Transcript network

Chapter 3
Internetworking
1
Problems
 In Chapter 2 we saw how to connect one node
to another, or to an existing network. How do
we build networks of global scale?
 How do we interconnect different types of
networks to build a large global network?
Chapter Outline
3.1 Switching and Bridging
3.2 Basic Interworking (IP)
3.3 Routing
3.4 Implementation and Performance
 Two limitations on the directly connected networks
 limit on how many hosts can be attached, examples
 only two hosts can be attached to a point-topoint link
 the Ethernet specification allows no more than
1,024 hosts
 limit on how large of a geographic area a single
network can serve, examples
 an Ethernet can span only 2,500 m
 wireless networks are limited by the ranges of
their radios
 point-to-point links can be quite long
 Goal
 build networks that can be global in scale
 Problem
 how to enable communication between hosts that
are not directly connected
 Solution
 computer networks use packet switches to enable
packets to travel from one host to another, even
when no direct connection exists between those
hosts
 Packet switch
 a device with several inputs and outputs leading to
and from the hosts that the switch interconnects
 Core job of a switch
 take packets that arrive on an input and forward (or
switch) them to the right output so that they will
reach their appropriate destination
 A key problem that a switch must deal with is the finite
bandwidth of its outputs
 if packets destined for a certain output arrive at a switch
and their arrival rate exceeds the capacity of that output,
then we have a problem of contention
 the switch queues (buffers) packets until the contention
subsides, but if it lasts too long, the switch will run out
of buffer space and be forced to discard packets
 when packets are discarded too frequently, the switch is
said to be congested
3.1 Switching and Bridging
 Switch
 a multi-input, multi-output device, which transfers
packets from an input to one or more outputs
 star topology
 switched networks are more scalable (i.e., growing
to large numbers of nodes) than shared-media
networks because of the ability to support many
hosts at full speed
A switch provides a star topology
Scalable Networks
 The figure shows the protocol graph that would run on
a switch that is connected to two T3 links and one
STS-1 SONET link
Example protocol graph running on a switch
 A switch forwards packets from input port to
output port
 Port selected based on address in packet header
 Advantages
 cover large geographic area (tolerate latency)
 support large numbers of hosts (scalable bandwidth)
Example switch with three input and output ports
 How does the switch decide on which output
port to place each packets?
 general answer
 it looks at the header of the packet for an identifier that
it uses to make the decision
 three common approaches
 datagram (or connectionless) approach
 virtual circuit (or connection-oriented approach)
 source routing
3.1.1 Datagram
 Sometimes called connectionless model
 Analogy: postal system
 No connection setup phase
 no round trip delay waiting for connection
setup
 a host can send data as soon as it is ready
 Each packet is forwarded independently of
previous packets that might have been sent to
the same destination
 two successive packets from host A to host
B may follow completely different paths
(perhaps because of a change in the
forwarding table at some switch in the
network)
 A switch or link failure might not have any
serious effect on communication if it is possible
to find an alternate route around the failure and
to update the forwarding table accordingly
 Since every packet must carry the full address
of the destination, the overhead per packet is
higher than for the connection-oriented model
 Source host has no way of knowing if the
network is capable of delivering a packet or if
the destination host is even up and running
 Each switch maintains a forwarding (routing)
table
 Example
 the hosts have addresses A, B, C, and so on
 a switch consults a forwarding table (routing table)
to decide how to forward a packet
Datagram forwarding: an example network
 The table shows the
forwarding information that
switch 2 needs to forward
datagrams
Destination
Port
A
3
B
0
C
3
D
3
E
2
F
1
G
0
H
0
3.1.2 Virtual Circuit Switching
 Sometimes called connection-oriented model
 Analogy: phone call
 Explicit connection setup (and tear-down)
phase
 it requires that a virtual connection from the
source host to the destination host is set up
before any data is sent
 Typically wait full RTT (Round Trip Time) for
connection setup before sending first data
packet
 If a switch or a link in a connection fails
 the connection is broken and a new one
needs to be established
 Subsequence packets follow same circuit
 Each switch maintains a Virtual Circuit (VC)
table
 Entry in the VC table on a single switch
contains
 a virtual circuit identifier (VCI)
 uniquely identifies the connection at this
switch
 which will be carried inside the header of
the packets that belong to this connection
Incoming
Interface
Incoming VCI
Outgoing
Interface
Outgoing VCI
2
5
1
11
Incoming
Interface
Incoming VCI
Outgoing
Interface
Outgoing VCI
3
11
2
7
Incoming
Interface
Incoming VCI
Outgoing
Interface
Outgoing VCI
0
7
1
4
 an incoming interface
 on which packets for this VC arrive at the
switch
 an outgoing interface
 in which packets for this VC leave the
switch
 a potentially different VCI that will be used
for outgoing packets
 Two classes of approaches to establish
connection state
 Permanent Virtual Circuit (PVC)
 Switched Virtual Circuit (SVC)
 Permanent Virtual Circuit (PVC)
 administrator configures the state, in which case the
virtual circuit is “permanent”
 administrator can also delete the state, so a
permanent virtual circuit (PVC) might be thought of
as a long-lived, or administratively configured VC
 Switched Virtual Circuit (SVC)
 a host may set up and delete a VC by sending
messages without the involvement of a network
administrator
 this is referred to as signaling, and the resulting
virtual circuits are said to be switched
 an SVC should more accurately be called a
“signaled” VC, since it uses signaling (not
switching) to distinguish an SVC from a PVC
 Example
 assume that a network administrator wants to
manually create a new virtual connection from host
A to host B
 two-stage process
 connection setup
 data transfer
(11)
(5)
(7)
(4)
An example of a virtual circuit network
 The administrator picks a VCI
value that is currently unused on
each link for the connection
 suppose
 VCI = 5, the link from
host A to switch 1
 VCI = 11, the link from
switch 1 to switch 2
 VCI = 7, the link from
switch 2 to switch 3
 VCI = 4, the link from
switch 3 to host B
Incoming
Interface
Incoming VCI
Outgoing
Interface
Outgoing
VCI
2
5
1
11
VC table entry at switch 1
Incoming
Interface
Incoming VCI
Outgoing
Interface
Outgoing
VCI
3
11
2
7
VC table entry at switch 2
Incoming
Interface
Incoming VCI
Outgoing
Interface
Outgoing
VCI
0
7
1
4
VC table entry at switch 3
A packet is sent into a virtual circuit network
A packet makes its way through a virtual circuit network
 Hop-by-hop flow control
 each node is ensured of having the buffers it needs
to queue the packets that arrive on that circuit
 example, an X.25 network-a packet-switched
network that uses the connection-oriented model
 X.25 network employs the following three-part strategy
1. buffers are allocated to each virtual circuit when the
circuit is initialized
2. the sliding window protocol is run between each pair of
nodes along the virtual circuit, and this protocol is
augmented with flow control to keep the sending node
from overrunning the buffers allocated at the receiving
node
3. the circuit is rejected by a given node if not enough
buffers are available at that node when the
connection request message is processed
 Examples of virtual circuit technologies
 Asynchronous Transfer Mode (ATM)
 Frame Relay, e.g., Virtual Private Network (VPN)
 Frame Relay operates only at the physical and
data link layers
ATM Cell Formats
 Two different cell formats
 User-Network Interface (UNI) format
 host-to-switch format
 interface between a telephone company and one of its
customers
 Network-Network Interface (NNI) format
 switch-to-switch format
 interface between a pair of telephone companies
Architecture of an ATM network
 User-Network Interface (UNI)






GFC (4 bits): Generic Flow Control
VPI (8 bits): Virtual Path Identifier
VCI (16 bits): Virtual Circuit Identifier
Type (3 bits): management, congestion control, AAL5
CLP (1 bit): Cell Loss Priority
HEC (8 bits): Header Error Check (CRC-8)
 Network-Network Interface (NNI)
 GFC becomes part of VPI field (no GFC and becomes 12-bit
VPI)
ATM cell format at the UNI
ATM Headers
ATM Virtual Path
 ATM uses a 24-bit identifier for vircuit circuits
 8-bit virtual path identifier (VPI)
 16-bit virtual circuit identifier (VCI)
 Example
 a corporation has two sites that connect to a public ATM
network, and that at each site the corporation has a network
of ATM switches
 we could establish a virtual path between two sites using only
the VPI field
 within the corporate sites, however, the full 24-bit space is
used for switching
Example of a virtual path
 Advantage of virtual path
 although there may be thousands or millions of
virtual connections across the public network, the
switches in the public network behave as if there is
only one connection
 there needs to be much less connection-state
information stored in the switches, avoiding the
need for big, expensive tables of per-VCI
information
TP、VPs、and VCs
Example of VPs and VCs
Connection Identifiers
Virtual Connection Identifiers in
UNIs and NNIs
ATM Cell
Routing with a Switch
3.1.3 Source Routing
 Neither virtual circuits nor conventional datagrams
 All the information about network topology that is
required to switch a packet across the network is
provided by the source host
 Various ways to implement source routing
 method1
 put an ordered list of switch ports in the header
and to rotate the list so that the next switch in the
path is always at the front of the list
 for each packet that arrives on an input, the
switch would read the port number in the header
and transmit the packet on that output
Source routing in a switched network (where the switch reads the rightmost number)
 method2
 example, rather than rotate the header, each
switch just strip the first element as it uses it
 method3
 have the header carry a pointer to the current
“next port” entry, so that each switch just updates
the pointer rather than rotating the header
Three ways to handle headers for source routing: (a) rotation, (b) stripping,
and (c) pointer. The labels are read right to left
3.1.4 Bridges and LAN Switches
 LANs have physical limitations (e.g., 2500m)
 Bridge
 connect two or more LANs
 Extended LAN
 a collection of LANs connected by one or more
bridges
 accept and forward strategy (accept all frames
transmitted on either of the Ethernets, so it could
forward them to the other)
Learning Bridges
 Do not forward when unnecessary
 whenever a frame from host A that is addressed to
host B arrives on port 1, there is no need for the
bridge to forward the frame out over port 2
Illustration of a learning bridge
Host
Port
A
1
B
1
C
1
X
2
Y
2
Z
2
 How does a bridge come to learn on which port
the various hosts reside?
 each bridge inspects the source address in all the
frames it receives
 when host A sends a frame to a host on either side
of the bridge, the bridge receives this frame and
records the fact that a frame from host A was just
received on port 1
 in this way, the bridge can build a table just like the
following table
Host
Port
A
1
B
1
C
1
X
2
Y
2
Z
2
Spanning Tree Algorithm
 Problem: extended LAN has a loop in it
 frames potentially loop through the extended LAN
forever
 example
 bridges B1, B4, and B6 form a loop
Extended LAN with loops
 Solution: bridges run a distributed spanning
tree algorithm
 spanning tree is a subgraph of a graph that covers
(spans) all the vertices, but contains no cycles
Example of (a) a cyclic graph; (b) a corresponding spanning tree
 Spanning tree algorithm (developed by Radia Perlman)
 each bridge has a unique identifier (e.g., B1, B2,
B3)
 the algorithm first elects the bridge with the
smallest ID as the root of the spanning tree
 the root bridge always forwards frames out over
all of its ports
 each bridge computes the shortest path to the root
and notes which of its ports is on this path
 this port is selected as the bridge’s preferred path
to the root
 finally, all the bridges connected to a given LAN
elect a single designated bridge that will be
responsible for forwarding frames toward the root
bridge
 each LAN’s designated bridge is the one that is
closest to the root, and if two or more bridges are
equally close to the root, then the bridges’
identifiers with the smallest ID wins
Spanning tree with some ports not selected
 Bridges have to exchange configuration messages with
each other and then decide whether or not they are the
root or a designated bridge based on these messages
 configuration messages contain
 the ID for the bridge that is sending the message
 the ID for what the sending bridge believes to be the
root bridge
 the distance, measured in hops, from the sending
bridge to the root bridge
 each bridge records current best configuration message
for each port
 initially, each bridge believes it is the root
 when learn not root, stop generating config messages
 in steady state, only root generates configuration
messages
 when learn not designated bridge, stop forwarding
config messages
 in steady state, only designated bridges forward
config messages
 root continues to periodically send config messages
 if any bridge does not receive config message after a
period of time, it starts generating config messages
claiming to be the root
 upon receiving a config message over a particular
port
 the bridge checks to see if that new message is
better than the current best configuration message
recorded for that
 the new configuration message is considered “better”
than the currently recorded information if
 it identifies a root with a smaller ID or
 it identifies a root with an equal ID but with a shorter
distance or
 the root ID and distance are equal, but the sending
bridge has a smaller ID
 Sequence of events
 assume all the bridges boot at about the same time
and all the bridges would start off by claiming to be
the root
 (Y, d, X) denotes a configuration message from
node X in which it claims to be distance d from root
node Y
 Sequence of events on the activity at node B3
1. B3 receives (B2, 0, B2)
2. since 2 < 3, B3 accepts B2 as root [(B2, 1, B3)]
3. B3 adds one to the distance advertised by B2 (0) and thus
sends (B2, 1, B3) toward B5 [(B2, 1, B3), (B2, 2, B5)]
4. meanwhile, B2 accepts B1 as root because it has the lower
ID, and it sends (B1, 1, B2) toward B3
[(B1, 1, B2), (B1, 2, B3)]
5. B5 accepts B1 as root and sends (B1, 1, B5) toward B3
[(B1, 1, B5), (B1, 2, B3)]
6. B3 accepts B1 as root, and it notes that both B2 and B5 are
closer to the root than it is
[(B1, 2, B3), (B1, 1, B2), (B1, 1, B5)]
7. B3 stops forwarding messages on both its interfaces (this
leaves B3 with both ports not selected)
[(B1, 1, B2), (B1, 1, B5)]
(2)
(7)
(1)
(3)
(4b)
(6)
(5b)
(5a)
(4a)
Spanning tree with some ports not selected
Broadcast and Multicast
 Since most LANs support both broadcast and
multicast, then bridges must also support these two
features
 Broadcast
 each bridge forwards a frame with a destination broadcast
address out on each active (selected) port other than the one
on which the frame was received
 Multicast
 implemented in exactly the same way, with each host
deciding itself whether or not to accept the message
Limitations of Bridges
 Do not scale
 Do not accommodate heterogeneity
Do not Scale
 It is not realistic to connect more than a few
(tens of) LANs by means of bridges
 the spanning tree algorithm scales linearly, i.e.,
there is no provision for imposing a hierarchy on
the extended LAN
 bridges forward all broadcast frames and broadcast
does not scale
 Virtual LAN (VLAN)
 used to increase the scalability of extended LANs
 allows a single extended LAN to be partitioned into
several seemingly separate LANs
 each virtual LAN is assigned an identifier (sometimes
called a color), and packets can only travel from one
segment to another if both segments have the same
identifier
 this limits the number of segments in an extended
LAN that will receive any given broadcast packet
 Example
 four hosts (W, X, Y, Z) on four different LAN segments
 in the absence of VLANs, any broadcast packet from any
host will reach all the other hosts
 suppose that we define the segments connected to hosts W
and X as being in one LAN, VLAN 100
 also define the segments that connect to hosts Y and Z as
being in VLAN 200
 to do his, we need to configure a VLAN ID on each port of
bridges B1 and B2
 the link between B1 and B2 is considered to be in both
VLANs
Two virtual LANs share a common backbone
 When a packet sent by host X arrives at bridge B2
 the bridge observes that it came in a port that was configured
as being in VLAN 100
 it inserts a VLAN header between the Ethernet header and its
Ethernet
VLAN
payload
Payload
header
header
 the bridge applies normal rules for forwarding to the packet,
with the extra restriction that the packet may not be sent out
an interface that is not part of VLAN 100
 thus, even a broadcast packet can’t be sent out the interface to
host Z, which is in VLAN 200
 An attractive feature of VLANs
 it is possible to change the logical topology without
moving any wires or changing any addresses
 example
 if we want to make the segment that connects to host Z
be part of VLAN 100, and thus enable X, W and Z be
on the same virtual LAN, we would just need to
change one piece of configuration on bridge B2
Do not Accommodate Heterogeneity
 Bridges are fairly limited in the kinds of networks they
can interconnect
 Bridges make use of the networks frame header and so
can support only networks that have exactly the same
format for addresses
 Bridges can be used to connect Ethernets to Ethernets,
802.5 (Token Ring) to 802.5, and Ethernets to 802.5
rings, since both networks support the same 48-bit
address format
 Bridges do not readily generalize to other kinds of
networks, such as ATM
3.2 Basic Internetworking (IP)
3.2.1 What is an Internework?
3.2.2 Service Model
3.2.3 Global Addresses
3.2.4 Datagram Forwarding in IP
3.2.5 Subnetting and Classless Addressing
3.2.6 Address Translation (ARP)
3.2.7 Host Configuration (DHCP)
3.2.8 Error Reporting (ICMP)
3.2.9 Virtual Networks and Tunnels
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3.2.1 What is an Internework?
 Concatenation of networks
A simple internetwork. Hn =host, Rn = router
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 An internetwork is a network of networks
 in the figure, we see Ethernets, an FDDI ring, and a
point-to-point link
 each of these is a single-technology network
 the nodes that interconnect the networks are called
routers (sometimes called gateways)
 The following figure shows how H1 and H8 are
logically connected by the internet, including
the protocol graph running on each node
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 A simple internetwork of protocol stack
Protocol layers used to connect H1 to H8.
ETH: the protocol that runs over Ethernet.
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3.2.2 Service Model
 Service model for an internetwork
 a host-to-host service only if this service can somehow be
provided over each of the underlying physical networks
 IP service model has two parts
 addressing scheme
 provides a way to identify all hosts in the internetwork
 datagram (conectionless) model of data delivery
 This service model is sometimes called best effort
 although IP makes every effort to deliver datagrams, it makes
no guarantees
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 Datagram
 a type of packet sent in a connectionless manner
over a network
 every datagram carry enough information to let
the network forward the packet to its correct
destination
 no need for any advance setup mechanism to tell
the network what to do when the packet arrives
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 Best-effort delivery (unreliable service)
 if something goes wrong and has the following
situations




packets are lost
packets are delivered out of order
duplicate copies of a packet are delivered
packets can be delayed for a long time
 the network does not make any attempt to recover
from the failure
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 Datagram format
98
 Datagram format
 a succession of 32-bit words
 the top word is transmitted first
 the leftmost byte of each word is transmitted first
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 1st word of the header
 Version: the version of IP
 the current version of IP is 4 (IPv4)
 HLen: the length of the header in 32-bit words
 most of the time, the header is 5 words (20
bytes) long
100
 TOS: the 8-bit type of service
 allow packets to be treated differently based
on application needs
 example, the TOS value might determine
whether or not a packet should be placed in a
special queue that receives low delay
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 Length: 16 bits of the header
 contain the length of the datagram, including
the header
 the field counts bytes rather than words
 the maximum size of an IP datagram is
65,535 bytes
 the physical network over which IP is
running may not support such long packets
 IP supports a fragmentation and
reassembly process
102
 2nd word of the header contains information about
fragmentation
 Offset: 12-bit counts 8-byte chunk, not bytes
 the distance (number of chunks) between the
start of the original data and the start of the
current fragment
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 3rd word of the header
 TTL: one-byte time to live
 a specific number of seconds that the packet
would be allowed to live
 routers along the path would decrement this
field until it reached 0
 Protocol: one-byte demultiplexing key
 identifies the higher-level protocol to which
this IP packet should be passed
 values defined for TCP (6), UDP (17)
104
 Checksum:
 calculated by considering the entire IP header
as a sequence of 16-bit words
 adding them up using ones complement
arithmetic, and taking the ones complement
of the result
105
 the fourth word of the header: SourceAddr
 the fifth word of the header: DestinationAddr
 there may be a number of options at the end of the
header
 the presence or absence of options may be determined
by examining the header length (HLen) field
106
Fragmentation and Reassembly
 Each network technology tends to have its own idea
of how large a packet can be, example,
 Ethernet can accept packets up to 1,500 bytes long
 FDDI packets may be 4,500 bytes long
 Every network type has a maximum transmission
unit (MTU)
 the largest IP datagram that it can carry in a frame
 this value is smaller than the largest packet size on that
network because the IP datagram needs to fit in the payload
of the link-layer frame
107
 Fragmentation
 typically occurs when necessary (MTU < Datagram)
 to enable these fragments to be reassembled at the
receiving host, they all carry the same identifier in
the Ident field
 this identifier is chosen by the sending host and is
intended to be unique among all the datagrams that
might arrive at the destination from this source over
some reasonable time period
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 since all fragments of the original datagram contain
this identifier, the reassembling host will be able to
recognize those fragments that go together
 should all the fragments not arrive at the receiving
host, the host gives up on the reassembly process
and discards the fragments that did arrive
 IP does not attempt to recover from missing
fragments
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 example
 consider what happens when host Hl sends a datagram
to host H8
 assuming that the MTU is 1,500 bytes for the two
Ethernets, 4,500 bytes for the FDDI network, and 532
bytes for the point-to-point network
 a 1,420-byte datagram (20-byte IP header plus 1,400 bytes
of data) sent from H1 makes it across the first Ethernet and
the FDDI network without fragmentation but must be
fragmented into three datagrams at router R2
 these three fragments are then forwarded by router R3
across the second Ethernet to the destination host
110
111
IP datagrams traversing the sequence of physical networks
112
 each fragment is itself a self-contained IP datagram
that is transmitted over a sequence of physical
networks, independent of the other fragments
 each IP datagram is reencapsulated for each
physical network over which it travels
113
(a)
(b)
Header fields used in IP fragmentation: (a) unfragmented packet; (b) fragmented packets.
114
 The unfragmented packet has 1,400 bytes of data and a
20-byte IP header
 when the packet arrives at router R2, which has an MTU of
532 bytes, it has to be fragmented
 a 532-byte MTU leaves 512 bytes for data after the 20-byte
IP header, so the first fragment contains 512 bytes of data
 the router sets the M bit in the Flags field, meaning that there
are more fragments to follow
 it sets the Offset to 0, since this fragment contains the first
part of the original datagram
115
 the data carried in the second fragment starts with
the 513th byte of the original data, so the field in
this header is set to 64 (= 512/8)
 the third fragment contains the last 376 bytes of
data, and the offset is now 2 × 512 / 8 = 128 (since
this is the last fragment, the M bit is not set)
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3.2.3 Global Addresses
 Ethernet addresses are globally unique
 that alone does not suffice for an addressing
scheme in a large internetwork
 Ethernet addresses are also flat
 they have no structure and provide very few clues
to routing protocols
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 IP addresses are hierarchical
 made up of two parts that correspond to some sort
of hierarchy in the internetwork
 network part
 identifies the network to which the host is
attached
 all hosts attached to the same network have the
same network part
 host part
 identifies each host uniquely on that particular
network
118
 example 1
 the addresses of the hosts on network 1 would all have the
same network part and different host parts
 example 2
 the routers are attached to two networks
 they need to have an address on each network, one for each
interface, e.g., router Rl
 an IP address on the interface to network 2 that has the same
network part as the hosts on network 2
 an IP address on the interface to network 3 that has the same
network part as the hosts on network 3
 IP addresses belong to interfaces than to hosts
119
 IP addresses are divided into three different
classes
 each of the following figure defines different-sized
network and host parts
 there are also class D addresses specify a multicast
group, and class E addresses that are currently
unused
 in all cases, the address is 32 bits long
120
7
A:
0
24
Network
Host
14
B:
1
0
16
Network
Host
21
C:
1
1
0
Network
8
Host
IP addresses: (a) class A; (b) class B; (c) class C
121
 the class of an IP address is identified in the most
significant few bits
 if the first bit is 0, it is a class A address
 if the first bit is 1 and the second is 0, it is a class B
 if the first two bits are 1 and the third is 0, it is a class
C address
 of the approximately 4 billion (= 232)possible IP
addresses
 one-half are class A
 one-quarter are class B
 one-eighth are class C
122
 Class A addresses
 7 bits for the network part and 24 bits for the host
part
 126 (= 27-2) class A networks (0 and 127 are
reserved)
 each network can accommodate up to 224-2 (about 16
million) hosts (again, two are reserved values)
 Class B addresses
 14 bits for the network part and 16 bits for the host
part
 65,534 (= 216-2) hosts
123
 Class C addresses
 21 bits for the network part and 8 bits for the
host part
 2,097,152 (= 22l) class C networks
 254 hosts (host identifier 255 is reserved for
broadcast, and 0 is not a valid host number)
124
 IP addresses are written as four decimal integers
separated by dots
 each integer represents the decimal value contained in
1 byte (= 0~255) of the address, starting at the most
significant
 Example, 171.69.210.245
 Internet domain names (DNS)
 also hierarchical
 domain names tend to be ASCII strings separated by
dots, e.g., cs.nccu.edu.tw
125
3.2.4 Datagram Forwarding in IP
 Forwarding
 the process of taking packet from an input and
sending it out on the appropriate output
 Routing
 the process of building up the tables that allow the
correct output for a packet to be determined
126
 Strategy
 every datagram contains destination’s address
 if connected to destination network
 then forward to host
 if not directly connected
 then forward to some router
 forwarding table maps network number
(NetworkNum) into next hop (NextHop)
 each host has a default router
 each router maintains a forwarding table
127
 Datagram forwarding algorithm
 if (NetworkNum of destination = NetworkNum of one
of my interfaces) then
deliver packet to destination over that interface
else
if (NetworkNum of destination is in my forwarding
table) then
deliver packet to NextHop route
else
deliver packet to default router
128
 For a host with only one interface and only one default
router in its forwarding table
 (simplified algorithm)
 if (NetworkNum of destination = my NetworkNum)
then
deliver packet to destination directly
else
deliver packet to default router
129
 Example1
 suppose H1 wants to send a datagram to H2
 since they are on the same physical network, H1
and H2 have the same network number in their IP
address
 H1 deduces that it can deliver the datagram directly
to H2 over the Ethernet
 the one that needs to be resolved is how Hl finds
out the correct Ethernet address for H2
130
 Example2
 suppose H1 wants to send a datagram to H8
 since they are on different physical networks
 H1 deduces that it needs to send the datagram to a
router
 Hl sends the datagram over the Ethernet to R1
 R1 knows that it cannot deliver a datagram directly
to H8 because neither of Rl’s interfaces is on the
same network as H8
131
 suppose R1’s default router is R2; R1 then sends
the datagram to R2 over the token ring network
 assume R2 has the forwarding table shown as
follows, it looks up H8’s network number
(network 1) and forwards the datagram to R3
132
Network
Number
Next Hop
1
R3
2
R1
3
Interface 1
4
Interface 0
Forwarding table for router R2
133
 R3 forwards the datagram directly to H8
 it is possible to include the information about directly
connected networks in the forwarding table
 example, we could label the network interfaces of router R2
as interface 0 for the point-to-point link (network 4) and
interface l for the token ring (network 3)
Network
Number
Next Hop
1
R3
2
R1
3
Interface 1
4
Interface 0
134
3.2.5 Subnetting and Classless Addressing
 Subnetting deals with address space utilization
 Original intent of IP addresses
 the network part would uniquely identify exactly one
physical network
 Problem of address assignment : inefficiency
 class C with 2 hosts (2/255 = 0.78% efficiency)
 class B with 256 hosts (256/65535 = 0.39% efficiency)
135
 Subnet
 add another level to address / routing hierarchy
 reduce the total number of network numbers that are
assigned
 idea
 take a single IP network number and allocate the IP
addresses with that network number to several
physical networks
 a perfect use of subnetting is a large campus or
corporation that has many physical networks
136
 Subnet mask
 define variable partition of host part
 a single network number can be shared among multiple
networks involves configuring all the nodes on each
subnet with a subnet mask
137
 subnet mask enables a subnet number
 hosts may be on different physical networks but
share a single network number
 example, to share a single class B address among
several physical networks, we could use a subnet
mask of 255.255.255.0 (all 1s in the upper 24 bits
and 0s in the lower 8 bits)
 the top 24 bits are network number
 the lower 8 bits are host number
 the top 16 bits identify the network in a class B
address
138
 three parts address
 network part (16 bits)
 subnet part (8 bits)
 host part (8 bits)
139
Subnetted Address
140
Subnet Example
Subnet mask: 255.255.255.128
Subnet number: 128.96.34.0
128.96.34.15
128.96.34.1
H1
R1
128.96.34.130
Subnet mask: 255.255.255.128
Subnet number: 128.96.34.128
128.96.34.139
128.96.34.129
H3
R2
H2
128.96.33.1
128.96.33.14
Subnet mask: 255.255.255.0
Subnet number: 128.96.33.0
141
 Exactly one subnet mask per subnet
 H1
 IP address: 128.96.34.15
 subnet mask: 255.255.255.128
 subnet number: 128.96.34.0
 Defines the subnet number of the host and of all other
hosts on the same subnet
 take bitwise AND of “IP address” and “subnet mask”
 example, 128.96.34.15 AND 255.255.255.128 equals
128.96.34.0
142
 When a host wants to send a packet to a certain
IP address
 perform a bitwise AND of its own subnet mask and
the destination IP address
 if the result equals the subnet number of the sending
host
 the destination host is on the same subnet and the
packet can be delivered directly over the subnet
143
 if the results are not equal
 the packet needs to be sent to a router to be forwarded
to another subnet
 example, if H1 is sending to H2, then H1 ANDs its
subnet mask (255.255.255.128) with the address for H2
(128.96.34.139) to obtain 128.96.34.128
 128.96.34.128 does not match the subnet number for
H1 (128.96.34.0), so H1 and H2 are on different
subnets
 H1 has to send packet to its default router R1 then to
H2
144
 Router with/without subnetting
 simple IP
 entries of forwarding tables is of the form
(NetworkNum, NextHop)
 support subnetting
 entries of forwarding tables is of the form
(SubnetNumber, SubnetMask, NextHop)
145
 find the right entry in the table
 the router ANDs the packet's destination address
with the SubnetMask for each entry in turn
 if the result matches the SubnetNumber of the
entry, then this is the right entry to use
 it forwards the packet to the next hop router
indicated
 router Rl of the “subnet example” would have
the following entries
146
147
 continuing with the example, a datagram from H1
being sent to H2
 Rl would AND H2's address (128.96.34.139)
with the subnet mask of the first entry
(255.255.255.128)
 compare the result (128.96.34.128) with the
network number for that entry (128.96.34.0)
 since this is not a match (the first entry), it
proceeds to the next entry
 this time a match does occur (the second entry),
so Rl delivers the datagram to H2 using interface
1, which is the interface connected to the same
network as H2
148
Datagram Forwarding Algorithm
D = destination IP address
for each entry (SubnetNum, SubnetMask, NextHop)
D1 = SubnetMask & D
if D1 = SubnetNum
if NextHop is an interface
deliver datagram directly to D
else
deliver datagram to NextHop (router)
149
Classless Routing (CIDR)
 Classless InterDomain Routing (CIDR, pronounced
"cider")
 CIDR addresses two scaling concerns in the Internet
 the growth of backbone routing tables as more and more
network numbers need to be stored
 the potential for the 32-bit IP address space to be exhausted
well before the 4 billionth (= 232) host is attached to the
Internet
 CIDR assigns block of contiguous network numbers to
nearby networks
150
 CIDR tries to balance the following
 minimize the number of routes that a router needs
to know
 the need to hand out addresses efficiently
 CIDR helps to aggregate routes
 uses a single entry in a forwarding table to reach a
lot of different networks by breaking the rigid
boundaries between address classes
151
 example, consider a hypothetical AS
(Autonomous System) with 16 class C network
numbers
 instead of handing out 16 addresses at
random, we can hand out a block of
contiguous class C addresses
 suppose we assign the class C network
numbers from 192.4.16 through 192.4.31
 the top 20 bits of all the addresses in this
range are the same (11000000 00000100
0001)
152
 what we have effectively created is a 20-bit
network number-something that is between a
class B network number and a class C number
153
7
A:
0
24
Network
Host
14
B:
1
0
16
Network
Host
21
C:
1
1
0
Network
8
Host
IP addresses: (a) class A; (b) class B; (c) class C
154
 CIDR allows the prefixes (network numbers) can be
of any length
 convention: place a /X after the prefix where X is the
prefix length in bits
 the example above, the 20-bit prefix for all the
networks 192.4.16 through 192.4.31 is represented as
192.4.16/20
 if we want to represent a single class C network
number, its prefix is 24 bits long, we would write it
192.4.16/24
155
 Routing protocol can use CIDR to deal with "classless"
addresses
 it must understand that a network number may be of
any length
 network numbers are represented by (length, value)
pairs
 length: gives the number of bits in the network
prefix, e.g., 20 in the above example
156
 Internet Service Provider (ISP) network has to
provide Internet connectivity to a large number of
corporations and campuses (customers)
 if we assign prefixes to the customers in such a way
that many different customer networks connected to
the provider network share a common, shorter address
prefix, then we can get even greater aggregation of
routes
157
 example, assume that eight customers served by the
provider network have each been assigned adjacent
24-bit network prefixes
 those prefixes all start with the same 21 bits
 all of the customer are reachable through the
same provider network
 it can advertise a single route to all of them by
just advertising the common 21-bit prefix they
share
158
128
1
0
0
0
0
0
0
0
135
1
0
0
0
0
1
1
1
Route aggregation with CIDR
159
IP Forwarding Revisited
 CIDR means that prefixes may be of any length, from
2 to 32 bits
 it is possible to have prefixes in the forwarding table that
"overlap," in the sense that some addresses may match
more than one prefix
 example1
 we might find both 171.69 (a 16-bit prefix) and 171.69.10
(a 24-bit prefix) in the forwarding table of a single router
 a packet destined to, say, 171.69.10.5, clearly matches
both prefixes
 171.69.10 would be the longest match in this case
160
 example2
 a packet destined to 171.69.20.5 would match 171.69
and not 171.69.10
 in the absence of any other matching entry in the
routing table, 171.69 would be the longest match
161
3.2.6 Address Translation (ARP)
 Issue
 IP datagrams contain IP addresses, but the physical
interface hardware on the host or router to which you
want to send the datagram only understands the
addressing scheme of that particular network
162
 Resolution
 translate the IP address to a link-level address that
makes sense on this network (e.g., a 48-bit Ethernet
address)
 encapsulate the IP datagram inside a frame that contains
that link-1evel address and send it either to the ultimate
destination or to a router that promises to forward the
datagram toward the ultimate destination
frame
link-level
address
IP datagram
Encapsulation
163
Network part
Host part
(physical address)
 Simple way to map an IP address into a physical network
address
 encode a host’s physical address in the host part of its
IP address
 example, a host with physical address 00100001
01001001 (the decimal value 33 in the upper byte and
73 in the lower byte) might be given the IP address
128.96.33.73
 it is limited in that the network’s physical addresses
can be no more than 16 bits long in this example
164
 More general solution
 each host maintains a table of address pairs (map IP
addresses into physical addresses)
 Alternative solution：Address Resolution Protocol
(ARP)
 enable each host on a network to build up a table of
mappings between IP addresses and link-level addresses
 since these mappings may over time (e.g. because an
Ethernet card in a host breaks and is replaced by a new one
with a new address), the entries are timed out periodically
and removed
165
 this happens on the order of every 15 minutes
 the set of mappings currently stored in a host is known as the
ARP cache or ARP table
166
 The ARP packet contains
 HardwareType
 the type of physical network (e.g., Ethernet)
 ProtocolType
 the higher-layer protocol (e.g., IP)
 HLen (“hardware” address length) and PLen
(“protocol” address length)
 the length of the link-layer address and higher-layer
protocol address
167
 Operation
 specifies whether this is a request or a response
 Addresses




source hardware (Ethernet) address (6 bytes)
source protocol (IP) address (4 bytes)
target hardware (Ethernet) address (6 bytes)
target protocol (IP) address (4 bytes)
168
0
8
16
Hardware type = 1
HLen = 48
31
ProtocolType = 0x0800
PLen = 32
Operation
SourceHardwareAddr (bytes 0-3)
SourceHardwareAddr (bytes 4-5)
SourceProtocolAddr (bytes 0-1)
SourceProtocolAddr (bytes 2-3)
TargetHardwareAddr (bytes 0-1)
TargetHardwareAddr (bytes 2-5)
TargetProtocolAddr (bytes 0-3)
ARP Packet Format
169
3.2.7 Host Configuration (DHCP)
 Dynamic Host Configuration Protocol (DHCP)
 relies on the existence of a DHCP server that is responsible
for providing configuration information to hosts
 there is at least one DHCP server for an administrative
domain
 at the simplest level, the DHCP server can function just as
a centralized repository for host configuration information
170
 a more sophisticated use of DHCP saves the network
administrator from even having to assign addresses to
individual hosts
 the DHCP server maintains a pool of available
addresses that it hands out to hosts on demand
 this considerably reduces the amount of configuration
an administrator must do by allocating a range of IP
addresses (all with the same network number) to each
network
171
 DHCP server discovery
 to contact a DHCP server, a newly booted or attached host
sends a DHCPDISCOVER message to a special IP
(broadcast) address (255.255.255.255)
 it will be received by all hosts and routers on that network
 in the simplest case, one of these nodes is the DHCP server
for the network
 the server would then reply to the host that generated the
discovery message (all the other nodes would ignore it)
172
 DHCP uses the concept of relay agent
 there is at least one relay agent on each network, and it is
configured with just one piece of information: the IP
address of the DHCP server
 when a relay agent receives a DHCPDISCOVER
message, it unicasts it to the DHCP server and awaits the
response, which it will then send back to the requesting
client
173
A DHCP relay agent receives a broadcast DHCPDISCOVER message from a host and sends
a unicast DHCPDISCOVER to a remote DHCP Server.
174
DHCP packet format
175
176
177
178
 Hardware address length (HLen): 8 bits
 Hop count (Hops): 8 bits
 used by relay agents
 Transaction ID (Xid): 32 bits
 a random number chosen by the client
 used by the client and server to associate messages and responses
between a client and a server
 Number of seconds (Secs): 16 bits
 the elapsed time in seconds since the client began an address
acquisition or renewal process
 Flags: 16 bits
 defined in RFC 1542
B (Broadcast): 1 bit
179





Client IP address (ciaddr): 32 bits
Your IP address (yiaddr): 32 bits
Server IP address (siaddr): 32 bits
Gateway IP address (giaddr): 32 bits
Client hardware address (chaddr): 16 bytes
180
 Server host name (sname): 64 bytes
 Boot filename (file): 128 bytes
 BOOTP/DHCP options: variable length
 the first four bytes contain the (decimal) values 99, 130,
83 and 99
 the remainder of the field consists of a list of tagged
parameters that are called options
 all of the vendor extensions used by BOOTP are also
DHCP options
181
3.2.8 Error Reporting (ICMP)
 Internet Control Message Protocol (ICMP)
 defines a collection of error messages that are sent
back to the source host whenever a router is unable
to process an IP datagram successfully
 ICMP segment structure
182
 ICMP header (starts at bit 160 of the IP header)
 Type
 ICMP type as specified above
 Code (see the following table)
 further specification of the ICMP type
 e.g. an ICMP Destination Unreachable might have this field
set to 1 through 15 each bearing different meaning
 Checksum
 contains error checking data calculated from the ICMP
header+data, with value 0 for this field
183
 ID
 contains an ID value, should be returned in case of
ECHO REPLY
 Sequence
 contains a sequence value, should be returned in case
of ECHO REPLY
184
List of permitted control messages
(incomplete list)
185
186
187
3.2.9 Virtual Networks and Tunnels
 Virtual Private Network (VPN)
 a more controlled connectivity
 corporations with many sites often build private networks by
leasing transmission lines from the phone companies and
using those lines to interconnect sites
 communication is restricted to take place only among the
sites of that corporation, which is often desirable for security
reasons
 to make a private network “virtual”, the leased transmission
lines - which are not shared with any other corporations would be replaced by some sort of shared network
188
An example of virtual private networks: (a) two separate private networks;
(b) two virtual private networks sharing common switches.
 In the above figure
 Frame Relay or ATM network is used to provide the
controlled connectivity among sites
 limited connectivity of a real private network is
maintained
 IP Tunnel
 a virtual point-to-point link between a pair of nodes that
are actually separated by an arbitrary number of
networks
190
A tunnel through an internetwork (the change in encapsulation
of the packet as it moves across the network)
191
 A tunnel has been configured from R1 to R2
and assigned a virtual interface number 0
 The forwarding table in R1 might therefore
look like the following table
 R1 has two physical interfaces
 interface 0 connects to network 1
 interface 1 connects to a large internetwork and is
thus the default for all traffic that does not match
something more specific in the forwarding table
192
 R1 has a virtual interface, which is the interface to the tunnel
 suppose R1 receives a packet from network 1 that contains an
address in network 2
 the forwarding table says this packet should be sent out
virtual interface 0
 in order to send a packet out this interface, the router
takes the packet, adds an IP header addressed to R2, and
then proceeds to forward the packet as it had just been
received
 R2’s address is 10.0.0.1
 since the network number of this address is 10, not 1 or 2,
a packet destined for R2 will be forwarded out the default
interface into the internetwork
193
NetworkNum
NextHop
1
Interface 0
2
Virtual
interface 0
Default
Interface 1
Forwarding table for router R1
194
3.3 Routing
3.3.1 Network as a Graph
3.3.2 Distance Vector (RIP)
3.3.3 Link State (OSPF)
3.3.4 Metrics
195
 Route
 a way or course taken in getting from a starting point to a
destination
 send or direct along a specified course
 Routing
 find the path or course of forwarding according to
information contained in packet (destination)
 Difference between network-layer and link-layer
 format of forwarding table
 way of updating the table
196
Link-layer
 Forwarding table
 mapping from destination physical address (MAC
address) to port of forwarding
 Update of the table
 manually configured
197
IP (Network) Layer
 Forwarding table
 mapping from destination network id (NetNum) to next-hop (or
interface) of forwarding
 Update the table
 manually configured (static route)
 dynamically learned from routing protocol
198
Forwarding vs. Routing
Forwarding
taking a packet  looking at its destination
address  consulting a table  sending the packet
in a direction determined by that table
locally done at a node
Routing
the process by which forwarding tables are built
depends on a distributed algorithm
199
Forwarding Table vs. Routing Table
 Forwarding table
 used when a packet is being forwarded and so must
contain enough information to accomplish the
forwarding function
 a row in the forwarding table contains the mapping
from a network number to an outgoing interface and
some MAC information, such as the Ethernet
address of the next hop
200
 Routing table
 the table that is built up by the routing algorithms as
a precursor to building the forwarding table
 it contains mappings from network numbers to
next-hops (IP addresses)
201
 Example, in the following tables
 the routing table tells us that network number 10 is
to be reached by a next hop router with the IP
address 171.69.245.10
 the forwarding table contains the information about
exactly how to forward a packet to that next hop
 send it out interface number 0 with a MAC address of
8:0:2b:e4:b:l:2 (the last piece of information is
provided by the Address Resolution Protocol)
202
Network Number
Next Hop
Network
Number
Interface
MAC address
10
171.69.245.10
10
if0
8:0:2b:e4:b:1:2
(a)
(b)
Example rows from (a) routing and (b) forwarding tables
203
3.3.1 Network as a Graph
204
 Basic problem of routing
 find the lowest-cost path between any two nodes, where
the cost of a path equals the sum of the costs of all the
edges that make up the path
205
 Solution
 routing is achieved in most practical networks by running
routing protocols among the nodes
 these protocols provide a distributed, dynamic way to solve
the problem of finding the lowest-cost path in the presence of
 node or link failure
 addition of new node or new link
 changes of link cost
 it is difficult to make centralized solutions scalable, so all the
widely used routing protocols use distributed algorithms
206
 Elements of a routing protocol
 local data structure
 the routing table
 format of messages for exchanging routing information
 Static vs. dynamic routing
 static
 manually set forwarding table
 not adaptive to changes in network topology
207
 dynamic
abstract: weighted graph
vertex: router
edge: link
weight: cost
criterion: best path from source to destination
“best”: path cost is minimum
metrics for the cost
hop
delay
loss
fee of charge
208
static
dynamic
R1
1
R2
2
R3
209
3.3.2 Distance Vector (RIP)
 Distance-Vector Algorithm (Bellman-Ford Algorithm)
 each node constructs a one-dimensional array (a vector)
containing the "distances" (costs) to all other nodes and
distributes that vector to its immediate neighbors
 response when receiving an announcement from a neighbor
for every entry in the announcement, store it if
the announced distance is shorter than what in the table
 a better route is found
the announcer is just the next-hop in the table
 the metric to destination has been changed
otherwise discard it
210
 assumption
 initially, each node knows the cost of the link to each
of its directly connected neighbors
 broken links are assigned an infinite cost, ∞
211
 Local data structure
 routing table
 destination
 cost to the destination
 corresponding next-hop
 TTL (Time to Live) of the route
212
 Messages exchanged among vertices
 Distance Vector (DV)
 C[n]: distance (cost) from current vertex to the
destination vertex, n
 periodically announced to all the neighbors
 DV is telling neighbors how far I am to all the others
213
Distance Vector Algorithm
 In this example
 the cost of each link is set to 1
 a least-cost path is simply the one with the fewest
hops
214
B
A
Initial State
C
E
D
F
Destination
Y
G
Node X’s Routing Table: Cost / Next-Hop
A’s
B’s
C’s
D’s
E’s
F’s
G’s
A
0
1/A
1/A
∞
1/A
1/A
∞
B
1/B
0
1/B
∞
∞
∞
∞
C
1/C
1/C
0
1/C
∞
∞
∞
D
∞
∞
1/D
0
∞
∞
1/D
E
1/E
∞
∞
∞
0
∞
∞
F
1/F
∞
∞
∞
∞
0
1/F
G
∞
∞
∞
1/G
∞
1/G
0
215
A’s routing table
Destination
Y
A
B
C
D
E
F
G
Cost/
Next-Hop
0/
1/B
1/C
2/C
1/E
1/F
2/F
B
A
C
E
D
F
G
Distance Vector sent by A
216
B
A
After One Step
C
E
D
F
Destination
Y
G
Node X’s Routing Table: Cost / Next-Hop
A’s
B’s
C’s
D’s
E’s
F’s
G’s
A
0
1/A
1/A
2/C
1/A
1/A
2/F
B
1/B
0
1/B
2/C
2/A
2/A
∞
C
1/C
1/C
0
1/C
2/A
2/A
2/D
D
2/C
2/C
1/D
0
∞
2/G
1/D
E
1/E
2/A
2/A
∞
0
2/A
∞
F
1/F
2/A
2/A
2/G
2/A
0
1/F
G
2/F
∞
2/D
1/G
∞
1/G
0
217
After Two
Steps
B
A
C
E
convergence: no more changes when
getting further announcement
Destination
Y
D
F
G
Node X’s Routing Table: Cost / Next-Hop
A
A’s
0
B’s
1/A
C’s
1/A
D’s
2/C
E’s
1/A
F’s
1/A
G’s
2/F
B
1/B
0
1/B
2/C
2/A
2/A
3/F
C
1/C
1/C
0
1/C
2/A
2/A
2/D
D
2/C
2/C
1/D
0
3/A
2/G
1/D
E
1/E
2/A
2/A
3/C
0
2/A
3/F
F
1/F
2/A
2/A
2/G
2/A
0
1/F
G
2/F
3/C
2/D
1/G
3/A
1/G
0
218
 Two different circumstances for a node to send a routing update
to its neighbors
 periodic update
 each node automatically sends an update message every so
often, even if nothing has changed
 triggered update
 happens whenever a node receives an update from one of its
neighbors that causes it to change one of the routes in its
routing table
 i.e., whenever a node's routing table changes, it sends an update
to its neighbors, which may lead to a change in their tables,
causing them to send an update to their neighbors
219
Link Failures
 Example 1 (stable)
Pattern：(Dest, Cost, NextHop)
 F detects that link to G has failed
 F sets distance to G to infinity and sends update to A
[F：(G, ∞, G)]
 A sets distance to G to infinity since it uses F to reach G
[A：(G, ∞, F)]
------------------------------------------------------------------------ A receives periodic update from C with 2-hop path to G
 A sets distance to G to 3 and sends update to F
[A：(G, 3, C)]
 F decides it can reach G in 4 hops via A
[F：(G, 4, A)]
220
∞
 Example 2 (count to infinity)
 link from A to E fails
 A advertises distance of infinity to E [A：(E, ∞, E)]
 B and C advertise a distance of 2 to E
[B：(E, 2, A)] ，[A：(E, 3, B)]，[C：(E, 2, A)]，[A：(E, 3, C)]
 B hears that E can be reached in 2 hops from C
 B decides it can reach E in 3 hops; advertises this to A
[B：(E, 3, C)]
 A decides it can reach E in 4 hops; advertises this to C
[A：(E, 4, B)]
 C decides that it can reach E in 5 hops… [C：(E, 5, A)]
221
 Loop-breaking heuristics (partial solutions)
 set infinity to 16
 split horizon
 split horizon with poison reverse
222
 Solution-1 (set infinity to 16)
 use some relatively small number as an approximation
of infinity, which at least bounds the amount of time
that it takes to count to infinity
 example, set the maximum number of hops to get across
a certain network is never going to be more than 16 (set
16 to be infinity value)
 drawback
 problem occurs if our network grew to a point where
some nodes were separated by more than 16 hops
223
 Solution-2 (split horizon)
 when a node sends a routing update to its neighbors, it
does not send those routes it learned from each neighbor
back to that neighbor
 example, if B has the route (E, 2, A) in its table, then it
knows it must have learned this route from A, and so
whenever B sends a routing update to A, it does not
include the route (E, 2, A) in that update
224
 Solution-3 (split horizon with poison reverse)
 (B actually sends that route back to A, but it puts
negative information in the route to ensure that A will
not eventually use B to get to E)
 Let B be a neighbor of A
 if in the routing table of B, the next hop entry for
destination Z is A, B informs A that its distance to Z
is infinite
[B：(Z, cost, A) → A：(Z, ∞, B)]
225
∞
 Solution 2 & 3 only work for routing loops that involve two
nodes
 example, for larger routing loops
 if B and C had waited for a while after hearing of the
link failure from A before advertising routes to E
 they would have found that neither of them really
had a route to E
226
(2,A)
B
(1,E)
A
(4,C)
D
E
(2,A)
(,-)
(3,B)
C
(3,F)
F
(,-)
F
(4,C)
D
(3,F)
G
(3,F)
F
(,-)
(3,B)
C
E
(2,A)
(4,C)
D
G
(2,A)
(,E)
A
(3,B)
C
E
G
B
B
(,E)
A
B
(,E)
A
(,-)
C
(4,C)
D
E
(,-)
F
(,-)
G
Routing Information Protocol (RIP)
 A DV (Distance Vector) routing protocol
 Rather than advertising the cost of reaching other routers,
the routers advertise the cost of reaching networks
 example, in the following figure, router C would
advertise to router A the fact that it can reach
 networks 2 and 3 at cost 0
[C：(Net2, 0, Net2)，C：(Net3, 0, Net3)]
 networks 5 and 6 at cost 1
[C：(Net5, 1, Net3)，C：(Net6, 1, Net3)]
 network 4 at cost 2
[C：(Net4, 2, Net3)]
228
Example network running RIP
229
 RIP packet format
 the majority of the packets is taken up with
(network-address, distance) pairs
 example
 if router A learns from router B that network X can be
reached at a lower cost via B than via the existing next
hop in the routing table, then
A updates the cost and next hop information for the
network number accordingly
230
RIP packet format
231
 RIP
 a fairly straightforward implementation of distancevector routing
 routers running RIP send their advertisements every 30
seconds
 a router also sends an update message whenever an
update from another router causes it to change its
routing table
232
 metrics or costs for routing
 all link costs being equal to 1
 always try to find the minimum hop route
 valid distances are 1 through 15, with 16 representing
infinity (this limits RIP to running on fairly small
networks-those with no paths longer than 15 hops)
233
3.3.3 Link State (OSPF)
 Distance-Vector approach
 “tell neighbors where I can go, and how far”
 Link-State approach
 “tell all which neighbors I have”
 key
 reliable dissemination of link-state information
 calculation of routes from sum of link-state
knowledge
234
 Link-state routing
 the second major class of intradomain routing protocol
 assumptions
 each node is assumed to be capable of finding out the
state of the link to its neighbors (up or down) and the
cost of each link
235
 basic idea
 every node knows how to reach its directly
connected neighbors, and if we make sure that the
totality of this knowledge is disseminated to
every node, then every node will have enough
knowledge of the network to build a complete
map of the network
 link-state routing protocols rely on two mechanisms
 reliable dissemination of link-state information
 calculation of routes from the sum of all the
accumulated link-state knowledge
236
Link-State Message Data Structure
 LSP (Link-State Packet)
 an update packet created by each node
 information for route calculation
 the ID of the node that created the LSP
 a list of directly connected neighbors of the node, with
the cost of the link to each one
237
 information for reliability
 a sequence number
 ensure having the most recent copy
 reset to zero when routing process restarted
 a time to live (TTL) for this packet
 toooooold packets are discarded
238
Reliable Flooding
 Send local LSP out on all of its directly connected
links
 Each node receiving the LSP forwards it out on all
of its links
 stores each node’s recent LSP
 forwards LSP to neighbors except the sender
itself
 makes confirmation and retransmission with
neighbors
239
 The following figure shows an LSP being
flooded in a small network
 each node becomes shaded as it stores the new LSP
(a) the LSP arrives at node X, which sends it to neighbors
A and C
(b) A and C do not send it back to X, but send it on to B
(c) B receives two identical copies of the LSP, it will
accept whichever arrived first and ignore the second as
a duplicate
(d) B passes the LSP onto D, who has no neighbors to
flood it to, and the process is complete
240
241
New LSP Generation
 Two circumstances to generate new LSP
 expiry of a periodic timer
 with period in tens minutes
 change in topology
 directly connected links go down
 detected by link-layer protocols
 immediate neighbors go down
 detected by periodic “hello” message
242
Calculation of Route
 Dijkstra’s Shortest Path Algorithm
 Notations
 N: vertex set of the graph
 l: l(i, j) is the (non-negative) cost of the edge (i, j)
 s: current vertex
 M: set of ever calculated vertices
 C(n): cost of path from s to n
243
 Calculate a minimum-cost tree from s
M = {s}
for each n in N-{s}
C(n) = l(s,n)
while (N != M)
M = M union {w} such that C(w) is the minimum for
all w in (N-M)
for each n in (N-M)
C(n) = MIN(C(n),C(w)+l(w,n))
244
 In practice, each switch computes its routing table directly
from the LSPs it has collected using a forward search
approach for Dijkstris algorithm
 each switch maintains two lists, known as Tentative
and Confirmed.
 each of these lists contains a set of entries of the form
(Destination, Cost, NextHop)
245
Forward Search Approach for
Dijkstra Algorithm
1. Initialize the Confirmed list with an entry for myself; this entry has a cost of 0.
2. For the node just added to the Confirmed list in the previous step, call it node
Next, select its LSP
3. For each neighbor (Neighbor) of Next, calculate the cost (Cost) to reach this
Neighbor as the sum of the cost from myself to Next and from Next to
Neighbor
 (a) If Neighbor is currently not on either the Confirmed or the Tentative
list, then add (Neighbor, Cost, NextHop) to the Tentative list, where
NextHop is the direction I go to reach Next
 (b) If Neighbor is currently on the Tentative list, and the Cost is less than
the currently listed cost for Neighbor, then replace the current entry with
(Neighbor, Cost, NextHop), where NextHop is the direction I go to reach
Next
4. If the Tentative list is empty, stop. Otherwise, pick the entry from the Tentative
list with the lowest cost, move it to the Confirmed list, and return to step 2
246
Example
Link-state routing: an example network
247
(B, 11, B) (C, 2, C)
(B, 5, C) (A, 12, C)
(A, 10, C)
Open Shortest Path First Protocol
(OSPF)
 OSPF
 one of the most widely used link-state routing protocols
 Open: refers to the fact that it is an open,
nonproprietary standard, created under the auspices
of the IETF
 SPF: comes from an alternative name for link-state
routing
249
 OSPF adds the following features to the basic link-state
algorithm
 authentication of routing messages
 additional hierarchy
 OSPF introduces another layer of hierarchy into routing
by allowing a domain to be partitioned into areas
 a router within a domain does not necessarily need to
know how to reach every network within that domain, but
know only how to get to the right area
 this reduces the amount of information that must be
transmitted to and stored in each node
250
 load balancing
 OSPF allows multiple routes to the same place to be
assigned the same cost and will cause traffic to be
distributed evenly over those routes
251
 There are several different types of OSPF messages, but all begin
with the same header
 OSPF header format
 Version: 2
 Type: 1 through 5
 SourceAddr: identifies the sender of the message
 AreaId: a 32-bit identifier of the area in which the node is
located
252
 Checksum
 the entire packet, except the authentication data, is
protected by a 16-bit checksum using the same algorithm
as the IP header
 Authentication type
 0: no authentication is used
 1: a simple password is used
 2: a cryptographic authentication checksum is used
253
OSPF header format
254
 Five OSPF message types
 Type 1: "hello" message, which a router sends to its
peers to notify them that it is still alive and connected
 Type 2~5: used to request, send, and acknowledge the
receipt of link-state messages
 Basic building block of link-state messages in OSPF is linkstate advertisement (LSA)
 one message may contain many LSAs
255
OSPF packet format for link-state advertisement (Type 1)
256
 OSPF link-state advertisement (LSA)
 Type 1 LSA: advertise the cost of links between routers
 Type 2 LSA: advertise networks to which the
advertising router is connected
 LS Age
 the equivalent of a time to live, except that it counts
up and the LSA expires when the age reaches a
defined maximum value
 Type
 tells us that this is a type 1 LSA
257
 Link-state ID & Advertising router
 in a type 1 LSA, these two fields are identical
 each carries a 32-bit identifier for the router that
created this LSA
 LS sequence number
 detect old or duplicate LSAs
 LS checksum
 verify that data has not been corrupted
 it covers all fields in the packet except LS Age
258
 Length
 the length in bytes of the complete LSA
 Link ID, Link Data, & metric
 each link in the LSA is represented by a Link ID,
some Link Data, and a metric
 TOS
 allow OSPF to choose different routes for IP packets
based on the value in their TOS field
259
3.3.4 Metrics
 Original ARPANET metric
 measures number of packets queued on each link
 took neither latency nor bandwidth into consideration
 New ARPANET metric
 stamp each incoming packet with its arrival time (AT)
 record departure time (DT)
 when link-level ACK arrives, the node compute the packet
delay
Delay = (DT-AT) + Transmit + Latency
 if timeout (ACK did not arrive), DT is reset to the time the
packet was retransmitted
 link cost = average delay over some time period
260
3.4 Implementation and Performance
3.4.1 Switch Basics
 A very simple way to build a switch
 buy a general-purpose workstation and equip it with
a number of network interfaces
 run suitable software to receive packets on one of
its interfaces
 perform any of the switching functions
 send packets out another of its interfaces
A workstation used as packet switch
 The figure shows a workstation with three
network interfaces used as a switch
 a path that a packet might take from the time it
arrives on interface 1 until it is output on interface 2
 we assume DMA (Direct Memory Access)
 the workstation has a mechanism to move data directly
from an interface to its main memory, i.e., direct
memory access (DMA)
 once the packet is in memory, the CPU examines its
header to determine on which interface the packet
should be out
 it then uses DMA to move the packet out to the
appropriate interface
 the packet does not go to the CPU because the CPU
inspects only the header of the packet
 Main problem with using a workstation as a switch
 its performance is limited by the fact that all packets
must pass through a single point of contention
 in the example shown, each packet crosses the I/O bus
twice and is written to and read from main memory
once
 the upper bound on aggregate throughput of such a
device is, thus, either half the main memory bandwidth
or half the I/O bus bandwidth, whichever is less (usually
it’s the I/O bus bandwidth)
 example
 a workstation with a 133-MHZ, 64-bit wide I/O bus can
transmit data at a peak rate of a little over 8 Gbps (= 133 ×
220 × 64)
 since forwarding a packet involves crossing the bus twice,
the actual limit is 4 Gbps
 this upper bound also assumes that moving data is the only
problem
 a fair approximation for long packets
 a bad one when packets are short
 the cost of processing each packet- (1) parsing its header
and (2) deciding which output link to transmit it on-is
likely to dominate
 example, a workstation can perform all the necessary
processing to switch 1 million packets each second (packet
per second (pps) rate)
 if the average packet is short, say, 64 bytes

throughput = pps × (bits per packet)
= 1 × 106 × 64 × 8 (bits per second)
= 512 × 106 (bits per second)
 this 512 Mbps would be shared by all users connected to the
switch
 example, a 10-port switch with this aggregate throughput
would only be able to cope with an average data rate of
51.2 Mbps on each port
Control
processor
Switch
fabric
 To address this problem
Input
port
Output
port
 a large array of switch designs that reduce the
amount of contention and provide high aggregate
throughput
 some contention is unavoidable
 if every input has data to send to a single output, then
they cannot all send it at once
 if data destined for different outputs is arriving at
different inputs, a well-designed switch will be able to
move data from inputs to outputs in parallel, thus
increasing the aggregate throughput
3.4.2 Ports
Control
processor
Switch
fabric
Input
Output
port
port
A 4 × 4 switch
Control
processor
Switch
fabric
 The 4 × 4 switch in the figure consists of
Input
port
Output
port
 ports (input ports and output ports)
 communicate with the outside world
 contain fiber-optic receivers and buffers to hold packets that
are waiting to be switched or transmitted, and often a
significant amount of other circuitry that enables the switch
to function
 switch fabric
 when presented with a packet, deliver it to the right output
port
 control processor (at least one)
 in charge of the whole switch
 Input port
 the first place to look for performance bottlenecks
 has to receive a steady stream of packets, analyze
information in the header of each one to determine
which output port (or ports) the packet must be sent
and pass the packet on to the fabric
 Another key function of ports: buffering
 it can happen in either the input or the output port
 it can also happen within the fabric (sometimes
called internal buffering)
 simple input buffering has some serious limitations
 example, an input buffer implemented as a FIFO
 as packets arrive at the switch, they are placed in the
input buffer
 the switch then tries to forward the packets at the front
of each FIFO to their appropriate output port
 if the packets at the front of several different input
ports are destined for the same output port at the same
time, then only one of them can be forwarded; the rest
must stay in their input buffers
Simple illustration of head-of-line blocking
 drawback (head-of-line blocking)
 occurs at input buffering
 those packets left at the front of the input buffer prevent
other packets further back in the buffer from getting a
chance to go to their chosen outputs
 buffering
 wherever contention is possible
 input port (contend for fabric)
 internal (contend for output port)
 output port (contend for link)
3.4.3 Fabrics
 Should be able to move packets from input
ports to output ports with minimal delay and in
a way that meets the throughput goals of the
switch
 Parallelism
 a high-performance fabric with n ports can often
move one packet from each of its n ports to one of
the output ports at the same time
Control
processor
Switch
 Types of fabric




shared bus
shared memory
crossbar
self-routing
fabric
Input
Output
port
port
 Shared bus
 found in a conventional workstation used as a switch
 the bus bandwidth determines the throughput of the switch,
high-performance switches usually have specially designed
busses rather than the standard busses found in PCs
 Shared memory
 packets are written into a memory location by an input port
and then read from memory by the output ports
 the memory bandwidth determines switch throughput, so
wide and fast memory is typically used in this sort of design
 it usually uses a specially designed, high-speed memory bus
 Crossbar
 a matrix of pathways that can be configured to
connect any input port to any output port
 in their simplest form, they require each output port
to be able to accept packets from all inputs at once
A 4 × 4 crossbar switches
000
001
010
011
 Self-routing
 rely on some information in the packet header to direct
each packet to its correct output
 usually a special “self-routing header” is appended to
the packet by the input port after it has determined
which output the packets needs to go to
 this extra header is removed before the packet leaves the
switch
 self-routing fabrics are often built from large numbers
of very simple 2×2 banyan switching fabrics
100
101
110
111
A self-routing header is applied
to a packet at input to enable the
fabric to send the packet to the
correct output, where it is
removed
(a) packet arrives at input port;
(b) input port attaches selfrouting header to direct packet to
correct output
(c) self-routing header is
removed at output port before
packet leaves switch
 Banyan Network
 constructed from simple 2 x 2 switching elements
 self-routing header attached to each packet
 elements arranged to route based on this header
 look at 1 bit in each self-routing header
 route packets toward the upper output if it is zero or
toward the lower output if it is one
000
001
010
011
100
101
110
111
 if two packets arrive at the same time and both have the
bit set to the same value, then they want to be routed to
the same output and a collision will occur
 the banyan network routes all packets to the correct
output without collisions if the packets are presented in
ascending order
000
001
010
011
100
101
110
111
Routing packets through a banyan network. The 3-bit numbers represent values
in the self-routing headers of four arriving packets.