Addressing IP v4 W.Lilakiatsakun.

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Presentation transcript:

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 (&) As a result, X & 0 – 0 X & 1 – X 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

Network Address = IP address & Netmask Example: IP address 192.168.10.20 /28 Network Address = IP address & Netmask 192.168.10.20 & 255.255.255.240 11000000 .10101000.00001010.00010100 11111111. 11111111.11111111.11110000 11000000 .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 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