiving host
sending host
datagram
frame
 sending side:



encapsulates datagram in
hardware frame
adds address info, error
checking bits, reliable data
transfer, flow control, etc.
sends to all hosts that are
directly connected
 receiving side



checks address
looks for errors, reliable
data transfer, flow control,
etc
extracts datagram from
hardware frame, and passes
it up to next layer
Chapter 5, slide: 9
Link Layer
 1 Introduction and services
 2 Error detection and correction
 3 Multiple access protocols
 4 Link-layer Addressing
 5 Ethernet
Chapter 5, slide: 10
Error Detection
D = Data protected by error checking, may include header fields
EDC= Error Detection and Correction bits (redundancy)
• error detection not 100% reliable!: may not detect errors (rarely)
• larger EDC field yields better detection and correction
• Parity check, Checksum, Cyclic redundancy check (CRC)
Chapter 5, slide: 11
Example: Parity Checking
Single Bit Parity:
Detect single bit errors
Two Dimensional Bit Parity:
Detect and correct single bit errors
1
In this example:
D = data bits
EDC = parity bit
0
0
Chapter 5, slide: 12
Checksumming: Cyclic Redundancy Check
Scheme:
Example:

D= 101110
 Fix r CRC bits

r = 3 bits
 Choose r+1 bit pattern (generator), G

G = 1001
 Sender: choose r CRC bits, R, such that

 View data bits, D, as a binary number

G exactly divides <D,R> (modulo 2)
 receiver knows G, divides <D,R> by G.
If non-zero remainder: error detected!
 widely used in practice (Ethernet, 802.11
WiFi, ATM)

Find R such that
G divides <D,R> (modulo 2)
equivalently
G divides <D.2r XOR R>
Chapter 5, slide: 13
CRC: modulo-2 arithmetic
 CRC calculations are done modulo-2 arithmetic
No carries in addition
 No borrows in subtraction
=> Addition = subtraction = bitwise-XOR
 XOR review:

0 XOR 0 = 0; 0 XOR 1 = 1; 1 XOR 0 = 1; 1 XOR 1 = 0
=> a XOR 0 = a & a XOR a = 0 for a=0,1
E.g.: 1010 XOR 0110 = 1110

Multiplications and divisions are same as in base-2
arithmetic, except all additions and subtractions
are done without carries or borrows
• That is: addition = subtraction = bitwise-XOR
Chapter 5, slide: 14
CRC Example
Want: G divides <D.2r XOR R>
equivalently: D.2r XOR R = nG
equivalently: (XOR both sides)
(D.2r XOR R) XOR R = nG XOR R
equivalently: (R XOR R = 0 & a XOR 0 = a)
D.2r = nG XOR R
equivalently: (division is modulo 2 too)
since remainder[(nG XOR R)/G] = R
(note: R < G), then dividing D.2r by G
(modulo 2) gives remainder = R
R = remainder[
D.2r
G
]
Chapter 5, slide: 15
Link Layer
 1 Introduction and services
 2 Error detection and correction
 3 Multiple access protocols
 4 Link-layer Addressing
 5 Ethernet
Chapter 5, slide: 16
Multiple Access Links and Protocols
Two types of “links”:
 point-to-point
 PPP for dial-up access
 point-to-point link between Ethernet switch and host
 broadcast (shared wire or medium)
 Ethernet
 802.11 wireless LAN
shared wire (e.g.,
cabled Ethernet)
shared RF
(e.g., 802.11 WiFi)
shared RF
(satellite)
humans at a
cocktail party
(shared air, acoustical)
Chapter 5, slide: 17
Multiple Access protocols
need for sharing of medium/channel
 single channel
needs be used by all nodes
 interference/collision
two or more simultaneous transmissions lead to collided signals
multiple access protocol
 allows multiple, concurrent access
algorithm that nodes use to share channel, i.e., determines when a
node can transmit
 no coordination, no out-of-band channel
agreeing about channel sharing must use channel itself!
Chapter 5, slide: 18
MAC Protocols: a taxonomy
Three broad classes:
 Channel Partitioning


divide channel into smaller “pieces” (time slots,
frequency)
allocate piece to node for exclusive use
 Random Access
 channel not divided, allow collisions
 need to know how to “recover” from collisions
 “Taking turns”
 nodes take turns, but nodes with more to send can take
longer turns
Chapter 5, slide: 19
Channel Partitioning MAC protocols: TDMA
TDMA: time division multiple access
 access to channel in "rounds"
 each station gets fixed length slot (length = pkt
trans time) in each round
 unused slots go idle
 E.g.: 6-station LAN, 1,3,4 have pkt, slots 2,5,6 idle
6-slot
frame
1
3
4
1
3
4
Chapter 5, slide: 20
Channel Partitioning MAC protocols: FDMA
FDMA: frequency division multiple access
 channel spectrum divided into frequency bands
 each station assigned fixed frequency band
 unused transmission time in frequency bands go idle
 E.g.: 6-station LAN, 1,3,4 have pkt, frequency bands
FDM cable
frequency bands
2,5,6 idle
Chapter 5, slide: 21
Random Access Protocols
 distributed
 unlike TDMA or FDMA
 no coordination among nodes
 one node at a time
 when a node transmits, it does so at full data rate R.
 collisions can occur
 two or more transmitting nodes ➜ “collision”,
 random access MAC protocol specifies:
 how to detect collisions
 how to recover from collisions
 examples of random access MAC protocols:
 ALOHA, CSMA/CD, CSMA/CA
(CSMA: Carrier Sense Multiple Access;
CD: Collision Detection; CA: Collision Avoidance)
Chapter 5, slide: 22
Slotted ALOHA
Assumptions:
 all frames same size
Operation:
 when node gets a fresh frame,
transmits in next slot
 time divided into slots

slot = time to transmit 1 frame

 start transmit at beginning
of slot only
 nodes are synchronized

if no collision: node can send
new frame in next slot
if collision: node retransmits
frame in each subsequent slot
with prob. p until success
 if multiple nodes transmit in
slot, all can detect collision
Chapter 5, slide: 23
Slotted ALOHA
Pros
 single active node can
continuously transmit
at full rate of channel
 highly decentralized:
only slots in nodes
need to be in sync
 simple
Legend:
C = collision
E = empty/idle
S = success
Cons
 collisions, wasting slots
 idle slots, wasting slots
 clock synchronization
Chapter 5, slide: 24
Slotted Aloha efficiency
Efficiency : long-run fraction of successful slots
(many nodes, all with many frames to send)
 suppose: N nodes with many frames to send,
each transmits in slot with probability p
 prob that given node has success in a slot ?
p(1-p)N-1
 prob that exactly one node of N nodes has a success ?
Np(1-p)N-1
 Efficiency = Np(1-p)N-1
Chapter 5, slide: 25
Slotted Aloha efficiency
 Efficiency = Np(1-p)N-1
 max efficiency: find p* that
maximizes Np(1-p)N-1
p* = 1/N
Max Eff = (1-1/N)N-1
 When N increases to
,
max eff = (1-1/N)N-1 goes
to 1/e = .37
At best: channel used for 37%
useful transmission time !
Chapter 5, slide: 26
CSMA (Carrier Sense Multiple Access)
CSMA: listen before transmit:
If channel sensed idle: transmit entire frame
 If channel sensed busy, defer transmission
 human analogy: don’t interrupt others!
Chapter 5, slide: 29
CSMA collisions
spatial layout of nodes
collisions can still occur:
propagation delay means
two nodes may not hear
each other’s transmission
collision:
entire packet transmission
time wasted
note:
role of distance & propagation
delay in determining collision
probability
Chapter 5, slide: 30
CSMA/CD (Collision Detection)
CD (collision detection):
easy in wired LANs: measure signal strengths,
compare transmitted, received signals
 difficult in wireless LANs: received signal strength
overwhelmed by local transmission strength
 human analogy: the polite conversationalist

CSMA/CD: (CSMA w/ Collision Detection)
collisions detectable
 colliding transmissions aborted
 reducing channel wastage

Chapter 5, slide: 31
CSMA/CD collision detection
Chapter 5, slide: 32
“Taking Turns” MAC protocols
Polling:
 master node
“invites” slave nodes
to transmit in turn
data
poll
master
 concerns:
 polling overhead
 latency
 single point of
failure (master)
data
slaves
Chapter 5, slide: 33
“Taking Turns” MAC protocols
Token passing:
 control token passed
from one node to next
sequentially.
 If token, then send
message
 concerns:
 token overhead
 latency
 single point of failure
(token)
T
(nothing
to send)
T
data
Chapter 5, slide: 34
Summary of MA protocols
 channel partitioning, by time or frequency

Time Division, Frequency Division
 random access (dynamic),
 Don’t start talking again right away
 Waiting for a random time before trying again
 carrier sensing: easy in some technologies
(wire), hard in others (wireless)
• CSMA/CD used in Ethernet
• CSMA/CA used in 802.11
 taking turns
 polling from central site, token passing
• Bluetooth, FDDI, IBM Token Ring
Chapter 5, slide: 35
Link Layer
 1 Introduction and services
 2 Error detection and correction
 3 Multiple access protocols
 4 Topologies and switches
 5 Link-layer Addressing
 6 Ethernet
Chapter 5, slide: 36
A little more about Physical Layer:
LAN topologies
 Networks may be classified by "shape"
 Three most popular:
 Star
 Ring
 Bus
 Networks may be classified by “shared
medium"
Cable
 Wireless
 Others …

Chapter 5, slide: 37
Star topology
 All computers attach to a central point:
 "Center" of star is usually a server or a
switched hub
Chapter 5, slide: 38
Star topology
 Don't be confused by a diagram.
 “Topology” refers to logical connections (not
physical layout)
 A likely diagram of a star topology LAN:
Chapter 5, slide: 39
Star topology
 If implemented with a hub, the hub must be
"programmable" (switched)


Switch “learns” connections
Can services multiple transmissions simultaneously
Chapter 5, slide: 40
Ring topology
 Computers connected in a closed loop
 First passes data to second, second passes
data to third, etc.
Chapter 5, slide: 41
Ring topology
 Refers to logical connections (not physical
layout)
 In practice, there is a short connector
cable from the computer to the ring
Chapter 5, slide: 42
Ring topology
 Can be implemented inside a “box"
 Don't confuse with Star topology
Chapter 5, slide: 43
Bus topology
 Single cable connects all computers
 Each computer has connector to shared cable
 Computers must synchronize and allow only one
computer to transmit at a time
Terminator
Terminator
Chapter 5, slide: 44
Bus topology
 Can be implemented inside a “box“
(unswitched hub)
 Don't confuse with Star topology
Chapter 5, slide: 45
Hubs
… physical-layer (“dumb”) repeaters:
 bits coming in one link go out all other links at
same rate
 all nodes connected to hub can collide with one
another
 no frame buffering
 no CSMA/CD at hub: host NICs detect
collisions
twisted pair
hub
Chapter 5, slide: 46
Switch
 link-layer device: smarter than hubs, take
active role
store, forward Ethernet frames
 examine incoming frame’s MAC address,
selectively forward frame to one-or-more
outgoing links when frame is to be forwarded on
segment, uses CSMA/CD to access segment

 transparent
 hosts are unaware of presence of switches
 plug-and-play, self-learning

switches do not need to be configured
Chapter 5, slide: 47
Switch Table
 Q: how does switch know that
A’ reachable via interface 4,
B’ reachable via interface 5?
 A: each switch has a switch
table, each entry:

C’
B
1 2
3
6
5 4
(MAC address of host, interface
to reach host, time stamp)
 looks like a routing table!
 Q: how are entries created,
maintained in switch table?

A
something like a routing
protocol?
C
B’
A’
switch with six interfaces
(1,2,3,4,5,6)
Chapter 5, slide: 48
Switch: self-learning
 switch learns which hosts
can be reached through
which interfaces


Source: A
Dest: A’
A A A’
C’
B
1 2
3
6
5 4
when frame received,
switch “learns” location of
sender: incoming LAN
segment
records sender/location
pair in switch table
C
B’
A’
MAC addr interface TTL
A
1
60 Switch table
(initially empty)
Chapter 5, slide: 49
Self-learning,
forwarding:
example
Source: A
Dest: A’
A A A’
C’
B
 frame destination
unknown: flood
1 2
3
A6 A’
5 4
 destination A
location known:
selective send
B’
C
A’ A
A’
MAC addr interface TTL
A
A’
1
4
60 Switch table
60 (initially empty)
Chapter 5, slide: 50
Interconnecting switches
 switches can be connected together
S4
S1
A
B
C
S3
S2
D
F
E
G
H
I
 Q: sending from A to G - how does S1 know to
forward frame destined to F via S4 and S3?
 A: self learning! (works exactly the same as in
single-switch case!)
Chapter 5, slide: 51
Institutional network
to external
network
mail server
router
web server
IP subnet
Chapter 5, slide: 52
Switches vs. Routers
 both store-and-forward devices
 routers: network layer devices (examine network layer
headers)
 switches are link layer devices
 routers maintain routing tables, implement routing
algorithms
 switches maintain switch tables, implement
filtering, learning algorithms
Switch
Chapter 5, slide: 53
Link Layer
 1 Introduction and services
 2 Error detection and correction
 3 Multiple access protocols
 4 Topologies and switches
 5 Link-layer Addressing
 6 Ethernet
Chapter 5, slide: 54
MAC Addresses
 32-bit IP address:


network-layer address
used to get datagram to destination IP subnet
 MAC (or LAN or physical or Ethernet) address:

48 bit MAC address (for most LANs)
• burned in NIC ROM, also sometimes software settable




function: get frame from one interface to another physicallyconnected interface (same network)
uses hexadecimal representation; i.e., base-16 (i.e., 4 bits)
it uses 16 distinct symbols: 0,1,…,9,A,B,C,D,E,F
4 bits needed for each symbol:  48/4 = 12 symbols
E.g.: 1A-2F-BB-76-09-AD
Chapter 5, slide: 55
MAC Addresses
Each adapter on LAN has unique MAC (LAN ) address
1A-2F-BB-76-09-AD
71-65-F7-2B-08-53
LAN
(wired or
wireless)
Broadcast address =
FF-FF-FF-FF-FF-FF
= adapter
58-23-D7-FA-20-B0
0C-C4-11-6F-E3-98
Chapter 5, slide: 56
MAC Addresses
Assume:
• A knows B’s MAC address
• A wants to send a frame to B
Node A
137.196.7.78
1A-2F-BB-76-09-AD
137.196.7.23
137.196.7.14
LAN
71-65-F7-2B-08-53
137.196.7.88
58-23-D7-FA-20-B0
0C-C4-11-6F-E3-98
Node B
 A encapsulates
B’MAC into the
frame
 A sends the frame
into the medium
 All nodes will hear
the frame
 Only B grabs the
frame
 All other nodes
discard the frame
Chapter 5, slide: 57
ARP: Address Resolution Protocol
Question: if A doesn’t know
B’s MAC address, how does it
determine this MAC address?
Node A
Solution: ARP
 Each IP node (host, router)
on LAN has ARP table
 ARP table: maps IP, MAC
address for some LAN nodes
137.196.7.78
< IP address; MAC address; TTL>
1A-2F-BB-76-09-AD
137.196.7.23
137.196.7.14

 A consults its table to
LAN
71-65-F7-2B-08-53
137.196.7.88
TTL (Time To Live): remove mapping
after TTL (typically 20 min)

58-23-D7-FA-20-B0
determine B’s MAC given it
knows B’s IP
Q: how to construct
these ARP tables?
0C-C4-11-6F-E3-98
Node B
Chapter 5, slide: 58
ARP: Case 1—Same LAN (network)
 A wants to send datagram
to B, and B’s MAC address
not in A’s ARP table.
 A broadcasts ARP packet,
containing B's IP address


dest MAC address = FFFF-FF-FF-FF-FF
all machines on LAN
receive ARP query
 B receives ARP packet,
replies to A with its (B's)
MAC address
 A sends frame to B since it
knows its MAC now
 A caches IP-to-MAC
address pair in its ARP
table until information
becomes old (times out)

soft state: information that
times out (goes away) unless
refreshed
 ARP is “plug-and-play”:

nodes create their ARP
tables without intervention
from net administrator
Chapter 5, slide: 59
ARP: Case 2—routing to another LAN
datagram needs to go from A to B via R
assume A knows B’s IP address
88-B2-2F-54-1A-0F
74-29-9C-E8-FF-55
A
111.111.111.111
E6-E9-00-17-BB-4B
1A-23-F9-CD-06-9B
222.222.222.220
111.111.111.110
111.111.111.112
R
222.222.222.221
222.222.222.222
B
49-BD-D2-C7-56-2A
CC-49-DE-D0-AB-7D
 how does ARP work now ?? Would the previous
scenario work?
Chapter 5, slide: 60
ARP: Case 2—routing to another LAN
What happens when A wants to send IP datagram to B
 A knows that B belongs to a different subnet by checking B’s IP address
 A also knows IP address of router R (routing table at network layer)
 If necessary, A uses ARP to get R’s MAC address for 111.111.111.110 (IP
address of router)
 A creates frame with R's MAC address as dest, frame contains A-to-B IP




datagram
A’s NIC sends frame and R’s NIC receives it
R removes IP datagram from Ethernet frame, sees its destined to B
R uses ARP to get B’s MAC address (R has B’s IP address, included in the
just received frame)
R creates frame containing A-to-B IP datagram and sends it to B
88-B2-2F-54-1A-0F
74-29-9C-E8-FF-55
A
111.111.111.111
E6-E9-00-17-BB-4B
1A-23-F9-CD-06-9B
222.222.222.220
111.111.111.110
111.111.111.112
CC-49-DE-D0-AB-7D
R
222.222.222.221
222.222.222.222
B
49-BD-D2-C7-56-2A
Chapter 5, slide: 61
Link Layer
 1 Introduction and services
 2 Error detection and correction
 3 Multiple access protocols
 4 Topologies and switches
 5 Link-layer Addressing
 6 Ethernet
Chapter 5, slide: 63
Ethernet
“dominant” wired LAN technology:
 cheap $20 for NIC (Network Interface Card)
 first widely used LAN technology
 simpler, cheaper than token LANs and ATM
 kept up with speed race: 10 Mbps – 10 Gbps
Metcalfe’s Ethernet
sketch
Chapter 5, slide: 64
Ethernet topology
 bus topology popular through mid 90s
 all nodes in same collision domain (can collide with each other)
 today: star topology prevails
 active switch in center
 each “spoke” runs a (separate) Ethernet protocol (nodes do
not collide with each other)
Ethernet cable (called a segment)
switch
bus: coaxial cable
star
Chapter 5, slide: 65
802.3 Ethernet Standards: Link & Physical Layers
 many different Ethernet standards
 common MAC protocol and frame format
 different speeds: 2 Mbps, 10 Mbps, 100 Mbps,
1Gbps, 10G bps
 different physical layer media: fiber, cable,
wireless
application
transport
network
link
physical
MAC protocol
and frame format
100BASE-TX
100BASE-T2
100BASE-FX
100BASE-T4
100BASE-SX
100BASE-BX
copper (twister
pair) physical layer
fiber physical layer
Chapter 5, slide: 66
Ethernet Frame Structure
Sending adapter encapsulates IP datagram (or other
network layer protocol packet) in Ethernet frame
Preamble:
 total 8 bytes: 7 bytes, each with pattern 10101010
followed by one byte with pattern 10101011
 used to synchronize receiver, sender clock rates
(sender’s data rate matches receiver’s data rate)
Chapter 5, slide: 67
Ethernet Frame Structure
 Addresses: 6 bytes
 if adapter receives frame with matching destination
address, or with broadcast address (eg ARP packet), it
passes data in frame to network layer protocol
 otherwise, adapter discards frame
 Type: indicates higher layer protocol (mostly IP
but others possible, e.g., Novell IPX, AppleTalk)
 CRC: checked at receiver, if error is detected,
frame is dropped
Chapter 5, slide: 68
Ethernet: Unreliable, connectionless
 connectionless: No handshaking between sending and
receiving NICs
 unreliable: receiving NIC doesn’t send acks or nacks
to sending NIC



stream of datagrams passed to network layer can have gaps
(missing datagrams)
gaps will be filled if app is using TCP
otherwise, app will see gaps
 Ethernet’s MAC protocol: CSMA/CD
(more on this next…)
Chapter 5, slide: 69
CSMA/CD vs. slotted ALOHA
CSMA/CD
Slotted ALOHA
1. Unsychronized:
1. Sychronized:
NIC (adapter) may transmit at
anytime; no notion of timeslots
2. Carrier-sense:
Never transmit if others are
transmitting
3. Collision detection:
stop transmitting as soon as
collision is detected
4. Random backoff:
transmit at beginning of a
timeslot only
2. No carrier-sense:
No check for whether
others transmit or not
3. No collision detection:
no stop during collision
4. No random backoff
wait a random time before
retransmitting (more later)
Chapter 5, slide: 70
Notion of bit time
Before describing CSMA/CD, let’s introduce:
bit time = time to transmit one bit on a Ethernet link
Example: consider a 10 Mbps Ethernet link
Q1: what is a “bit time”
A1: 1/(10x10^6) second = 0.1 microsecond
Q2: how much time is “96 bit time”
A2: 96 x 0.1 = 9.6 microsecond
Chapter 5, slide: 71
Ethernet CSMA/CD algorithm
1. adapter receives datagram from 5. If adapter detects another
network layer, creates frame
transmission while transmitting,
aborts and sends a 48-bit jam
signal
2. If adapter senses channel idle
(to make sure all nodes are aware
for 96 bit time, starts frame
of collision)
transmission
(gap to allow interface recovery)
6. After aborting (after sending
jam signal), adapter enters
3. If adapter senses channel busy,
exponential backoff:
waits until channel idle (plus 96
 adapter chooses K at random
bit time), then transmits
4. If adapter transmits entire
frame without detecting another
transmission, adapter is done
with frame !
(next slide is explained how)
 adapter waits K·512 bit times,
 returns to Step 2
Chapter 5, slide: 72
Ethernet’s CSMA/CD (more)
Jam Signal:
make sure all other transmitters are aware of collision; 48 bits
Exponential Backoff:
 Goal: adapt retrans. attempts to estimated current load
heavy load: random wait will be longer
 Light load: random wait will be shorter
first collision: choose K from {0,1}
after 2nd collision: choose K from {0,1,2,3}
after 3rd collision: choose K from {0,1,2,3,4,5,6,7}
after 10th collision, choose K from {0,1,2,3,4,…,1023}
after 10 collision, choose K from {0,1,2,…,1023}
Then, delay transmission until K· 512 bit times







Chapter 5, slide: 73
Example
A
10Mbps
B
 A and B are connected via an Ethernet link of 10 Mbps
 Propagation delay between them = 224 bit time
 At time t=0, both transmit which results in collision
 After collision, A chooses K=0 and B chooses K=1
 Q1: how much is “bit time”: =1/10 = 0.1 microsecond
 Q2: at what time does collision occur?
=(224/2) x 0.1 = 11.2 microsecond
 Q3: at what time does bus become idle?
=224 (both A & B detect collision) + 48 (A & B finish sending jam signals) +
224 (last bit of B’s jam signal arrives at A) = 496 bit time
=496x0.1 (bit time)= 49.6 micseccond
 Q4: at what time does A begin retransmission?
[496 (bus becomes idle) + 96]x0.1 (bit time)= 59.2 micsecond
Chapter 5, slide: 74