Here are the solutions: 1. (C) 200.1.1.0, Break into 40 networks Subnet mask: 255.255.255.248 Increment: 8 Network ranges: 200.1.1.0 - 200.1.1.7 200.1.1.8 - 200.1.1.15 ... 200.1.1.240 - 200.1.1.247 2. (C) 199.9.10.0, Break into 14 networks Subnet mask: 255.255.255.224 Increment: 16 Network ranges: 199.9.10.0 - 199.9.10.

1. DIT
Dar es Salaam institute of Technology (DIT)
ETU 08102
Digital Networks
Ally, J
jumannea@gmail.com

2. DIT
IP Network

3. What we have today is beyond
any of the inventors’
imagination …

4. InterNetwork
 Millions of end points (you, me, and toasters)
connected across a mesh of links
 Many end points can be addressed by numbers
 Many others lie behind a virtual end point
 Many networks form a bigger network
 The overall structure called the Internet
 With a capital I
 Defined as a network of networks

5. Organizing the Giant Structure
Networks are complex!
 many “pieces”:
 hosts
 routers
 links of various
media
 applications
 protocols
 hardware
 software
Question:
Is there any hope of
organizing structure
of network?

6. Turn to analogies in air travel
 A series of steps
ticket (purchase)
baggage (check)
gates (load)
runway takeoff
airplane routing
ticket (complain)
baggage (claim)
gates (unload)
runway landing
airplane routing
airplane routing

7. ticket (purchase)
baggage (check)
gates (load)
runway (takeoff)
airplane routing
departure
airport
arrival
airport
intermediate air-traffic
control centers
airplane routing airplane routing
ticket (complain)
baggage (claim
gates (unload)
runway (land)
airplane routing
ticket
baggage
gate
takeoff/landing
airplane routing
Layering of Airline Functionality
Layers: each layer implements a service
 layers communicate with peer layers
 rely on services provided by layer below

8. Internet Protocol Stack
 Application: supporting network applications
 FTP, SMTP, HTTP
 Transport: host-host data transfer
 TCP, UDP
 Network: routing of datagrams from source to
destination
 IP, routing protocols
 Link: data transfer between neighboring
network elements
 PPP, Ethernet, WiFi, Bluetooth
 Physical: bits “on the wire”
application
transport
network
link
physical

9. DIT
Legacy Network Architecture
DataData
(TCP / UDP)(TCP / UDP)
DataData
(TCP / UDP)(TCP / UDP)
VoiceVoiceVoiceVoice
VideoVideoVideoVideo
Computer Comm.Computer Comm. Human Comm.Human Comm.
Photonic Network (WDM)Photonic Network (WDM)Photonic Network (WDM)Photonic Network (WDM)
Network LayerNetwork Layer
Data Link LayerData Link Layer
Physical LayerPhysical Layer
SONET / SDHSONET / SDHSONET / SDHSONET / SDHATMATMATMATM
IPIP v4 / v6v4 / v6IPIP v4 / v6v4 / v6 SS-7SS-7SS-7SS-7 DedicatedDedicatedDedicatedDedicated
ＯＩＦＯＩＦＯＩＦＯＩＦ
Photonic Network InterfacePhotonic Network InterfacePhotonic Network InterfacePhotonic Network Interface
OIF: Optical Internetworking Forum

10. DIT
Network Architecture from Today
Photonic Network (WDM)Photonic Network (WDM)Photonic Network (WDM)Photonic Network (WDM)
Data Link LayerData Link Layer
Physical LayerPhysical Layer
SONET / SDHSONET / SDHSONET / SDHSONET / SDHATMATMATMATM
ＯＩＦＯＩＦＯＩＦＯＩＦ
Photonic Network InterfacePhotonic Network InterfacePhotonic Network InterfacePhotonic Network Interface
Ｅｔｈｅｒｎｅｔ （Ｅｔｈｅｒｎｅｔ （ as a standard interfaceas a standard interface ））
IPIP v4 / v6v4 / v6IPIP v4 / v6v4 / v6Network LayerNetwork Layer
VoiceVoiceVoiceVoice
VoiceVoiceVoiceVoice
DataData
(TCP / UDP)(TCP / UDP)
DataData
(TCP / UDP)(TCP / UDP)
VideoVideoVideoVideo
DedicatedDedicatedDedicatedDedicated
VOIPVOIP L3-VPNL3-VPN
L2-VPNL2-VPN
etwork Applicationetwork Application
VPN: Virtual Private Network

11. DIT
Structure of IP Network
End station
End station
End station
End station
Segment;
Ethernet Switching
Ethernet SW
Segment;
Ethernet Switching
Ethernet SW
Ethernet SW
1) Ethernet Switches conform Ethernet switching segments
- “Mac-address” are in use only in each segment
2) Routers & IP-functions in each end station realize routing plane
- “IP-address” are in use over inter-segment routing
IPIPIPIP IPIPIPIP
IPIPIPIP
IPIPIPIP
Router Router
IP IP
Router Router
IP
3) TCP in end station control the quality and flow of the session
TCPTCPTCPTCP
TCPTCPTCPTCP TCPTCPTCPTCP
TCPTCPTCPTCP
TCP session
4) Applications in end stations communicate each other over TCP
Apl.Apl.Apl.Apl.
SocketSocket Apl.Apl.Apl.Apl.
SocketSocket
Apl.Apl.Apl.Apl.
SocketSocket
Apl.Apl.Apl.Apl.
SocketSocket
Application session

12. DIT
Evolution of IP Network
 Improving the capacity of network
 Transmission Speed: Ethernet
 Switching capacity: Ethernet Switch
 Routing capacity: Routing Engine
 Re-constructing the network
 Management System based on IP-plane
 MPLS & G-MPLS
 Services on IP Network
 VPN
 IP Telephony
 NGN (Next Generation Network)

13. DIT
Network and Host Addressing
Using the IP address of the
destination network, a router
can deliver a packet to the
correct network.
When the packet arrives at a
router connected to the
destination network, the router
uses the IP address to locate
the particular computer
connected to that network.
Accordingly, every IP address
has two parts.

14. DIT
Network Layer Communication Path
A router forwards packets from the originating
network to the destination network using the IP
protocol. The packets must include an identifier for
both the source and destination networks.

15. DIT
Internet Addresses
IP Addressing is a hierarchical structure. An IP address
combines two identifiers into one number. This number must be
a unique number, because duplicate addresses would make
routing impossible. The first part identifies the system's network
address. The second part, called the host part, identifies which
particular machine it is on the network.

16. DIT
IP Address Classes
Address
Class
Number of
Networks
Number of Hosts
per Network
A 126 16,777,214
B 16,384 65,534
C 2,097,152 254
IP addresses are divided into classes to define the large,
medium, and small networks.
Class A addresses are assigned to larger networks.
Class B addresses are used for medium-sized
networks.
Class C for small networks.

17. DIT
Identifying Address Classes

18. DIT
Network and Host Division
Each complete 32-bit IP address is broken down into a
network part and a host part. A bit or bit sequence at the
start of each address determines the class of the address.
There are 5 IP address classes.

19. DIT
Class A Addresses
The Class A address was designed to support extremely
large networks, with more than 16 million host addresses
available. Class A IP addresses use only the first octet to
indicate the network address. The remaining three octets
provide for host addresses.

20. DIT
Class B Addresses
The Class B address was designed to support the needs
of moderate to large-sized networks. A Class B IP
address uses the first two of the four octets to indicate the
network address. The other two octets specify host
addresses.

21. DIT
Class C Addresses
The Class C address space is the most commonly
used of the original address classes. This address
space was intended to support small networks with a
maximum of 254 hosts.

22. DIT
Class D Addresses
The Class D address class was created to enable
multicasting in an IP address. A multicast address is a
unique network address that directs packets with that
destination address to predefined groups of IP addresses.
Therefore, a single station can simultaneously transmit a
single stream of data to multiple recipients.

23. DIT
Class E Addresses
A Class E address has been defined. However, the
Internet Engineering Task Force (IETF) reserves
these addresses for its own research. Therefore, no
Class E addresses have been released for use in the
Internet.

24. Converting Between Decimal Numbers
and Binary
 In any given octet of an IP address, the 8 bits can be defined as
follows:
 To convert a decimal number into binary
187 = 10111011 = 128+32+16+8+2+1, 224 = 11100000 = 128+64+32
 To convert a binary number into decimal
10101010 = 128+32+8+2 = 170, 11110000 = 128+64+32+16 = 240
 The IP address 138.101.114.250 is represented in binary as
10001010.01100101.01110010.11111010
 The subnet mask of 255.255.255.192 is represented in binary as
11111111.11111111.11111111.11000000
DIT
27
26
25
24
23
22
21
20
128 64 32 16 8 4 2 1

25. DIT
IP Address Ranges
The graphic below shows the IP address range of the
first octet both in decimal and binary for each IP
address class.

26. DIT
Finding the Network Address with ANDing
By ANDing the Host address of 192.168.10.2 with 255.255.255.0
(its network mask) we obtain the network address of 192.168.10.0

27. DIT
Network Address

28. DIT
Public IP Addresses
Unique addresses are required for each device on a network.
Originally, an organization known as the Internet Network
Information Center (InterNIC) handled this procedure.
InterNIC no longer exists and has been succeeded by the
Internet Assigned Numbers Authority (IANA).
No two machines that connect to a public network can have the
same IP address because public IP addresses are global and
standardized.
All machines connected to the Internet agree to conform to the
system.
Public IP addresses must be obtained from an Internet service
provider (ISP) or a registry at some expense.

29. DIT
Private IP Addresses
Private IP addresses are another solution to the problem of
the impending exhaustion of public IP addresses. As
mentioned, public networks require hosts to have unique IP
addresses.
However, private networks that are not connected to the
Internet may use any host addresses, as long as each host
within the private network is unique.

30. DIT
Introduction to Subnetting
Subnetting a network means to use the subnet mask to
divide the network and break a large network up into
smaller, more efficient and manageable segments, or
subnets.
With subnetting, the network is not limited to the default,
Class A, B, or C network masks and there is more
flexibility in the network design.
Subnet addresses include the network portion, plus a
subnet field and a host field. The ability to decide how to
divide the original host portion into the new subnet and
host fields provides addressing flexibility for the network
administrator.

31. DIT
The 32-Bit Binary IP Address

32. DIT
Numbers That Show Up In Subnet Masks
(Memorize Them!)

33. DIT
Addressing with Subnetworks

34. Subnet Example 1
The DIT has purchased the class C address 216.21.5.0 and want to
use it for five (5) networks.
Determine the number of networks and convert to binary
5 in binary is 00000101
We need to borrow 3 bits from host and use them as network bits
Reserve bits in subnet mask and find your increment
Subnet mask for class C
255.255.255.0 = 11111111.11111111.11111111.00000000
The new subnet mask for class C (Add 3 bits in host octet)
255.255.255.224 = 11111111.11111111.11111111.11100000
The increment is 100000 = 32
Use increment to find your network ranges
216.21.5.0 – 216.21.5.31, 216.21.5.32 – 216.21.5.63
216.21.2.64 – 216.21.5.95 ….. 216.21.5.192 – 216.21.5.223
DIT

35. Subnet Example 2
The DIT has purchased the class C address 195.5.20.0 and want to
use it for fifty (50) networks.
Determine the number of networks and convert to binary
50 in binary is 00110010
We need to borrow 6 bits from host and use them as network bits
Reserve bits in subnet mask and find your increment
Subnet mask for class C
255.255.255.0 = 11111111.11111111.11111111.00000000
The new subnet mask for class C (Add 6 bits in host octet)
255.255.255.252 = 11111111.11111111.11111111.11111100
The increment is 100 = 4
Use increment to find your network ranges
195.5.20.0 – 195.5.20.3, 195.5.20.4 – 195.5.20.7
195.5.20.8 – 195.5.20.11 …. 195.5.20.248 – 195.5.20.251
DIT

36. Subnet Example 3
The DIT has purchased the class B address 150.5.0.0 and want to
use it for 100 networks.
Determine the number of networks and convert to binary
100 in binary is 01100100
We need to borrow 7 bits from host and use them as network bits
Reserve bits in subnet mask and find your increment
Subnet mask for class B
255.255.0.0 = 11111111.11111111.00000000.00000000
The new subnet mask for class B (Add 7 bits in host octet)
255.255.254.0 = 11111111.11111111.11111110.00000000
The increment is 10 = 2
Use increment to find your network ranges
150.5.0.0 – 150.5.1.255, 150.5.2.0 – 150.5.3.255
150.5.4.0 – 150.5.5.255, …. 150.5.252.0 – 150.5.253.255
DIT

37. Subnet Example 4
The DIT has purchased the class A address 10.0.0.0 and want to
use it for 500 networks.
Determine the number of networks and convert to binary
500 in binary is 0111110100
We need to borrow 9 bits from host and use them as network bits
Reserve bits in subnet mask and find your increment
Subnet mask for class A
255.0.0.0 = 11111111.00000000.00000000.00000000
The new subnet mask for class A (Add 9 bits in host octet)
255.255.128.0 = 11111111.11111111.10000000.00000000
The increment is 10000000 = 128
Use increment to find your network ranges
10.0.0.0 – 10.0.127.255, 10.0.128.0 – 10.0.255.255
10.1.0.0 – 10.1.127.255, 10.1.128.0 – 10.1.255.255 ……..
10.254.128.0 – 10.254.255.255DIT

38. Exercise 1
1. (C) 200.1.1.0, Break into 40 networks
2. (C) 199.9.10.0, Break into 14 networks
3. (B) 170.50.0.0, Break into 1000 networks
4. (A) 12.0.0.0, Break into 25 networks
Also determine the total number of hosts per
networks
DIT

39. Host Example 1
DIT has purchased class C address 216.21.5.0 and would like to
use it to create networks of 30 hosts each
Determine the number of hosts and convert to binary
30 in binary is 00011110
We need to save 5 bits for host, use 3 bits remain for network
Reserve bits in subnet mask and find your increment
Subnet mask for class C
255.255.255.0 = 11111111.11111111.11111111.00000000
The new subnet mask for class C (Add 3 bits in host octet)
255.255.255.224 = 11111111.11111111.11111111.11100000
The increment is 100000 = 32
Use increment to find your network ranges
216.21.5.0 – 216.21.5.31, 216.21.5.32 – 216.21.5.63
216.21.2.64 – 216.21.5.95 ….. 216.21.5.224 – 216.21.5.255
DIT

40. Host Example 2
The DIT has purchased the class C address 195.5.20.0 and want to
use it for 50 hosts each.
Determine the number of hosts and convert to binary
50 in binary is 00110010
We need to save 6 bits from host, use 2 bits remain for network
Reserve bits in subnet mask and find your increment
Subnet mask for class C
255.255.255.0 = 11111111.11111111.11111111.00000000
The new subnet mask for class C (Add 2 bits in host octet)
255.255.255.192 = 11111111.11111111.11111111.11000000
The increment is 1000000 = 64
Use increment to find your network ranges
195.5.20.0 – 195.5.20.63, 195.5.20.64 – 195.5.20.127
195.5.20.128 – 195.5.20.191, 195.5.20.192 – 195.5.20.255
DIT

41. Host Example 3
The DIT has purchased the class B address 150.5.0.0 and want to
use it for 500 hosts each.
Determine the number of hosts and convert to binary
500 in binary is 0111110100
We need to save 9 bits from host, use 7 bits remain for network
Reserve bits in subnet mask and find your increment
Subnet mask for class B
255.255.0.0 = 11111111.11111111.00000000.00000000
The new subnet mask for class B (Add 7 bits in host octet)
255.255.254.0 = 11111111.11111111.11111110.00000000
The increment is 10 = 2
Use increment to find your network ranges
150.5.0.0 – 150.5.1.255, 150.5.2.0 – 150.5.3.255
150.5.4.0 – 150.5.5.255, …. 150.5.252.0 – 150.5.253.255
DIT

42. Host Example 4
The DIT has purchased the class A address 10.0.0.0 and want to
use it for 100 networks.
Determine the number of hosts and convert to binary
100 in binary is 01100100
We need to save 7 bits from host, use 17 bits remain for network
Reserve bits in subnet mask and find your increment
Subnet mask for class A
255.0.0.0 = 11111111.00000000.00000000.00000000
The new subnet mask for class A (Add 17 bits in host octet)
255.255.225.128 = 11111111.11111111.10000000.00000000
The increment is 10000000 = 128
Use increment to find your network ranges
10.0.0.0 – 10.0.127.255, 10.0.128.0 – 10.0.255.255
10.1.0.0 – 10.1.127.255, 10.1.128.0 – 10.1.255.255 ……..
10.254.128.0 – 10.254.255.255DIT

43. Exercise 2
1. (C) 200.1.1.0, Break into networks of 40 hosts
each
2. (C) 199.9.10.0, Break into networks of 12
hosts each
3. (B) 170.50.0.0, Break into networks of 1000
hosts each
4. (A) 12.0.0.0, Break into networks of 100 hosts
each
Also determine the total number of networks
DIT

44. Exercise 3
1) The host has an IP and mask address of
192.168.1.127 and 255.255.255.224
respectively. What is the network address of
the host, and state that if the IP address of the
host is assigned correct.
2) The host has an IP, mask and gateway
address of 172.16.68.65, 255.255.255.240
and 172.16.68.62 respectively, is connected to
the network router of IP and mask address of
172.16.68.62 and 255.255.255.240 respectively.
Determine that if the above configuration is correct.
DIT

45. DIT
Why not IPv4
 IPv4 has been extremely successful.
 It is beginning to show its age as the internet
grows.
 In order to meet the challenges of the rapidly
growing internet, new features and scalability
measures will be needed.
 IPv6 is an evolutionary step to the current
IPv4.
 It uses the best of IPv4 and takes into
account all of the lessons that have been
learned over the years of its use.

46. DIT
IPv6 Development History
 1991: Work starts on next generation Internet
protocols
 More than 6 different proposals were developed
 1993: IETF forms IPng Directorate
 To select the new protocol by consensus
 1995: IPv6 selected
 Evolutionary (not revolutionary) step from IPv4
 1996: 6Bone started
 1998: IPv6 standardized
 Today: Initial products and deployments

47. Necessity for IPv6 Addresses
 32-bit addressing structure of IPv4 provides only
4,294,967,296 IP numbers
 In order to use this address space more efficiently,
technologies such as CIDR, DHCP, Pvt IP, NAT etc.
were developed
 These interim solutions helped only to postpone
exhaustion of IPv4 address space.
 Exponential growth of Internet, Wireless Subscribers and
deployment of NGN Technology etc. demand still a large
amount of address space
 IPv6 is meticulously designed to correct some problems
of IPv4 and to provide various enhancements with respect
to security, routing addresses, auto configuration, mobility
and Quality of Service (QoS) etc.
DIT

48. DIT
Larger IP Address Space
 IPv4 address space is 32 bits long
 4,294,967,296 possible hosts
 New types of devices need to be addressed
 Mobile/wireless devices
 Desktop devices
 NAT works, but is not ideal
 IPv6 address space is 128 bits long
 340,282,366,920,938,463,463,374,607,431,768,211,456 possible
hosts
= 67 billion billion addresses per cm2
of the planet surface
 End-to-end addressing
 No need for Network Address Translation (NAT)

49. DIT
New Rationale behind IPv6
 IP Everywhere for Data, Voice, Audio, Video
integration
 ~300 millions Mobile Phone Users in 1998, 700 millions
in 2004
 3G will support IP
 1 billion Cars in 2010 with GPS & Yellow Page services
 PDA’s, Toaster’s, Fridges, ...
 Emerging Internet Countries
 China, India, Russia, …
 Internet in every school,...
 New Technologies/Applications for Home users
 Cable, xDSL, Wireless,...

50. IPv6 deployment
 The existing pool of IPv4 addresses is expected to be
exhausted by August-2012
 All service providers and other stakeholders will gradually
transit to IPv6 in a phased manner
 The co-existence of IPv4 & Ipv6 will be there for some
more years to come.
 There are 2 operating situations –
(a) IPv6 nodes have to communicate with IPv4 nodes.
This problem is solved using Dual Stack technique.
(b) Isolated islands of IPv6 will have to communicate
with each other using the widely available IPv4
networks. This problem is solved using Tunneling
technique.
DIT

51. DIT
IPv6 Main Features
 Expanded Address Space
 Header Format Simplification
 Improved host and router discovery
 Auto-configuration
 Multi-Homing
 Class of Service/Multimedia support
 Improved Mobile IP support
 Authentication and Privacy Capabilities
 No more broadcast  Multicast
 Anycast

52. IPv6 Address Syntax
 IPv6 address in binary form
0010000000000001000011011011100000000000000000000010111100111011
0000001010101010000000001111111111111110001010001001110001011010
 Divided along 16-bit boundaries
0010000000000001 0000110110111000 0000000000000000 0010111100111011
0000001010101010 0000000011111111 1111111000101000 1001110001011010
 Each 16-bit block is converted to hexadecimal
and delimited with colons
2001:0DB8:0000:2F3B:02AA:00FF:FE28:9C5A
 Suppress leading zeros within each block
2001:DB8:0:2F3B:2AA:FF:FE28:9C5A
DIT

53. DIT
IPv6 Packet Header
 IPv4 header fields
are very detailed.
 Some of the
information is rarely
used or poorly
defined.
 Example: Type of
Service
 Other information is
no longer needed.
 Example: Header
Checksum
 IPv6 has a
simplified header
with only the
minimum number of
necessary fields.
IPv4 Header
IPv6 Header
IHLIHL Type of ServiceType of Service
OptionsOptions
Total LengthTotal Length
IdentificationIdentification FlagsFlags FragmentFragment OffsetOffset
ProtocolProtocol Header ChecksumHeader Checksum
Source Address
Destination Address
PaddingPadding
TrafficTraffic ClassClass Flow LabelFlow Label
Payload LengthPayload Length Next HeaderNext Header Hop LimitHop Limit
Source AddressSource Address
Destination AddressDestination Address
VersionVersion
Time to LiveTime to Live
VersionVersion

54. IPv6 Packet Header (2)
 Version - A 4-bit field, set to the number six for
IPv6
 Traffic Class - Also called priority. Similar to the
type of service (ToS) field in IPv4, this 8-bit field
describes relative priority and is used for quality of
service (QoS)
 Flow Label - The 20-bit flow label allows traffic to
be tagged so that it can be handled faster, on a
per-flow basis; this field can also be used to
associate flows with traffic classes
 Payload Length - This 16-bit field is the length of
the data in the packet.
DIT

55. IPv6 Packet Header (3)
 Next Header - Like the protocol field in the IPv4 header,
this 8-bit field indicates how the fields after the IPv6 basic
header should be interpreted.
It could indicate that the following field is (TCP) or
(UDP) transport layer information, or it could indicate
that an extension header is present.
 Hop Limit - Similar to the time to live (TTL) field of IPv4,
this 8-bit field is decremented by intermediate routers and,
to prevent looping, the packet is discarded and a message
is sent back to the source if this field reaches zero
 Source Address and Destination Address - These 128-
bit fields are the IPv6 source and destination addresses of
the communicating devices.
DIT

56. DIT
IPv6 Addressing
 IPv6 Addressing rules are covered by multiples
RFC’s
 Architecture defined by RFC 2373
 Address Types are :
 Unicast : One to One (Global, Link local, Site local,
Compatible)
 Anycast : One to Nearest (Allocated from Unicast)
 Multicast : One to Many
 No Broadcast Address -> Use Multicast
 Reserved
 A single interface may be assigned multiple IPv6
addresses of any type (unicast, anycast, multicast)

57. DIT
IPv6 Addressing (Cont.)
 Prefix Format (PF) Allocation
 PF = 0000 0000 : Reserved
 PF = 0000 001 : Reserved for OSI NSAP Allocation (see
RFC 1888), so far only way to embedded E.164
addresses (VoIP)
 PF = 0000 010 : Reserved for IPX Allocation (under
Study)
 PF = 001 : Aggregatable Global Unicast Address
 PF = 1111 1110 10 : Link Local Use Addresses
 PF = 1111 1110 11 : Site Local Use Addresses
 PF = 1111 1111 : Multicast Addresses
 Other values are currently Unassigned (approx. 7/8th of
total)
 All Prefix Formats have to have EUI-64 bits Interface ID
 But Multicast

58. DIT
IPv6 Addressing Examples
 Global unicast address(es) is :
 2001:304:101:1::E0:F726:4E58,
 subnet is 2001:304:101:1::0/64
 link-local address is FE80::E0:F726:4E58
 Unspecified Address is 0:0:0:0:0:0:0:0 or ::
 Loopback Address is 0:0:0:0:0:0:0:1 or ::1
 Group Addresses (Multicast) is FF02::9 for RIPv6

59. DIT
Text Representation of IPv6
Addresses
 “Preferred” form:
1080:0:FF:0:8:800:200C:417A
 Compressed form: FF01:0:0:0:0:0:0:43
becomes FF01::43
 IPv4-compatible: 0:0:0:0:0:0:13.1.68.3
or ::13.1.68.3
 RFC 2732: Preferred format for literal IPv6
address in URL

60. DIT
Benefits of IPv6 Addresses
 Enough for stable, unique addresses for all
devices
 Note: stable does not mean permanent!
 Allow continued growth of the Internet (for
centuries to come)
 Restore end-to-end transparency of the Internet
 Additional benefits:
 Plug-and-play (no need for configuration servers)
 Verifiable end-to-end packet integrity (no need for
NATs)
 Simpler mobility (no need for “foreign agent”
function)

61. DIT
Configuring Interface IDs
 There are several choices for configuring the
interface ID of an address:
 Manual configuration
 DHCPv6 (configures whole address)
 Automatic derivation from MAC address or other
hardware serial number
 Pseudo-random generation (for client privacy)
 The latter two choices enable “serverless” or
“stateless” autoconfiguration, when combined
with high-order part of the address learned
via Router Advertisements

62. EUI-64 Interface ID Example
Host A has the MAC address of 00-AA-00-3F-2A-1C
1. Convert MAC address to EUI-64 format
00-AA-00-FF-FE-3F-2A-1C
2. Complement the U/L bit (seventh bit of first byte)
The first byte in binary form is 00000000. When
the seventh bit is complemented, it becomes
00000010 (0x02).
02-AA-00-FF-FE-3F-2A-1C
3. Convert to colon hexadecimal notation
::2AA:FF:FE3F:2A1C
Link-local address for node with the MAC address of 00-
AA00-3F-2A-1C is FE80::2AA:FF:FE3F:2A1C
DIT

63. DIT
An example of IPv6 addresses
LAN: 3ffe:b00:c18:1::/64
Ethernet0
MAC address: 0060.3e47.1530
router# show ipv6 interface Ethernet0
Ethernet0 is up, line protocol is up
IPv6 is enabled, link-local address is FE80::260:3EFF:FE47:1530
Global unicast address(es):
2001:410:213:1:260:3EFF:FE47:1530, subnet is 2001:410:213:1::/64
Joined group address(es):
FF02::1:FF47:1530
FF02::1
FF02::2
MTU is 1500 bytes
interface Ethernet0
ipv6 address 2001:410:213:1::/64 eui-64

64. DIT
IPv4-IPv6 Co-Existence/Transition
 A wide range of techniques have been identified and
implemented, basically falling into three categories:
 Dual-stack techniques, to allow IPv4 and IPv6 to co-
exist in the same devices and networks
 Tunneling techniques, to avoid order dependencies
when upgrading hosts, routers, or regions
 Translation techniques, to allow IPv6-only devices to
communicate with IPv4-only devices
 Expect all of these to be used, in combination
 RFC 2893, Transition Mechanisms for IPv6 Hosts
and Routers, August 2000.

65. DIT
Native IPv6-Only Backbone
 Requires:
 IPv4 over IPv6
Tunnels for
IPv4 traffic
 Hardware
forwarding for
IPv6
 Network
management
over IPv6
IPv6 Intranet
IPv4 Tunnel
IPv4/v6 Intranet
Mobile IPv6
IPv4 Intranet
IPv6 Intranet
Translating
Gateway
Translating
Gateway
IPv6 Backbone

66. DIT
Dual Stack IPv4-IPv6 Backbone

67. DIT
Diversity of Today's Available Mobile
Devices

68. DIT
Thanks!
Technology changes but communication lasts.