lecture 11: ipv6 - university of california, san...

Lecture 11:IPv6

CSE 123: Computer NetworksAlex C. Snoeren

HW 2 due FRIDAY

Lecture 11 Overview• Subnetting example

• Classless Inter-domain Routing (CIDR)

• Forwarding example

• IPv6u Addressingu Packet format

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CSE 123 – Lecture 12: Aggregation

CSE 123 – Lecture 11: IPv6

Subnet Addresses● Every (sub)network has an address and a netmask

u Netmask tells which bits of the network address is importantu Convention suggests it be a proper prefix

● Netmask written as an all-ones (bits) IP addressu E.g., Class B netmask is 255.255.0.0 (i.e, 11111111111111110000000000000000)u Sometimes expressed in terms of number of 1s, e.g., /16

● Need to size subnet appropriately for each LANu Only have remaining bits to specify host addresses

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What is the netmask for a Class A network with 16 subnets?

A. 255.15.0.0B. 255.0.0.15C. 255.240.0.0D. 255.255.240.0

IP Address Problem (1991)● Address space depletion

u In danger of running out of classes A and B

● Why?u Class C too small for most organizations (only ~250 addresses)u Very few class A – very careful about giving them out (who has 16M hosts anyway?)u Class B – greatest problem

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● Classless Inter-Domain Routing (1993)u Networks described by variable-length prefix and lengthu Allows arbitrary allocation between network and host address

u e.g. 10.95.1.2 contained within 10.64.0.0/10:» 10.64.0.0 is network and remainder (31.1.2) is host» 0000101001 / 0111110000000100000010

● Pro: Finer grained allocation; aggregation● Con: More expensive lookup: longest prefix match

Network Host

Prefix Mask=# significant bits representing prefix

CIDR

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Longest Matching Prefix• Forwarding table contains many prefix/length tuples

u They need not be disjoint!u E.g. 200.23.16.0/20 and 200.23.18.0/23u What to do if a packet arrives for destination 200.23.18.1?u Need to find the longest prefix in the table which matches it (200.23.18.0/23)

• Not a simple table, requires multiple memory lookupsu Lots and lots of research done on this problemu Lots of this work was historically done by UCSD faculty

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Route Aggregation● Combine adjacent networks in forwarding tables

u Helps keep forwarding table size down

“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

...

...

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● But what if address range is not contiguous?

“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

...

...

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Most Specific Route



Forwarding example

Provider

• Packet to 10.1.1.6 arrives from Provider

10

H2

H3

H4

R1

10.1.16/24

10.1.1.6

10.1.2/23

R2

10.1.0.2

10.1.8.4

10.1.2.2

10.1.16.2

H1

10.1.1.101

10.1.1.5

10.1.1.1

10.1.0.1

10.1.1.4/30

10.1.2.1

10.1.8.1

10.1.1/24

10.1.8/24

10.1.0/24



What is the correct path?

A. R2 – R1 – H3B. R2 – R1 – H2C. R2 – R1 – H1 – H2D. No path to desintation

Forwarding example (2)

Forwarding table at R2Destination Next Hop

127.0.0.1 loopback

Default or 0/0 10.1.16.1

10.1.8.0/24 interface1


10.1.0.0/23 10.1.2.2

• Packet to 10.1.1.6• Matches 10.1.0.0/23

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H2

H3

H4

R1

10.1.16/24

10.1.1.6

10.1.2/23

R2

10.1.0.2

10.1.8.4

10.1.2.2

10.1.16.2

H1

10.1.1.5

10.1.1.1

10.1.0.1

10.1.1.4/30

10.1.2.1

10.1.8.1

10.1.1/24

10.1.8/24

10.1.0/24

10.1.16.0/24 interface3

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3

10.1.1.101




Routing table at R1Destination Next Hop

127.0.0.1 loopback

Default or 0/0 10.1.2.1

10.1.0.0/24



• Packet to 10.1.1.6• Matches 10.1.1.4/30

• Longest prefix match

10.1.1.4/30 10.1.1.101

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H2

H3

H4

R1

10.1.16/24

10.1.1.6

10.1.2/23

R2

10.1.0.2

10.1.8.4

10.1.2.2

10.1.16.2

H1

10.1.1.101

10.1.1.5

10.1.1.1

10.1.0.1

10.1.1.4/30

10.1.2.1

10.1.8.1

10.1.1/24

10.1.8/24

10.1.0/24

interface1

1

2

3




H2

H3

H4

R1

10.1.1/24

10.1.16/24 10.1.8/24

10.1.0/24

10.1.1.6

10.1.2/23

R2

10.1.0.2

10.1.8.4

10.1.2.2

10.1.16.2

H1

Routing table at H1Destination Next Hop

127.0.0.1 loopback

Default or 0/0 10.1.1.1



• Packet to 10.1.1.6• Direct route

• Longest prefix match

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10.1.1.5

10.1.1.1

10.1.0.1

10.1.1.4/30

10.1.2.1

10.1.8.1

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10.1.1.101



The space crunch…● Still running out of IP addresses… what to do?● Two solutions

u Network Address Translation – multiple multiple hosts on a single IP address (future class)

u Get bigger addresses -> IPv6● IPv6: 128bit addresses… we won’t run out

u 64bit routing prefix, 64bit host id

CSE 123 – Lecture 11: IPv6 14


IPv6 Addresses● Colon-Hex notation

u 8 groups of four HEX digits separated by colons, e.g.» FEDC:0000:0000:0065:4321:0000:DEAD:BEEF

u Can drop leading zeros:» FEDC:0000:0000:65:4321:0000:DEAD:BEEF

u Can even skip first sequence of all zeros w/ ::» FEDC::65:4321:0000:DEAD:BEEF

u Every IPv4 address is a IPv6 address:» E.g., ::FFFF:222.173.190.239 (prepended w/zeros and FFFF)

● Network names expressed as prefix/length:» FEDC::65:43/64

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How many (hex) zeros between FEDC and 65 in the last address?

A. NoneB. 4C. 6D. 8

For Next Time

• Homework 2 due FRIDAY

• Midterm a WEEK FROM TODAY

• Last discussion before Midterm TODAY

• Read 3.2 and 9.3 for next Lecture

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lecture 11: ipv6 - university of california, san...

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