This presentation of CyberLab explains how Transmission Control Protocol(TCP) Internet Protocol(IP) works and it also describes what is IP addressing and it's various classes. at the end of Presentation an overview of subnatting also given.
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Internet Layer - IP Addresses
Hierarchical addressing scheme
32 bit address
Dotted decimal notation
Used for addressing and routing
Divided into 2 main sections
Network identity
Host identity
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IP version 4
4 bytes divided into 4 sections
Each section can range from 0 to 255
If any portion of the IP address contains a number
greater than 255 it is ILEGAL !
Presented with dots in between 4 bytes
e.g. 193.60.61.243
11000001.00111100.00111101.11110011
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For subnetting use BINARY
Always work in Binary
All solutions can be discovered in Binary
Eventually you will see shortcuts
Use Binary first
It will take time to understand
Practice makes perfect
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Address Classes (a history
lesson)
Classes A, B, C, D, E were created
IP numbers that began with :
1 to 126 are class A addresses
128 to 191 are class B addresses
192 to 223 are class C addresses
224 to 239 are class D addresses
240 to 254 are class E addresses.
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Exercise 1 – ‘class’
identification
To which class do the following IP numbers legally belong ?
193.45.67.123
22.23.24.25
131.74.59.12
210.217.12,98
193.260.50.17
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The end of Address class
system
Classes wasted addresses through inflexibility
A better way of dividing network space was sought
This was known as Variable Length Subnet Masking or
VLSM
Now we have to explicitly show the boundary of the
host and network portion of an address
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Caveat to classless addressing
Many people and textbooks still refer to the address class system
No networks EVER use the class system in a production environment
Be aware of the translations between the classful and classless system
Remember ALWAYS to use the classless system despite the terminology
of the question
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Subnet Mask identifies the
boundary
Sometimes called the Extended Network Prefix
Identifies the two sections of the IP address, network
and host fields
Used by routers to work out the network a particular
host belongs to
Written either as dotted decimal or slash notation
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Subnet Mask – two notations
Dotted decimal is one of the notations
255.255.255.0
In binary this is
11111111.11111111.11111111.00000000
Slash is the second notation
Count the number of ‘1’s
There are 24 ‘1’s
So another way of expressing this subnet
mask is /24
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Network (or wire) Address
This is the identity of the network and is used by routers
to deliver packets across networks
No host is ever given this address
To discover the network address, convert IP address from
decimal to binary
Replace the host section with 0s
Convert back to decimal
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Finding the Network Identity
175.13.155.121 /16
10101111.00001101. 10011011. 01111001
What address class was this ?
Which is the host field ?
Replace with 0s
Convert back to dotted decimal
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Broadcast Address RFC 919
Each network needs a broadcast address
No host is ever given this address
A broadcast can message all hosts in a LAN
Useful when a host needs to find information without
knowing exactly what other host can supply it e.g. ARP
When a host wants to provide information to a large set
of hosts in a timely manner
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Finding the Broadcast
Number
Substitute the host section of the binary version of the
IP address with 1s
Convert back to dotted decimal
What is the broadcast address for the IP
175.13.155.121 /16
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Reserved addressing numbers
Host bits set to 0 are Network address
Host bits set to 1 are Broadcast address
127 network numbers reserved for loopback
127.0.0.1 is ‘my NIC’
192.168.x.x & 172.16-31.x.x & 10.x.x.x are non-
routable private addresses – for NAT and private
networks only
defined in RFC 1918
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Host numbers on networks
Original assignment of classes wasted many IP addresses
( 3 bears problem )
Class A for governments with 224 hosts
Class B can have 65 534 hosts
more than most companies require
Class C can have 254 hosts
less than most companies require
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Subnetting
Take a /24 (class C) address
24 network bits & 8 host bits
Company may need different departments
We can borrow bits from host field to augment network field
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Subnetting /24 (Class C)
address
e.g. 198.234.125.0
11000110.11101010.01111101.00000000
Write network number
10111100.11101010.00000000.00000000
Borrow 4 bits from host field
10111100.11101010.00000000.00000000
4 bits remain for host identity
30/0
9/20
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Subnetted /24 (Class C)
How many hosts does this allow on each subnet ?
How many subnetworks can we use ?
First and last subnet numbers are reserved
First and last host numbers are reserved
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More Subnetting
Can borrow 2 host bits minimum
Maximum host bits that can be borrowed is all but the
rightmost 2 bits of the IP address
Why is this so ?
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Calculating subnet mask
1. Express the subnetwork IP address in binary form
2. Replace the network and subnet portion of the address
with all 1s
3. Replace the host portion of the address with all 0s
4. Now convert the binary expression back to dotted-
decimal notation
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Example of subnet mask
Calculate the subnet mask for the Subnetted
class C address 199.177.166.34 that has
borrowed 3 bits from the host field
11000111.10110001.10100110.00100010
27 bits for network identity - substitute these
27 bits for 1s
Change remaining 5 host bits to 0s
Convert to dotted decimal
255.255.255.224
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Exercise 3
Calculate subnet masks for class B that has borrowed 4
bits from host field
155.233.2.13
Calculate subnet masks for class C that has borrowed 5
bits from host field
200.123.23.3
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Use of Subnet Mask
Mask is ANDed with IP address of host
This gives network identity
Can now be used to route the message
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Example
What is the network number for the IP address
199.177.166.34 given that the subnet mask is
255.255.255.224
Convert to binary
AND both numbers together
Result is network identity
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Exercise 3
IP network address 213.72.83.0. You require at
least 17 subnets. Calculate
1. The required subnet mask
2. The number of total subnets
3. The number of available host IP per subnet
4. The first four subnet addresses
5. The valid range of host addresses for the first
four subnets
6. The Broadcast address for each subnet
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Conclusion
3 Classes available, A, B, C
Strictly not used any longer but well understood
Flexibility gained by subnetting
Subnets concealed from outside networks using subnet masks
Masks are ANDed with IP address of host to discover network identity
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Hosts on class A
There are a possible 224 = 16 777 216 combinations
available for the last 24 bits
Subtract the two combinations that cannot be used as
host addresses
A class A address can have 16 777 216 - 2 = 16 777 214
different hosts
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Hosts on class B
There are a possible 216 = 65 536
combinations available for the last 16
bits
Subtract the two combinations that
cannot be used as host addresses
A class B address can have
65 536 - 2 = 65 534 different hosts
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Hosts on class C
There are a possible 28 = 256
combinations available for the last 8
bits
Subtract the two combinations that
cannot be used as host addresses
A class B address can have
256 - 2 = 254 different hosts
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Classes & numbers
12.123.14.0 12.123.14.255 (looks like Class A)
177.177.0.0 177.177.255.255 (Class B)
220.17.124.0 220.17.124.255 (Class C)
277.14.13.86 Illegal number
And illegal subnet mask
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Exercise 3
Need 17 subnets. Calculate bits required for 17 subnets
= 5
Leaves 3 bits for host identity on subnets
When calculating the first four subnet addresses, you
will begin with:
00000000 = 0
00001000 = 8
00010000 = 16
00011000 = 24
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Exercise 3 contd
3 bits for the host leaves 23 - 2 = 6 host addresses per
subnet
Valid host range
213.72.83.1 213.72.83.2
213.72.83.3 213.72.83.4
213.72.83.5 213.72.83.6
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