CyberLab TCP/IP and IP Addressing & Subnetting

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Introduction to
Computer
Networking
IP Addressing and subnetting

In This Session. . . .
 IP addressing
 Reserved addresses
 Network identity
 Host identity
 Subnetworking
 Subnet masks
2

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
3

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
4

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
5

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.
6

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
7

8
Class C addresses
 First 3 bytes is network portion
 Last byte for hosts on network
 How many hosts can class C have ?
C

Class B addresses
 First 2 bytes is network portion
 Last 2 bytes for hosts on network
 How many hosts can class B have ?
9
B

Class A addresses
 First byte is network portion
 last 3 bytes for hosts on network
 How many hosts can class A have ?
10
A

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
11

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
12

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
13

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
14

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
15

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
16

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
17

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
18

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
19

Exercise 2
 Write the network & broadcast number
12.123.14.235 /24
177.177.177.177 /16
220.17.124.14 /24
277.14.13.86 /33
20

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
21

Subnetworking
 Solution is to subdivide networks
 Take a network number
 Divide it into smaller networks
 These are called subnetworks
(subnets)
22

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
23

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
24

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
25

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 ?
26

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
27

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
28

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
29

Use of Subnet Mask
 Mask is ANDed with IP address of host
 This gives network identity
 Can now be used to route the message
30

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
31

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
32

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
33

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
34

35
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

36
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

Network Address
 10101111.00001101. 00000000.00000000
 175.13.0.0
 It means ‘ this network ’
37

Broadcast number
 10101111.00001101. 11111111.11111111
 175.13.255.255
 Reaches all hosts on a network
 Will not reach other hosts on different network
38

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
39

Subnetted Class C
 24 = 16 -2 = 14 hosts
 24 = 16 subnetworks
 Note that 14 hosts x 16 subnets is less than 256
 Why is this ?
40

Subnet masks
11111111.11111111.11110000.00000000
255.255.240.0
11111111.11111111.11111111.11111000
255.255.255.248
41

Network mask discovery
Convert to dotted decimal
199.177.166.32
How many hosts on this subnet ?
42

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
43

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
44

Thank You

CyberLab TCP/IP and IP Addressing & Subnetting

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