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1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16
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Page 1: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

1

Announcements

• Review session next Friday 03/11• Homework 5 due on Friday 03/04• Project 3 due Wednesday 03/16

Page 2: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

The Network Layer

Page 3: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

3

Purpose of Network layer

• Given a packet, send it across the network to destination• 2 key issues:

– Portability: • connect different technologies

– Scalability• To the Internet scale

networkdata linkphysical

networkdata linkphysical

networkdata linkphysical

networkdata linkphysical

networkdata linkphysical

networkdata linkphysical

networkdata linkphysical

networkdata linkphysical

application

transportnetworkdata linkphysical

application

transportnetworkdata linkphysical

Page 4: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

4

What does it involve?

Two important functions:• routing: determine path from source to dest. • forwarding: move packets from router’s input to output

T1T3

Sts-1

T3

T1

Page 5: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

5

Network service model

Q: What service model for “channel” transporting packets from sender to receiver?

• guaranteed bandwidth?• preservation of inter-packet

timing (no jitter)?• loss-free delivery?• in-order delivery?• congestion feedback to

sender?

? ??virtual circuit

or datagram?

The most important abstraction provided

by network layer:

serv

ice a

bst

ract

ion

Which things can be “faked” at the transport layer?

Page 6: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

6

Two connection models• Connectionless (or “datagram”):

– each packet contains enough information that routers can decide how to get it to its final destination

• Connection-oriented (or “virtual circuit”)– first set up a connection between two nodes– label it (called a virtual circuit identifier (VCI))– all packets carry label

BAb b

C

BA1 1

C

1

Page 7: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

7

Virtual circuits: signaling protocols

• used to setup, maintain teardown VC• setup gives opportunity to reserve resources• used in ATM, frame-relay, X.25• not used in today’s Internet

application

transportnetworkdata linkphysical

application

transportnetworkdata linkphysical

1. Initiate call 2. incoming call

3. Accept call4. Call connected5. Data flow begins 6. Receive data

Page 8: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

8

Virtual circuit switching• Forming a circuit:

– send a connection request from A to B. Contains VCI + address of B– rule: VCI must be unique on the link its used on– switch creates an entry mapping input messages with VCI to output

port– switch picks a new VCI unique between it and next switch

a b

25

21

c12

1

Page 9: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

9

(Input link,VCI) (output link, new VCI)(1, 2) (?, ?)(1, 5) (?, ?)

Virtual circuit forwarding• For each VCI switch has a table which maps input link to

output link and gives the new VCI to use– if a’s messages come into switch 1 on link 2 and go out on link 3 then

the table will be:

a b

25

21

c12

1

Switch 1

Switch 2

Switch 3

Page 10: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

10

Virtual Circuits: Discussion

• Plusses: easy to associate resources with VC– Easy to provide QoS guarantees (bandwidth, delay)– Very little state in packet

• Minuses:– Not good in case of crashes

• Requires explicit connect and teardown phases

– What if teardown does not get to all routers?– What if one switch crashes?

• Will have to teardown and rebuild route

Page 11: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

11

Datagram networks• no call setup at network layer• routers: no state about end-to-end connections

– no network-level concept of “connection”

• packets typically routed using destination host ID– packets between same source-dest pair may take different paths

• Best effort: data corruption, packet drops, route loops

application

transportnetworkdata linkphysical

application

transportnetworkdata linkphysical

1. Send data 2. Receive data

Page 12: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

12

Datagrams: Forwarding

How does packet get to the destination?• switch creates a “forwarding table”, mapping destinations to output

port (ignores input ports)• when a packet with a destination address in the table arrives, it

pushes it out on the appropriate output port• when a packet with a destination address not in the table arrives, it

must find out more routing information (next problem)

a b

c1d

22

00S1

S2

S3

1

01

Page 13: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

13

Datagrams

• Plusses:– No round trip connection setup time– No explicit route teardown– No resource reservation each flow could get max bandwidth– Easily handles switch failures; routes around it

• Minuses– Difficult to provide resource guarantees– Higher per packet overhead

• Internet uses datagrams: IP (Internet Protocol)

Page 14: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

14

Datagrams Forwarding

• How to build forwarding tables?– Manually enter it

• What if nodes crashed

• What about scale?

• The graph-theoretic routing problem– Given a graph, with vertices (switches), edges (links), and edge

costs (cost of sending on that link)– Find the least cost path between any two nodes

• Path cost = (cost of edges in path)

Page 15: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

15

Simple Routing Algorithm

• Choose a central node– All nodes send their (nbr, cost) information to this node– Central node uses info to learn entire topology of the network– It then computes shortest paths between all pairs of nodes

• Using All Pair Shortest Path Algorithm

– Sends the new matrix to every node

• Nice, simple, elegant!• What is the problem?

– Scalability: centralization hurts scalability– Central node is “crushed” with traffic

Page 16: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

16

Link State Routing

• Basic idea:– Every node propagates its (nbr, cost) information– This information at all nodes is enough to construct topology– Can use a graph algorithm to find the shortest routes

• Mechanisms required:– Reliable flooding of link information– Method to calculate shortest route (Dijkstra’s algorithm)

• Example link state update packet:– [node id, (nbr, cost) list, seq. no., ttl]

• Seq. no. to identify latest updates, ttl specifies when to stop msg.

Page 17: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

17

Reliable flooding

receive(pkt) If already have a copy of LSP from pkt.ID

if pkt’s sequence number <= copy’sdiscard pkt

elsedecrement pkt.TTLreplace copy with pktforward pkt to all links besides the one that we received it on

# done every 10 minutes or sogen_LSP()

increment node’s sequence # by onerecompute cost vectorsend created LSP to all neighbors

Page 18: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

18

Discussion: Link-State Routing

• Plusses: – Simple, determines the optimal route most of the time– Used by OSPF

• Minuses:– Might have oscillations

– Avoid using load as cost metric, reduce herding effect

A

D

C

B1 1+e

e0

e

1 1

0 0

A

D

C

B2+e 0

001+e1

A

D

C

B0 2+e

1+e10 0

A

D

C

B2+e 0

e01+e1

Initially start withalmost equal routes

… everyone goes with least loaded

… recomputeLeast loaded =>

Most loaded

… recompute

Page 19: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

19

Is our routing algo scalable?

• Route table size grows with size of network– Because our address structure is flat!

• Solution: have a hierarchical structure– Used by OSPF– Divide the network into areas, each area has unique number

• Nodes carry their area number in the address 1.A, 2.B, etc.

– Nodes know complete topology in their area– Area border routers (ABR) know how to get to any other area

Page 20: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

20

Hierarchical Addressing

1.a3.b

2.b11.b2

2

00S1

S2

S3

1

01

2.a

3

3.a

2

Zone 3

Zone 2

Forwarding table for switch 1Destination switch port

2. ?3. ? 1.b ? 1.a ?

Page 21: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

21

IP has 2-layer addressing

• Each IP address is 32 bits– Network part: which network the host is on?– Host part: identifies the host.

• All hosts on same network have the same network part

• 3 classes of addresses: A, B and C

18.26.0.1

network 32-bits host

0 net host

1 7 24 bits

1 0 net host

2 14 16 bits

110 net host

3 21 8 bits

Page 22: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

22

IP addressing

• The different classes:

• Problems: inefficient, address space exhaustion

0network host

10 network host

110 network host

1110 multicast address

A

B

C

D

class1.0.0.0 to127.255.255.255

128.0.0.0 to191.255.255.255

192.0.0.0 to223.255.255.255

224.0.0.0 to239.255.255.255

32 bits

Unicast

Multicast

1111 reservedE 240.0.0.0 to255.255.255.255Reserved

Page 23: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

23

IP addressing: CIDR

• Classless InterDomain Routing– network portion of address of arbitrary length– address format: a.b.c.d/x, where x is # bits in network portion

– Examples:• Class A: /8• Class B: /16• Class C: /24

11001000 00010111 00010000 00000000

networkpart

hostpart

200.23.16.0/23

Page 24: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

24

Internet Protocol Datagram

ver length

32 bits

data (variable length,typically a TCP

or UDP segment)

16-bit identifier

Internet checksum

time tolive

32 bit source IP address

IP protocol versionNumber

header length

max numberremaining hops

(decremented at each router)

forfragmentation/reassembly

total datagramlength (bytes)

upper layer protocolto deliver payload to

head.len

type ofservice

“type” of data flgsfragment

offsetupper layer

32 bit destination IP address

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

Page 25: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

25

Datagram Portability

• IP Goal: To create one logical network from multiple physical networks– All intermediate routers should understand IP– IP header information sufficient to carry the packet to destination– Goal: Run over anything!

• Problem:– Physical networks have different MTUs– “max. transmission unit”: 1500 for Ethernet, 48 for ATM

• Solution 1:– Fit everything in the MTU (!)

Page 26: 1 Announcements Review session next Friday 03/11 Homework 5 due on Friday 03/04 Project 3 due Wednesday 03/16.

26

IP Fragmentation & Reassembly

• Solution 2: (the one used)– If packet size > MTU of network, then fragment into pieces

• Each fragment is less than MTU size

• Each has IP headers + frag bit set + frag id + offset

– Packets may get refragmented on the way to destination– Reassembly only done at the destination– What is a good initial packet size?

fragmentation: in: one large datagramout: 3 smaller datagrams

reassembly


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