Chapters 1-4 Midterm Review

1
Chapters 1-4Midterm Review

• Professor Rick Han
• University of Colorado at Boulder
• rhan_at_cs.colorado.edu

2
Announcements

• HW 2 up front
• Solutions to HW 2 and 3 available by March 12
  evening
• Programming Assignment 2 due Friday March 22 by
  1159 pm
• Midterm March 14
• Through Chapter 4 Network Layer
• Excluding 2.9 (network adaptors), 3.4 (switching
  hardware), and 4.4 (multicast)
• Next, review

3
Format of Midterm

• 75 minutes
• In class
• Closed book
• Calculator OK
• 4-5 long questions (multiple sections)
• maybe some short-answer/multiple-choice questions
• About 15 minutes for each long question

4
Potential Topics for Midterm

• List is not all-inclusive some topics may appear
  on midterm not listed here
• How does ___ work? Why is it used?
• Physical
• Propagation delay, bandwidthdelay product
• Byte stuffing
• Checksums
• Data-link
• Reliable protocols
• Stop-and-Wait
• Sliding Window
• Go-Back-N cumulative ACKs
• Selective Repeat selective ACKs

5
Potential Topics for Midterm (2)

• MAC
• Aloha, Slotted Aloha
• CSMA Listen before talk
• CSMA/CD Ethernet Listen while talk
• exponential backoff
• CSMA/CA wireless Ethernet
• Token Ring
• Ethernet Hubs/Repeaters
• Ethernet Bridges/Switches
• How do bridges build their forwarding tables?
• How are Spanning Trees constructed?

6
Potential Topics for Midterm (3)

• Network
• Virtual circuit routing
• IP
• Addressing, CIDR
• ARP
• First, Broadcast on Ethernet
• DHCP
• DHCP Relay to a DHCP Server
• ICMP
• How do ping and traceroute use ICMP?

7
Potential Topics for Midterm (4)

• Network
• IP
• Distance-vector routing (RIP)
• Distributed Bellman-Ford
• Bouncing effect, counting-to-infinity, split
  horizon
• Link-state routing (OSPF)
• Flooding of LSPs Dijkstra
• Link-state cost metrics
• Inter-domain routing (BGP)
• Path vector exchanged between ASs

8
Problems With Bridges

• Bridges can interconnect LANs and have multiple
  paths between every node
• Inadvertent

Layer 2 Bridge
Bridge

• Purposely for robustness, in case highest tier
  fails

Bridge

• Problem Frames can cycle forever in a loop and
  multiply to crash LAN!

Bridge
9
Problems With Bridges Packet Multiplication
Effect

• Suppose all bridges have just booted
• Suppose A wants to send to Z

Bridge 4
Bridge 1

• Bridge 1 sends As frame to LAN 5 4
• These two frames propagate to Bridge 3, where
  they multiply into 4 copies

LAN4
Z
LAN1
A
LAN3
LAN5
Bridge 2
LAN2
Bridge 3

• Exponentially multiplying copies!

10
Problems With BridgesEndless Looping

• Suppose all bridges have just booted
• Suppose A wants to send to Z

Bridge 4
Bridge 1

• Bridge 2 sends frame to LAN 2
• Bridge 3 sends frame to LAN 3
• Bridge 4 -gt LAN 4
• Back to LAN 1

LAN4
A
Z
LAN1
LAN3
Bridge 2
LAN2
Bridge 3

• Frames can cycle forever!

11
Solution Spanning Tree Algorithm

• Invented by Radia Perlman, modified into 802.1d
  spanning tree standard
• Bridges communicate with each other to set up a
  spanning tree that has no loops

Bridge 4
Bridge 1

• Disconnect some interfaces, though physical link
  exists
• Some frames may take long route though shorter
  direct route exists

LAN4
A
Z
LAN3
LAN1
Bridge 2
LAN2
Bridge 3

• Some bridges may become orphans

12
Rules to Build Spanning Tree

1. Elect a root bridge with the smallest global id
2. Each bridge computes its shortest distance to
   root
3. Each LAN selects a forwarding/designated bridge
   closest to root

Bridge 4
Bridge 3

• Spanning tree root forwarding bridges
• Root forwards frames on all outgoing ports
• If dest. not on LAN, send via forwarding bridge
• Eliminates loops!

LAN A
LAN C
LAN D
LAN E
Bridge1
Bridge 2
Root
13
More Intuition on Constructing Spanning Trees

• The 3 basic mechanisms for constructing 802.1d
  spanning trees (as defined in standard)
• A Bridge that believes itself to be the Root
  (all Bridges start by believing themselves to be
  the Root until they discover otherwise)
  originates Configuration Messages on all the LANs
  to which it is attached, at regular intervals.
• A Bridge that receives a Configuration Message
  on what it decides is its Root Port conveying
  better information (i.e., highest priority Root
  Identifier, lowest Root Path Cost, highest
  priority transmit-ting Bridge and Port), passes
  that information on to all the LANs for which it
  believes itself to be the Designated Bridge.
• A Bridge that receives inferior information, on
  a Port it considers to be the Designated Port on
  the LAN to which it is attached, transmits its
  own information in reply, for all other Bridges
  attached to that LAN to hear.

14
Control Messages to Build Spanning Tree

• Each bridge creates a configuration message
• ltbridge source id, distance to root, root bridge
  idgt
• Each bridge floods its initial configuration
  message on each of its ports/LANs
• ltsrcmy id, dist.0, rootmy idgt
• Each bridge stores best config msg for each
  port/LAN
• A config msg C1 is better than stored config msg
  C2 if
• Root id of C1 lt root id of C2
• Root ids equal and distance of C1 lt distance of
  C2
• If root ids and distances equal, C1 is better
  than C2 if transmitting bridge on C1 is lower
  than C2

15
First, Elect the Root

• If advertised root of new config msg C1 has
  smaller id, then
• Stop sending out its own bridge id config msgs
• Forward new smaller id on all outgoing ports

• Higher id config messages are discarded.
• Eventually, lowest ID bridge suppresses all other
  bridges config msgs
• Root bridge knows it is the root because the
  lowest ID is its own

Bridge 4
Bridge 3
LAN A
LAN C
LAN D
LAN E
Bridge1
Bridge 2
16
First, Elect the Root (2)

• Example
• Regardless of the config msgs exchanged by
  Bridges 2,3, and 4, as soon as Bridge 1 floods
  its config msg to Bridge 2 and 4, they both

• stop sending out their own bridge id config msgs
  and
• Begin forwarding Bridge 1s config msg on all
  outgoing ports
• Eventually, Bridge 3 also stops sending its
  config msgs

Bridge 4
Bridge 3
LAN A
LAN C
LAN D
LAN E
Bridge1
Bridge 2
17
More Intuition on Constructing Spanning Trees (1)

• Each bridge thinks its the root initially.
• Each bridge begins by periodically broadcasting
  on its attached LANs a config message saying I
  know who the root is and Im closest to the
  root.
• On each LAN, all connected bridges exchange these
  messages and try to convince their neighbors that
  theyre correct.

18
More Intuition on Constructing Spanning Trees (2)

• Think of each bridge as a person connected to
  many rooms (LANs).
• If a person P in room 1 said I know who the root
  is. to you, then youd want to verify this
  statement by comparing it to what youd heard
  everyone else say in all the other rooms that
  youre connected/listening to.
• If person P was correct, then youd remember what
  they said.
• If another room had a better root, then youd
  correct person P by sending a new config message
  saying I really know who the root is.

19
More Intuition on Constructing Spanning Trees (3)

• At any given time, each person (bridge) will be
  able to identify the lowest ID root heard from
  all of the announcements in all of its attached
  rooms, and will remember from which room X (LAN)
  the lowest ID announcement came
• The person will think that room X offers the
  shortest path to the root.
• The door into room X (LAN port) is called the
  root port.

20
More Intuition on Constructing Spanning Trees (4)

• As the true root bridge (lowest ID) broadcasts
  onto its attached LANs, its immediate radius-1
  neighbors are quickly convinced that this is the
  root.
• Each bridge between a radius-1 and radius-2 LAN
  will initially hear periodic false announcements
  on each radius-2 LAN claiming to know the root,
  and will reply with a new config message saying
  I really know who the root is, and Im one hop
  from the root (and therefore closest on this
  LAN).
• As the radius-2 bridges hear this, theyll stop
  sending their false claims. Also, theyll
  respond to and stop the false claims of radius-3
  bridges, propagating knowledge of the root bridge
  outward.

21
Next, Build Shortest-Path Forwarding Tree to Root

• Conceptually, build shortest-path forwarding tree
  after electing the root
• But, as the roots config msg floods the network,
  notice that the shortest-path tree can
  simultaneously be calculated
• Thus, piggyback on Bridge 1s config msg flooding
  to set up the shortest path tree to root
• Each bridge increments by one the distance, as it
  receives Bridge 1s config msg, and forwards
  config msg with Bridge 1 as root to all outgoing
  ports

22
Next, Build Shortest-Path Forwarding Tree to Root
(2)

• When a bridge receives a config message from
  another bridge on same LAN with Bridge 1 as root,
  it stops sending config messages on that port/LAN
  if
• Other bridge is closer to root
• Other bridge is same distance from root, but has
  a lower ID
• Thus, a bridge de-selects itself as the
  designated forwarding bridge for that port/LAN

23
Next, Build Shortest-Path Forwarding Tree to Root
(3)

• Bridge 1 floods its config message
• Bridge D is part of a loop, and will receive
  multiple config msgs from Bridge 1
• Bridge D deselects itself from both LANs because
  Bridges 2 3 are closer to root Bridge 1

Bridge 1
Bridge 2
Bridge 3
Bridge D
24
Next, Build Shortest-Path Forwarding Tree to Root
(4)

• Bridge 4 is designated forwarding bridge for LAN
  A, since it closer to root than Bridge 3 on LAN A
• Bridge 3 removes itself
• For LAN B, Bridge 2 is designated forwarding
  bridge
• Bridge 3 removes itself

Bridge 4
Bridge 3
LAN A
LAN C
LAN D
LAN E
Bridge1
Bridge 2
25
More Intuition on Constructing Spanning Trees (5)

• As a result, the roots lowest ID will propagate
  outward from the root bridge until all bridges
  become aware of the global roots lowest ID
• Thus far, weve only identified the root bridge.
  The shortest-path spanning tree still needs to be
  built.
• Remember, the full claim is I know who the root
  is, and Im closest to the root.
• Resolving the 2nd claim Im closest to the root
  is the key to building the shortest path spanning
  tree

26
More Intuition on Constructing Spanning Trees (6)

• When a bridge A hears the claim on a LAN from a
  bridge B that Im closest to the root., it must
  verify this claim.
• Recall that at any given time, each bridge can
  calculate what it believes is the root and the
  LAN from which the announcement came
• The root announcement also contained a distance
  from the root. Thus, each bridge A knows its
  distance from the root dist(A,root) is the root
  announcements advertised distance 1.
• When a new claim arrives from a bridge B that
  Im closest to the root., and contains a
  distance dist(B,root), then bridge A compares
  dist(B,root) to dist(A,root) to verify the
  truthfulness of the claim

27
More Intuition on Constructing Spanning Trees (7)

• If bridge Bs claim Im closest to the root.
• is correct, then bridge B should be the
  designated bridge forwarding towards the root for
  that LAN and bridge A will abstain from
  forwarding data to/from LAN
• is false, then bridge A will send a reply saying,
  No, Im closest to the root., and assign itself
  to be the designated forwarding bridge for that
  LAN

28
More Intuition on Constructing Spanning Trees (8)

• The root ID will propagate outwards, in response
  to periodic false claims of root ID. The
  distance-to-root will piggyback and will be
  incremented by one at each bridge.
• all bridges will have a correct distance estimate
  to the root
• Bridges with multiple paths from root will
  receive conflicting claims, and will follow rules
  of slide (7) to decide who is the designated
  bridge for that LAN

29
More Intuition on Constructing Spanning Trees (9)

• The method of correcting false distance claims
  will propagate outward from the root bridge,
  until
• Each bridge will have a root port facing towards
  the root, and some (possibly zero) ports facing
  away from the root on which it is the designated
  forwarding LAN
• The root port will periodically be flooding its
  config messages to the root ports of radius-1
  bridges, who in turn will be flooding the config
  messages with distance incremented by one out of
  their designated ports toward the root ports of
  the radius-2 bridges in the spanning tree

30
Distance Vector Routing

• Book assumes that only routing table is kept at
  each router
• I (and Kuroses text and Perlmans text) assume
  that the entire distance table is kept at each
  router
• For midterm, use textbooks assumption
• For programming assignment 2, use
  my/Kurose/Perlman assumption of distance table in
  every router

31
Distance Vector Routing

• Compare the following examples to see how the two
  interpretations of distance vector react
  differently to the same link failure

32
Distance Vector Routing (1)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 2
B 27 0 2
C 26 inf 1
A
B
Distance Table
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 2
B inf 25 2
C inf 26 1
1
1
C
A 2
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A 1 3 inf
B 3 1 inf
C 2 2 0
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
33
Distance Vector Routing (2)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 2
B 27 0 2
C 26 inf 1
A
B
Distance Table
A B notice link failure and should reset columns
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 2
B inf 25 2
C inf 26 1
1
1
C
A 2
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A 1 3 inf
B 3 1 inf
C 2 2 0
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
34
Distance Vector Routing (3)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A inf inf 2
B inf 0 2
C inf inf 1
A
B
Distance Table
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 inf 2
B inf inf 2
C inf inf 1
1
1
C
A 2
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A 1 3 inf
B 3 1 inf
C 2 2 0
Same!
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
Same!
Same!
35
Books Distance Vector Routing
25
A
Cost Next hop
A 2 C
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B 2 C
C 1 C
1
1
C
A 2
B 0
C 1
Cost Next hop
A 1 A
B 1 B
C 0 C
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
36
Books Distance Vector Routing (2)
25
A
Cost Next hop
A 2 C
B 0 B
C 1 C
B
Routing Table
Neither A nor B change their routing tables
because none of the next-hop paths go through the
broken link
Cost Next hop
A 0 A
B 2 C
C 1 C
1
1
C
Same
A 2
B 0
C 1
Cost Next hop
A 1 A
B 1 B
C 0 C
Same
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
Same
37
Books Distance Vector Routing (3)
25
A
Cost Next hop
A 2 C
B 0 B
C 1 C
B
Routing Table
A C should change routing tables because some
next-hop paths go through the broken link
Cost Next hop
A 0 A
B 2 C
C 1 C
1
1
C
A 2
B 0
C 1
Cost Next hop
A 1 A
B 1 B
C 0 C
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
38
Books Distance Vector Routing (4)
25
A
Cost Next hop
A 2 C
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B inf -
C inf -
1
1
C
A 2
B 0
C 1
Cost Next hop
A inf -
B 1 B
C 0 C
Distance vector
A inf
B 1
C 0
A 0
B inf
C inf
39
Books Distance Vector Routing (5)
25
A
Cost Next hop
A 2 C
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B inf -
C inf -
1
1
C
Same
A 2
B 0
C 1
Cost Next hop
A inf -
B 1 B
C 0 C
Distance vector
A inf
B 1
C 0
A 0
B Inf
C inf
40
Books Distance Vector Routing (6)
25
A
Cost Next hop
A inf -
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B inf -
C inf -
1
1
C
A inf
B 0
C 1
Cost Next hop
A inf -
B 1 B
C 0 C
Distance vector
A inf
B 1
C 0
A 0
B Inf
C inf
41
Books Distance Vector Routing (7)
25
A
Cost Next hop
A 25 A
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B inf -
C inf -
1
1
C
A 25
B 0
C 1
Cost Next hop
A inf -
B 1 B
C 0 C
Distance vector
A inf
B 1
C 0
A 0
B Inf
C inf
42
Books Distance Vector Routing (8)
25
A
Cost Next hop
A 25 A
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B inf -
C inf -
1
1
C
Same
A 25
B 0
C 1
Cost Next hop
A 26 B
B 1 B
C 0 C
Distance vector
A 26
B 1
C 0
A 0
B Inf
C inf
43
Books Distance Vector Routing (9)
25
A
Cost Next hop
A 25 A
B 0 B
C 1 C
B
Routing Table
Cost Next hop
A 0 A
B 25 B
C 26 B
1
1
C
Same
A 25
B 0
C 1
Cost Next hop
A 26 B
B 1 B
C 0 C
Distance vector
A 26
B 1
C 0
A 0
B 25
C 26
44
Distance Vector Routing (4)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 2
B 27 0 2
C 26 inf 1
A
B
Distance Table
A C should update their distance tables
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 2
B inf 25 2
C inf 26 1
1
1
C
A 2
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A 1 3 inf
B 3 1 inf
C 2 2 0
Distance vector
A 1
B 1
C 0
A 0
B 2
C 1
45
Distance Vector Routing (5)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 2
B 27 0 2
C 26 inf 1
A
B
Distance Table
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 inf
B inf 25 inf
C inf 26 inf
1
1
C
A 2
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A inf 3 inf
B inf 1 inf
C inf 2 0
Distance vector
A 3
B 1
C 0
A 0
B 25
C 26
46
Distance Vector Routing (6)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 2
B 50 0 2
C 51 inf 1
A
B
Distance Table
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 inf
B inf 25 inf
C inf 26 inf
1
1
C
A 2
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A inf 3 inf
B inf 1 inf
C inf 2 0
Distance vector
A 3
B 1
C 0
A 0
B 25
C 26
47
Distance Vector Routing (7)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 4
B 50 0 2
C 51 inf 1
A
B
Distance Table
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 inf
B inf 25 inf
C inf 26 inf
1
1
C
A 4
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A inf 3 inf
B inf 1 inf
C inf 2 0
Distance vector
A 3
B 1
C 0
A 0
B 25
C 26
48
Distance Vector Routing (8)
25
B Via neighbor Via neighbor Via neighbor
dest A B C
A 25 inf 4
B 50 0 2
C 51 inf 1
A
B
Distance Table
A Via neighbor Via neighbor Via neighbor
dest A B C
A 0 27 inf
B inf 25 inf
C inf 26 inf
1
1
C
A 4
B 0
C 1
C Via neighbor Via neighbor Via neighbor
dest A B C
A inf 5 inf
B inf 1 inf
C inf 2 0
Distance vector
A 5
B 1
C 0
A 0
B 25
C 26
Bouncing Effect