Digital Network - Lecturer2
Digital Network - Lecturer2
Digital Network - Lecturer2
(DIT)
ETU 08102
Digital Networks
Ally, J
jumannea@gmail.com
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IP Network
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InterNetwork
With a capital I
Defined as a network of networks
many pieces:
hosts
Question:
routers
Is
there
any
hope
of
links of various media
organizing structure
applications
of network?
protocols
hardware
software
ticket (complain)
baggage (check)
baggage (claim)
gates (load)
gates (unload)
runway takeoff
runway landing
airplane routing
airplane routing
airplane routing
A series of steps
Layering of Airline
Functionality
ticket (purchase)
ticket (complain)
ticket
baggage (check)
baggage (claim
baggage
gates (load)
gates (unload)
gate
runway (takeoff)
runway (land)
takeoff/landing
airplane routing
airplane routing
airplane routing
departure
airport
airplane routing
airplane routing
intermediate air-traffic
control centers
arrival
airport
TCP, UDP
application
transport
network
link
physical
Computer Comm.
Voice
Voice
Data
Data
Video
Video
(TCP
(TCP // UDP)
UDP)
SS-7
SS-7
IP
v4 // v6
v6
IP v4
Network Layer
ATM
ATM
Dedicated
Dedicated
SONET
SONET // SDH
SDH
Photonic
Photonic Network
Network Interface
Interface
Photonic
Photonic Network
Network (WDM)
(WDM)
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Data
Data
(TCP
(TCP // UDP)
UDP)
Network Layer
Video
Video
VOIP
IP
Dedicated
Dedicated
L3-VPN
L2-VPN
v4
v4 // v6
v6
as a standard interface
Data Link Layer
Physical Layer
SONET
ATM
SONET // SDH
SDH
ATM
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Structure of IP Network
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
3) TCP in end station control the quality and flow of the session
4) Applications in end stations communicate each other over TCP
Application session
Apl.
Apl.
Socket
TCP
TCP
Apl.
Apl.
TCP session
IP
IP
IP
IP
Router
Ethernet SW
Socket
TCP
TCP
IP
IP
IP
Router
End station
End station
Socket
Apl.
Apl.
Ethernet SW
Segment;
Ethernet SW
Ethernet Switching
Apl.
Apl.
Segment;
TCP
TCP
TCP
TCP
IP
IP
End station
Socket
Ethernet Switching
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End station
IP
IP
Evolution of IP Network
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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.
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IP Address Classes
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.
Address
Class
A
B
C
Number of
Networks
126
16,384
2,097,152
Number of Hosts
per Network
16,777,214
65,534
254
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Identifying Address
Classes
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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.
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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.
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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.
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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.
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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.
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26
25
24
23
22
21
20
128
64
32
16
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IP Address Ranges
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Network Address
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Public IP Addresses
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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.
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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.
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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
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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
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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
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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 10.254.255.255
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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
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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
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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
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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
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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 10.254.255.255
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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
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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.
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Mobile/wireless devices
Desktop devices
End-to-end addressing
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IPv6 deployment
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Other information is
no longer needed.
Example: Type of
Service
Example: Header
Checksum
IPv4 Header
Version IHL Type of Service
Flags
Identification
Time to Live
Total Length
Protocol
Fragment Offset
Header Checksum
Source Address
Destination Address
Options
Padding
IPv6 Header
Version Traffic Class
Payload Length
Flow Label
Next Header Hop Limit
Source Address
Destination Address
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IPv6 Addressing
IPv6 Addressing
Global unicast address(es)
is :
Examples
2001:304:101:1::E0:F726:4E58,
subnet is 2001:304:101:1::0/64
Preferred form:
1080:0:FF:0:8:800:200C:417A
IPv4-compatible: 0:0:0:0:0:0:13.1.68.3
or ::13.1.68.3
Benefits of IPv6
Enough for stable, unique addresses for all
Addresses
devices
Additional benefits:
Configuring Interface
There are severalIDs
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)
An example of IPv6
LAN: 3ffe:b00:c18:1::/64
addresses
Ethernet0
interface Ethernet0
ipv6 address 2001:410:213:1::/64 eui-64
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Dual-stack techniques, to allow IPv4 and IPv6 to coexist 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
Native IPv6-Only
Backbone
Requires:
IPv4 Intranet
Translating
Gateway
IPv6 Backbone
Translating
Gateway
Mobile IPv6
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IPv6 Intranet
IPv4/v6 Intranet
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Thanks!
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