· Web view172.16.255.255, which is all subnets and hosts on network 17.16.0.0 10.255.255.255, which broadcasts to all subnets and hosts on Network 10.0.0.0. The Hierarchical IP Addressing

An IP address is a numeric identifier assigned to each machine on an IP network.

IP TermsBit (Binary digit) either a 1 or a 0.

Byte = 8 bits.

Octet - Always 8 bits. Base-8 addressing scheme.

Network address The designation used in routing to send packets to a remote network,

Ex.10.0.0.0, 172.16.0.0, and 192.168.10.0.

Broadcast address Used by applications and hosts to send information to all nodes on a

Network.

Ex. 255.255.255.255, which is all networks, all nodes;

172.16.255.255, which is all subnets and hosts on network 17.16.0.0

10.255.255.255, which broadcasts to all subnets and hosts on

Network 10.0.0.0.

The Hierarchical IP Addressing Scheme

An IP address consists of 32 bits of information. These bits are divided into four sections, referred to as octets or bytes, each containing 1 byte (8 bits).

You can depict an IP address using one of three methods:

1) Dotted-decimal, as in 172.16.30.56

2) Binary, as in 10101100.00010000.00011110.00111000

3) Hexadecimal, as in 82 39 1E 38

IMP Points (About IP Address)

1) Software address or logical address

2) Combination of Network Address & Host Address

3) Consists of 32 Bits

8bit . 8bit . 8bit . 8bit

xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx

00000000.00000000.00000000.00000000 = 0.0.0.0

11111111.11111111.11111111.11111111 = 255.255.255.255

Dotted Binary Dotted Decimal

Classes of Networks

8 bits 8 bits 8 bits 8 bits

Class A Network Host Host Host

Class B Network Network Host Host

Class C Network Network Network Host

Class D Multicast

Class E Research

FIGURE: Classes of Networks

Network Address Range: Class A

(To calculate n/w address range)

Rule: The first bit of the first byte must always be 0.

Class A Network Host Host Host

0xxxxxxx

00000000=0 (Turn the other 7 bits all off)

01111111=127 (Turn the other 7 bits all on)

So, a Class A network range in the first octet between 0 and 127.

IP address range 0.0.0.0

127.255.255.255

Class B

Rule: The first two bit of the first byte must always be 1 0

Class B Network Network Host Host

10xxxxxx



Class B network address range is from 128 to 191 in first octet.


191.255.255.255

Class C

Rule: The first three bit of the first byte must always be 1 1 0

Class C Network Network Network Host

110xxxxx

11 0 00000=192 (Turn the other 5 bits all off)

11 0 11111=223 (Turn the other 5 bits all on)

Class C network address range is from 192 to 223 in first octet.


223.255.255.255

Class D

Rule: The first four bit of the first byte must always be 1 1 1 0

Class D 8bit 8bit 8bit 8bit

1110xxxx



Class D network address range is from 224 to 239 in first octet.


239.255.255.255

Class E

Rule: The first four bit of the first byte must always be 1 1 1 1

Class E 8bit 8bit 8bit 8bit

1111xxxx



Class E network address range is from 240 to 255 in first octet.


255.255.255.255

Network Addresses: Special PurposeSome IP addresses are reserved for special purposes, and network administrators shouldn’t assign these addresses to hosts.

FUNCTION ADDRESS

Network address of all 0s Mean “this network or segment.”

Network address of all 1s Mean “all network

Host address of all 0s Mean “this host”

Host address of all 1s Mean “all host”

Network 127.0.0.1 Reserved for loopback test

Entire IP address set to all 0s ( 0.0.0.0)

Used by Cisco routers to define the default route.

Entire IP address set to all 1s (255.255.255.255) Broadcast

TABLE: Reserved IP Addresses

Actual Range of IP Addresses1)Class A

Network

Host Host Host

0.0.0.0

127.255.255.255

Network Address Range

Earlier Range is 0-127

But, 0 is reserved to designate the default route and

127 is reserved for loopback test.

So, the actual range of network addresses 1 to 126

Host Address Range

Earlier Range is 0.0.0

255.255.255

But, host address of all 0s and all 1s are reserved,

So, the actual range of host addresses 0.0.1

255.255.254

Actual IP Address Range

1.0.0.1

126.255.255.254

2)Class B

Network

Network

Host Host

128.0.0.0

192.255.255.255

Network Address Range

Earlier Range is 128.0

192.255

But, Nothing is reserved

So, the actual range of network addresses 128.0

192.255

Host Address Range


255.255


So, the actual range of host addresses 0.1

255.254


128.0.0.1

191.255.255.254

3)Class C

Network

Network

Network

Host

192.0.0.0

223.255.255.255

Network Address RangeEarlier Range is 128.0.0

192.255.255

But, nothing is reserved

So, the actual range of network addresses 192.0.0

223.255.255

Host Address Range


255.255


So, the actual range of host addresses 1 - 254


192.0.0.1

223.255.255.254

Class Format No. of

N/W Bits

No. of

Host Bits

No. of Reserved

Bits

N/W Address Range

Host Address Range

IP Address Range

No. of Networks

No. of Hosts

No. of IP Addresses

A N.H.H.H 8 24 1(0) 1-126 0.0.1 -255.255.25

1.0.0.1 - 126 224 -2=16777214

126*16777214

4 126.255.255.254

B N.N.H.H 16 16 2(10) 128.0 -191.255

0.1 – 255.254

128.0.0.1 -191.255.255.254

16384 216 -2=65534 16384*65534

C N.N.N.H 24 8 3(110) 192.0.0 -223.255.255

1 - 254 192.0.0.1 -223.255.255.254

2097150 28-2=254 2097150*254

SubnettingA one larger network breaks into many smaller networks is known as subneting.

Benefits of Subnetting:

1) Simplified management

It’s easier to identify and isolate network problems in a group of smaller connected networks than within one gigantic network

2) Optimized network performance

This is a result of reduced network traffic.

3) Reduced network traffic

The smaller broadcast domains you create the less network traffic on that network segment.

Subnet MasksA subnet mask is a 32-bit value that allows the recipient of IP packets to distinguish the network

ID portion of the IP address from the host ID portion of the IP address.

Class Format Default Subnet Mask

A N.H.H.H 255.0.0.0

B N.N.H.H 255.255.0.0

C N.N.N.H 255.255.255.0

TABLE: Default Subnet Mask

Note: You can define any subnet mask below any class range.

Subnetting Class C Addresses

In a Class C address, only 8 bits is available for defining the hosts.

Rule:

1) Subnet bits start at the left and go to the right, without skipping any single bit. This means that subnet masks can be

XXXXXXXX

10000000=128 Illegal /25

11000000=192 /26

11100000=224 /27

11110000=240 /28

11111000=248 /29

11111100=252 /30

11111110=254 Illegal /31

2) You cannot have only single bit for subnetting, which would be illegal.

Then valid subnet masks are

255.255.255.192

255.255.255.224

255.255.255.240

255.255.255.248

255.255.255.252

A] The Binary Method:

Subnetting a Class C Address

Ex. 255.255.255.192.

192=11000000

2 bits for subnetting,

6 bits for defining the hosts in each subnet.

Rule: The subnet bits can’t be both off or on at the same time,

00000000=0 (all host bits off) Illegal (According to rule)

01000000=64 (all host bits off)

10000000=128 (all host bits off)

11000000=192 (all host bits off) Illegal (According to rule)

Then valid subnets are 0,64,128,192

TABLE: Subnet 64

Subnet Bit Host Bit Meaning

00 000000=0 The Network Address

00 000001=1 First Valid Host

00 111110=62 Last Valid Host

00 111111=63 The Broadcast Address

TABLE: Subnet 64

TABLE: Subnet 64






TABLE: Subnet 128






TABLE: Subnet 192






B] Alternate Method:

Subnetting a Class C Address

To solve this, you need to do answer only four questions:

Questions:

1. How many subnets?

2. What are the valid subnets?

3. How many hosts? (In each subnet)

4. What are the valid hosts? (In each subnet)

Answers:

1. No. of Subnets =2x

where X = No. of subnet bits

2. 256–Subnet Mask=Base Number (First valid subnet)

Keep adding the base no. to itself until you reach the subnet mask

i.e. Add On

3. No. of Hosts = 2y–2

where Y = No. of Host bits

4. Valid hosts are the numbers between the subnets, minus all 0s and all

1s.

Ex. Class C Addresses

1] 255.255.255.192

192 = 11000000

No. of subnet bits X=2, No. of host bits Y=6

1) No. of subnets = 22

= 4

2) 256–192=64, (first valid subnet) base number

0.0+64=64,64+64=128. 128+64=192

Then valid subnets are 0.64,128 and 192

3) No. of hosts (per subnet) = 26–2

= 62

4) Valid Hosts

2] 255.255.255.224

224= 111 00000



= 8


Then valid subnets are 0,32, 32+32=64, 64+32=96, 96+32=128,

128+32=160, 160+32=192, 192+64=224


= 30

4) Valid Hosts

3] 255.255.255.240

240= 1111 0000



= 16


Then valid subnets are

0,16. 16+16=32. 32+16=48. 48+16=64. 64+16=80. 80+16=96. 96+16=112. 112+16=128. 128+16=144. 144+16=160. 160+16=176. 176+16=192. 192+16=208. 208+16=224. 224+16=240


= 14

4) Valid Hosts

4] 255.255.255.128

Earlier stated that only one subnet bit was illegal and not to use it.

Breaking the rule, this mask can be used when you need two subnets, each with 126 hosts.

The standard questions don’t work here,

To use this, use ip subnet-zero command on the global configuration mode to tell your router to break the rules and use a 1-bit subnet mask.

128 = 1000000

No. of subnet bits X=1, i.e. either 0 or 1

Therefore valid subnets are

00000000 = 0

10000000 = 128

And No. of host bits Y=7

Then Valid Hosts are in each subnet

Subnetting Class B Addresses

In a Class B address, 16 bits is available for defining the hosts.

Net.Net.Host.Host Subnet Mask Bits

10000000.00000000=255.255.128.0 Illegal /17

11000000.00000000=255.255.192.0 /18

11100000.00000000=255.255.224.0 /19

11110000.00000000=255.255.240.0 /20

11111000.00000000=255.255.248.0 /21

11111100.00000000=255.255.252.0 /22

11111110.00000000=255.255.254.0 /23

11111111.00000000=255.255.255.0 /24

11111111.10000000=255.255.255.128 /25

11111111.11000000=255.255.255.192 /26

11111111.11100000=255.255.255.224 /27

11111111.11110000=255.255.255.240 /28

11111111.11111000=255.255.255.248 /29

11111111.11111100=255.255.255.252 /30

11111111.11111110=255.255.255.254 Illegal /31

11111111.11111111=255.255.255.255 Broadcast /32

Ex: Class B Addresses

1] 255.255.192.0

192.0 = 11000000.00000000



= 4


64+64=128. 128+64=192 (Invalid-it is subnet mask),

Then valid subnets are 0.0,64.0,128.0 and 192.0


= 16382

4) Valid Hosts

2] 255.255.240.0

240.0 = 11110000.00000000



= 16


Then valid subnets are 0.0,16.0, 32.0,48.0 and so on up to 224.0


= 4096

4) Valid Hosts (for first 3 subnets)

3] 255.255.255.192

255.192 = 11111111.11000000

1. 210=1024 subnets.

2. 256–192=64 and 128. However, as long as all the subnet bits on the

third are not all off, then subnet 0 in the fourth octet is valid. Also, as

long as all the subnet bits in the third octet are not all on, 192 is valid

in the fourth octet as a subnet.

The subnets are 0.64, 0.128, 0.192, 1.64, 1.128, and 1.192 …

3. 26–2=62 hosts.

Subnetting Class A Addresses

Net.Host.Host.Host Subnet Mask Bits

10000000.00000000.00000000=255.128.0.0 Illegal /9

11000000.00000000.00000000=255.192.0.0 /10

11100000.00000000.00000000=255.224.0.0 /11

11110000.00000000.00000000=255.240.0.0 /12

11111000.00000000.00000000=255.248.0.0 /13

11111100.00000000.00000000=255.252.0.0 /14

11111110.00000000.00000000=255.254.0.0 /15

11111111.00000000.00000000=255.255.0.0 /16

11111111.10000000.00000000=255.255.128.0 /17

11111111.11000000.00000000=255.255.192.0 /18

11111111.11100000.00000000=255.255.224.0 /19

11111111.11110000.00000000=255.255.240.0 /20

11111111.11111000.00000000=255.255.248.0 /21

11111111.11111100.00000000=255.255.252.0 /22

11111111.11111110.00000000=255.255.254.0 /23

11111111.11111111.00000000=255.255.255.0 /24

11111111.11111111.10000000=255.255.255.128 /25

11111111.11111111.11000000=255.255.255.192 /26

11111111.11111111.11100000=255.255.255.224 /27

11111111.11111111.11110000=255.255.255.240 /28

11111111.11111111.11111000=255.255.255.248 /29

11111111.11111111.11111100=255.255.255.252 /30

11111111.11111111.11111110=255.255.255.254 Illegal /31

11111111.11111111.11111111=255.255.255.255 B.C /32

Ex. Class A Addresses

1] 255.255.0.0

255.0.0= 111111111.00000000.00000000


1) No. of subnets = 28 – 2

= 254

2) 256–255=1 (first valid subnet) base number

1, 2, 3, 4, 5, 6……………254.

255 (Invalid-it is subnet mask),

Then Valid subnets are 1.0.0, 2.0.0 and so on

3) No. of hosts (per subnet) = 216 –2

= 65534

4) Valid Hosts (for first & last subnet)

*********************************************************************************

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· Web view172.16.255.255, which is all subnets and hosts on network 17.16.0.0 10.255.255.255, which broadcasts to all subnets and hosts on Network 10.0.0.0. The Hierarchical IP Addressing