Addressing IP v4 W.Lilakiatsakun. Anatomy of IPv4 (1) Dotted Decimal Address Network Address Host Address.

Addressing IP v4

W.Lilakiatsakun

Anatomy of IPv4 (1)

Dotted Decimal Address

Network Address

Host Address

Anatomy of IPv4 (2)

Octet

32 Bit Address

Binary to Decimal Conversion (1)

Binary to Decimal Conversion (2)

Decimal to Binary Conversion (1)

Decimal to Binary Conversion (2)

Decimal to Binary Conversion (3)

Type of IPv4 Address-Network Address

Type of IPv4 Address – Broadcast Address

Type of IPv4 Address – Host Address

Netmask (1)

• To identify the bit for the same position in the IP address whether it is Network bit or Host bit

• If the value of Netmask is– 1 is the Network bit– 0 is the Host bit

• Example : IP address – 192.168.10.10 Netmask- 255.255.255.0 – Network bit = the most 24 significant bits – Host bit = the least 8 significant bits

Netmask (2)

• Netmask can be represented as “prefix” (slash format)

• The number of 1 is counted – Netmask is 255.255.255.0 = /24– 255 = 1111 1111 (8 bits)

• We can find Network Address (Network ID) from IP Address

• Network Address = IP adresss & Netmask

Netmask (3)

• Logic AND (&)– 0 & 0 – 0 – 0 & 1 – 0– 1 & 0 – 0– 1 & 1 – 1

As a result,• X & 0 – 0• X & 1 – X

• Logic OR (|)– 0 & 0 – 0 – 0 & 1 – 1– 1 & 0 – 1– 1 & 1 – 1

As a result,• X | 1 – 1• X | 0 – X

Netmask (4)

• Example: IP address 192.168.10.20 /28Network Address = IP address & Netmask

192.168.10.20 & 255.255.255.240 11000000 .10101000.00001010.0001010011111111. 11111111.11111111.1111000011000000 .10101000.00001010.00010000

• Question: IP address = 192.168.10.20 /22 • Network ID = ?

Netmask (5)

Network Address & Broadcast Address

Network ID = IP address that all host bits are 0s– 192.168.10.20/24 – Network ID = 192.168.10.0

Broadcast ID = IP address that all host bits are 1s– 192.168.10.20/24– Broadcast ID = 192.168.10.255

Different Prefixes (1)

Different Prefixes (2)

Assigning Addresses

Calculating Addresses

Quiz

Basic Subnetting (1)

Basic Subnetting (2)

Basic Subnetting (3)

Basic Subnetting (4)

Basic Subnetting (5)

Basic Subnetting (6)

VLSM (Subnetting a subnet)

Problem

• This scenario has the following requirements:– IP address : 192.168.15.0/24– AtlantaHQ 58 host addresses– PerthHQ 26 host addresses– SydneyHQ 10 host addresses– CorpusHQ 10 host addresses– WAN links 2 host addresses (3 Links)

Answer