ip addressing: cidr

Network Layer 4-40

IP addressing: CIDRCIDR: Classless InterDomain Routing

m subnet portion of address of arbitrary lengthm address format: a.b.c.d/x, where x is # bits in

subnet portion of address

11001000 00010111 00010000 00000000

subnetpart

hostpart

200.23.16.0/23

Network Layer 4-41

IP addresses: how to get one?

Q: How does host get IP address?

r hard-coded by system admin in a filem Wintel: control-panel->network->configuration-

>tcp/ip->propertiesm UNIX: /etc/rc.config

r DHCP: Dynamic Host Configuration Protocol: dynamically get address from serverm “plug-and-play”

(more in next chapter)

Network Layer 4-42

IP addresses: how to get one?Q: How does network get subnet part of IP

addr?A: gets allocated portion of its provider ISP’s

address space

ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20

Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/23 Organization 1 11001000 00010111 00010010 00000000 200.23.18.0/23 Organization 2 11001000 00010111 00010100 00000000 200.23.20.0/23

... ….. …. ….Organization 7 11001000 00010111 00011110 00000000 200.23.30.0/23

Network Layer 4-44

Hierarchical addressing: route aggregation

“Send me anythingwith addresses beginning 200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7Internet

Organization 1

ISPs-R-Us “Send me anythingwith addresses beginning 199.31.0.0/16”

200.23.20.0/23Organization 2

...

...

Hierarchical addressing allows efficient advertisement of routing information:

Note This

Network Layer 4-45

Hierarchical addressing: more specific routes

ISPs-R-Us has a more specific route to Organization 1

“Send me anythingwith addresses beginning 200.23.16.0/20”

200.23.16.0/23

200.23.18.0/23

200.23.30.0/23

Fly-By-Night-ISP

Organization 0

Organization 7Internet

Organization 1

ISPs-R-Us “Send me anythingwith addresses beginning 199.31.0.0/16or 200.23.18.0/23”

200.23.20.0/23Organization 2

...

...

Network Layer 4-46

IP addressing: the last word...

Q: How does an ISP get block of addresses?A: ICANN: Internet Corporation for Assigned

Names and Numbersm allocates addressesm manages DNSm assigns domain names, resolves disputes

Network Layer 4-47

NAT: Network Address Translation

10.0.0.1

10.0.0.2

10.0.0.3

10.0.0.4

138.76.29.7

local network(e.g., home network)

10.0.0/24

rest ofInternet

Datagrams with source or destination in this networkhave 10.0.0/24 address for source, destination (as usual)

All datagrams leaving localnetwork have same single source

NAT IP address: 138.76.29.7,different source port numbers

Network Layer 4-48


r Motivation: local network uses just one IP address as far as outside world is concerned:m range of addresses not needed from ISP: just one IP

address for all devicesm can change addresses of devices in local network

without notifying outside worldm can change ISP without changing addresses of

devices in local networkm devices inside local net NOT explicitly addressable,

visible by outside world (a security plus).

Network Layer 4-49

NAT: Network Address TranslationImplementation: NAT router must:

m outgoing datagrams: replace (source IP address, port #) of every outgoing datagram to (NAT IP address, new port #). . . remote clients/servers will respond using (NAT IP

address, new port #) as destination addr.

m remember (in NAT translation table) every (source IP address, port #) to (NAT IP address, new port #) translation pair

m incoming datagrams: replace (NAT IP address, new port #) in dest fields of every incoming datagram with corresponding (source IP address, port #) stored in NAT table

Network Layer 4-50


10.0.0.1

10.0.0.2

10.0.0.3

S: 10.0.0.1, 3345D: 128.119.40.186, 80

110.0.0.4

138.76.29.7

1: host 10.0.0.1 sends datagram to 128.119.40.186, 80

NAT translation tableWAN side addr LAN side addr

138.76.29.7, 5001 10.0.0.1, 3345…… ……

S: 128.119.40.186, 80 D: 10.0.0.1, 3345 4

S: 138.76.29.7, 5001D: 128.119.40.186, 802

2: NAT routerchanges datagramsource addr from10.0.0.1, 3345 to138.76.29.7, 5001,updates table

S: 128.119.40.186, 80 D: 138.76.29.7, 5001 3

3: Reply arrivesdest. address:138.76.29.7, 5001

4: NAT routerchanges datagramdest addr from138.76.29.7, 5001 to 10.0.0.1, 3345

Network Layer 4-51


r 16-bit port-number field: m 60,000 simultaneous connections with a single

LAN-side address!r NAT is controversial:

m routers should only process up to layer 3m violates end-to-end argument

• NAT possibility must be taken into account by app designers, eg, P2P applications

m address shortage should instead be solved by IPv6

Network Layer 4-52

Chapter 4: Network Layer

r 4. 1 Introductionr 4.2 Virtual circuit and

datagram networksr 4.3 What’s inside a

routerr 4.4 IP: Internet

Protocolm Datagram formatm IPv4 addressingm ICMPm IPv6

r 4.5 Routing algorithmsm Link statem Distance Vectorm Hierarchical routing

r 4.6 Routing in the Internetm RIPm OSPFm BGP

r 4.7 Broadcast and multicast routing

Network Layer 4-53

ICMP: Internet Control Message Protocol

r used by hosts & routers to communicate network-level informationm error reporting:

unreachable host, network, port, protocol

m echo request/reply (used by ping)

r network-layer “above” IP:m ICMP msgs carried in IP

datagramsr ICMP message: type, code plus

first 8 bytes of IP datagram causing error

Type Code description0 0 echo reply (ping)3 0 dest. network unreachable3 1 dest host unreachable3 2 dest protocol unreachable3 3 dest port unreachable3 6 dest network unknown3 7 dest host unknown4 0 source quench (congestion

control - not used)8 0 echo request (ping)9 0 route advertisement10 0 router discovery11 0 TTL expired12 0 bad IP header

Network Layer 4-55

Traceroute and ICMP

r Source sends series of UDP segments to destm First has TTL =1m Second has TTL=2, etc.

r When nth datagram arrives to nth router:m Router discards datagramm And sends to source an

ICMP message (type 11, code 0)

m Message includes name of router& IP address

r When ICMP message arrives, source calculates RTT

r Traceroute does this 3 times

Stopping criterionr UDP segment eventually

arrives at destination hostr Destination returns ICMP

“host unreachable” packet (type 3, code 3)

r When source gets this ICMP, stops.

Network Layer 4-56









Network Layer 4-57

IPv6r Initial motivation: 32-bit address space soon

to be completely allocated. r Additional motivation:

m header format helps speed processing/forwardingm header changes to facilitate QoS IPv6 datagram format:m fixed-length 40 byte headerm no fragmentation allowed

Network Layer 4-58

IPv6 Header (Cont)Priority: identify priority among datagrams in flowFlow Label: identify datagrams in same “flow.”

(concept of“flow” not well defined).Next header: identify upper layer protocol for data

Network Layer 4-59

Other Changes from IPv4

r Checksum: removed entirely to reduce processing time at each hop

r Options: allowed, but outside of header, indicated by “Next Header” field

r ICMPv6: new version of ICMPm additional message types, e.g. “Packet Too Big”m multicast group management functions

Network Layer 4-60

Transition From IPv4 To IPv6

r Not all routers can be upgraded simultaneousm no “flag days”m How will the network operate with mixed IPv4 and

IPv6 routers? r Tunneling: IPv6 carried as payload in IPv4

datagram among IPv4 routers

Network Layer 4-61

TunnelingA B E F

IPv6 IPv6 IPv6 IPv6

tunnelLogical view:

Physical view:A B E F

IPv6 IPv6 IPv6 IPv6IPv4 IPv4

Network Layer 4-62

TunnelingA B E F

IPv6 IPv6 IPv6 IPv6

tunnelLogical view:

Physical view:A B E F

IPv6 IPv6 IPv6 IPv6

C D

IPv4 IPv4

Flow: XSrc: ADest: F

data


data


data

Src:BDest: E


data

Src:BDest: E

A-to-B:IPv6

E-to-F:IPv6B-to-C:

IPv6 insideIPv4

B-to-C:IPv6 inside

IPv4

Network Layer 4-63









Network Layer 4-64

1

23

0111

value in arrivingpacket’s header

routing algorithm

local forwarding tableheader value output link

0100010101111001

3221

Interplay between routing, forwarding

Network Layer 4-65

u

yx

wv

z2

21

3

1

1

2

53

5

Graph: G = (N,E)

N = set of routers = { u, v, w, x, y, z }

E = set of links ={ (u,v), (u,x), (v,x), (v,w), (x,w), (x,y), (w,y), (w,z), (y,z) }

Graph abstraction

Remark: Graph abstraction is useful in other network contexts

Example: P2P, where N is set of peers and E is set of TCP connections

Network Layer 4-66

Graph abstraction: costs

u

yx

wv

z2

21

3

1

1

2

53

5 • c(x,x’) = cost of link (x,x’)

- e.g., c(w,z) = 5

• cost could always be 1, or inversely related to bandwidth,or inversely related to congestion

Cost of path (x1, x2, x3,…, xp) = c(x1,x2) + c(x2,x3) + … + c(xp-1,xp)

Question: What’s the least-cost path between u and z ?

Routing algorithm: algorithm that finds least-cost path

Network Layer 4-67

Routing Algorithm classificationGlobal or decentralized

information?Global:r all routers have complete

topology, link cost infor “link state” algorithmsDecentralized:r router knows physically-

connected neighbors, link costs to neighbors

r iterative process of computation, exchange of info with neighbors

r “distance vector” algorithms

Static or dynamic?Static:r routes change slowly

over timeDynamic:r routes change more

quicklym periodic updatem in response to link

cost changes

Network Layer 4-68









Network Layer 4-69

A Link-State Routing Algorithm

Dijkstra’s algorithmr net topology, link costs

known to all nodesm accomplished via “link

state broadcast” m all nodes have same info

r computes least cost paths from one node (‘source”) to all other nodesm gives forwarding table

for that noder iterative: after k

iterations, know least cost path to k dest.’s

Notation:r c(x,y): link cost from node

x to y; = ∞ if not direct neighbors

r D(v): current value of cost of path from source to dest. v

r p(v): predecessor node along path from source to v

r N': set of nodes whose least cost path definitively known

Network Layer 4-70

Dijsktra’s Algorithm1 Initialization:2 N' = {u} 3 for all nodes v 4 if v adjacent to u 5 then D(v) = c(u,v) 6 else D(v) = ∞ 7 8 Loop9 find w not in N' such that D(w) is a minimum 10 add w to N' 11 update D(v) for all v adjacent to w and not in N' : 12 D(v) = min( D(v), D(w) + c(w,v) ) 13 /* new cost to v is either old cost to v or known 14 shortest path cost to w plus cost from w to v */ 15 until all nodes in N'

ip addressing: cidr

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