EEC-484/584 Computer Networks

EEC-484/584EEC-484/584Computer NetworksComputer Networks

Lecture 10Lecture 10

Wenbing ZhaoWenbing Zhao

[email protected](Part of the slides are based on Drs. Kurose & (Part of the slides are based on Drs. Kurose &

RossRoss’’s slides for their s slides for their Computer Networking Computer Networking book)book)

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04/19/2304/19/23 EEC-484/584: Computer NetworksEEC-484/584: Computer Networks Wenbing ZhaoWenbing Zhao

OutlineOutline• Distance vector routing• Hierarchical routing• Internet protocol

– Header– Fragmentation

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Distance Vector RoutingDistance Vector Routing• Also called Bellman-Ford or Ford-Fulkerson• Each router maintains a table (a vector), giving

best known distance to each destination and which line to use to get there– Table is updated by exchanging info with neighbors– Table contains one entry for each router in network with

• Preferred outgoing line to that destination• Estimate of time or distance to that destination

– Once every T msec, router sends to each neighbor a list of estimated delays to each destination and receives same from those neighbors

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Distance Vector Routing:Distance Vector Routing:How each entry is updatedHow each entry is updated

d(A,X)

d(A,Y)A

X Z

d(Y,Z)

d(X,Z)

At router A, for ZCompute d(A,X) + d(X,Z) and d(A,Y) + d(Y,Z), take minimum

Y

d(A,Z) = min {d(A,v) + d(v,Z) }

where min is taken over all neighbors v of A

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x y z

xyz

0 2 7

∞ ∞ ∞∞ ∞ ∞

from

cost to

from

from

x y z

xyz

0

from

cost to

x y z

xyz

∞ ∞

∞ ∞ ∞

cost to

x y z

xyz

∞ ∞ ∞7 1 0

cost to

∞2 0 1

∞ ∞ ∞

2 0 17 1 0

time

x z12

7

y

node x table

node y table

node z table

d(x,y) = min{d(x,y) + d(y,y), d(x,z) + d(z,y)} = min{2+0 , 7+1} = 2

d(x,z) = min{d(x,y) + d(y,z), d(x,z) + d(z,z)} = min{2+1 , 7+0} = 3

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x y z

xyz

0 2 7

∞ ∞ ∞∞ ∞ ∞

from

cost to

from

from

x y z

xyz

0 2 3

from

cost tox y z

xyz

0 2 3

from

cost to

x y z

xyz

∞ ∞

∞ ∞ ∞

cost tox y z

xyz

0 2 7

from

cost to

x y z

xyz

0 2 3

from

cost to

x y z

xyz

0 2 3

from

cost tox y z

xyz

0 2 7

from

cost to

x y z

xyz

∞ ∞ ∞7 1 0

cost to

∞2 0 1

∞ ∞ ∞

2 0 17 1 0

2 0 17 1 0

2 0 13 1 0

2 0 13 1 0

2 0 1

3 1 0

2 0 1

3 1 0

time

x z12

7

y

node x table

node y table

node z table

d(x,y) = min{d(x,y) + d(y,y), d(x,z) + d(z,y)} = min{2+0 , 7+1} = 2

d(x,z) = min{d(x,y) + d(y,z), d(x,z) + d(z,z)} = min{2+1 , 7+0} = 3

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Distance Vector RoutingDistance Vector Routing

• Distance from A to B 12ms, to C 25ms, to D 40ms, to G 18ms

• Distance from J to A 8ms, to I 10ms, to H 12ms, to K 6ms

• Distance from J to A to G 8+18 = 26msto I to G 10+31 = 41msto H to G 12+6=18msto K to G 6+31=37ms

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Distance Vector RoutingDistance Vector Routing• Good news travels fast• Bad news travels slow• Count to infinity problem: Takes too long to converge

upon router failure

×

Routers’ knowledge about the cost to A

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Hierarchical RoutingHierarchical Routing

• Problem: Bigger network => bigger routing table– As network size increases, more router memory used to

store routing table, more time to process routing tables, more bandwidth to transmit states reports

• Use hierarchical structure to solve the problem– Regions: router knows details of how to route packets

within its region, does not know internals of other regions– Clusters of regions, zones of clusters, groups of zones

• Tradeoff: savings in memory space may result in longer path

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Hierarchical RoutingHierarchical Routing

1111


Collection of SubnetworksCollection of SubnetworksThe Internet is an interconnected collection of many networks,

or Autonomous Systems (ASes)

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The Network Layer in Internet The Network Layer in Internet

forwardingtable

Host, router network layer functions:

Routing protocols•path selection•RIP, OSPF, BGP

IP protocol•addressing conventions•datagram format•packet handling conventions

ICMP protocol•error reporting•router “signaling”

Transport layer: TCP, UDP

Link layer

physical layer

Networklayer

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IP Datagram FormatIP Datagram Format

ver Totallength

32 bits

data (variable length,typically a TCP

or UDP segment)

16-bit identifier

header checksum

time tolive

32 bit source IP address

IP protocol versionnumber

header length (bytes)

max numberremaining hops

(decremented at each router)

forfragmentation/reassembly

total datagramlength (bytes)

upper layer protocolto deliver payload to

IHL type ofservice

“type” of data flgsfragment

offsetprotocol

32 bit destination IP address

Options (if any) E.g. timestamp,record routetaken, specifylist of routers to visit.

How much overhead with TCP?

• 20 bytes of TCP

• 20 bytes of IP

• = 40 bytes + app layer overhead

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The IPv4 HeaderThe IPv4 Header• Version – 4• IHL – length of header in 32-bit words

– Min 5, max 15 – i.e., 60 bytes

• Type of service - to distinguish different classes of service– To accommodate differentiated services (which class this

packet belongs to)

• Total length – header and data 65,535 (216-1) bytes• Identification – allows destination to determine which

datagram a fragment belongs to

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The IPv4 HeaderThe IPv4 Header• Time to live – counter to limit packet lifetimes

– Max lifetime 255sec– Packet is destroyed when counter becomes 0

• Protocol – which transport layer protocols being used

• Header checksum – verifies header

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The IPv4 HeaderThe IPv4 Header

• Options – security, error reporting, etc.– Some of the IP options

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IP FragmentationIP Fragmentation• Fragmentation Flags

– DF – tells routers “Don’t Fragment”– MF – More Fragments. All fragments except last have

this set. Used as check against total length

• Fragment offset – where in datagram this fragment belongs– All fragments (payload in the IP packet) except last

must be multiples of 8 bytes

– The number of 8 byte blocks is called Number of Fragment Blocks (NFB)

– The unit of the offset is NFB

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IP Fragmentation & ReassemblyIP Fragmentation & Reassembly• Network links have MTU

(max.transfer size) - largest possible link-level frame.– different link types,

different MTUs • Large IP datagram divided

(“fragmented”) within net– one datagram becomes

several datagrams– “reassembled” only at

final destination– IP header bits used to

identify, order related fragments

fragmentation: in: one large datagramout: 3 smaller datagrams

reassembly

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IP Fragmentation and ReassemblyIP Fragmentation and Reassembly

ID=x

offset=0

MF=0

length=4000

ID=x

offset=0

MF=1

length=1500

ID=x

offset=185

MF=1

length=1500

ID=x

offset=370

MF=0

length=1040

One large datagram becomesseveral smaller datagrams

Example• 4000 byte

datagram• MTU = 1500

bytes1480 bytes in data field

offset =1480/8

Fragment should be as large as possible

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Distance Vector Routing: ExerciseDistance Vector Routing: Exercise

• Consider the subnet shown below. Distance vector routing is used, and the following vectors have just come in to router C: from B: (5, 0, 8, 12, 6, 2); from D: (16, 12, 6, 0, 9, 10); and from E: (7, 6, 3, 9, 0, 4). The measured delays to B, D, and E, are 6, 3, and 5, respectively. What is C's new routing table? Give both the outgoing line to use and the expected delay.

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Exercise: IP FragmentationExercise: IP Fragmentation• Suppose that host A is connected to a router R 1, R 1 is

connected to another router, R 2, and R 2 is connected to host B. Suppose that a TCP message that contains 900 bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Total length, Identification, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header, and link R2-B can support a maximum frame size of 512 bytes including a 12-byte frame header.

EEC-484/584 Computer Networks

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