ipand subnet

[email protected] 1

Insight IP and Subnet

I – IPv4

II – Subnet

[email protected] 2

I – IPv4

[email protected] 3

[email protected] 4

Application

PresentationSession

Transport

Network

Data link

Physical

IP – Internet Protocol

OSI

[email protected] 5

- 32- bit and consists of 2 parts:

- Write it down: 4 blocks of 8 bits

IPv4

Network Host

8-bit 8-bit 8-bit 8-bit

192. 168. 1. 1.

1100000

10101000

0000001

00000001

[email protected] 6

Class A:

5 classes of IPv4

0xxxxxx

Host Host Host IP range: 0.0.0.0 – 127.255.255.255

Reserved: 10.0.0.0 – 10.255.255.255 (Private IP)

Localhost: 127.0.0.0 - 127.255.255.255

[email protected] 7

Class B:

5 classes of IPv4

10xxxxx

Network Host Host

IP range: 128.0.0.0 - 191.255.255.255

Reserved: 172.16.0.0 - 172.31.255.255 (Private IP)

[email protected] 8

Class C:

5 classes of IPv4

110xxxx

Network

Network Host

IP range: 192.0.0.0 - 223.255.255.255

Reserved: 192.168.0.0 - 192.168.255.255 (Private IP)

[email protected] 9

Class D:

5 classes of IPv4

IP range: 224.0.0.0 - 239.255.255.255

1110xxx

Network

Network Host

[email protected] 10

Class E:

5 classes of IPv4

IP range: 240.0.0.0 - 255.255.255.255

http://tools.ietf.org/html/rfc6890


Write down start and end IP addresses which use to assign to PC for class A, B and C.

Challenge 1


- Class A: x.0.0.1 – x.255.255.254- Class B: x.x.0.1 – x.x.255.254- Class C: x.x.x.1 – x.x.x.254

Challenge 1 - Solution


Network address: when we set all bit of our host part to 0.-Class A: 0.0.0.0- Class B: 128.0.0.0- Class C: 192.0.0.0

Broadcast Address: when we set all bit of our host part to 1.- Class A: 126.255.255.255- Class B: 191.255.255.255- Class C: 223.255.255.255

Network/Broadcast Address


II - SUBNET


How to plan IP for this network???


Has 3 networks need 2 more bits for Network part

192.168.1.XXYYYYYY/26

IP Planning - Solution 1 VLMS: Variable Length Subnet Mask

XX Network address Broadcast Address

00 192.168.1.0 192.168.1.63

01 192.168.1.64 192.168.1.127

10 192.168.1.128 192.168.1.191

11 192.168.1.192 192.168.1.255


60 hosts need 6 bits at host part (2^6 = 64) 192.168.1.0/26 28 hosts need 5 bits at host part (2^5=32)192.168.1.0/27 2 router need 2 bits at host part (2^2 = 4)192.168.1.0/30

IP Planning - Solution 2VLMS: Variable Length Subnet Mask


Write down:• Network IP,• Start IP,• End IP,• Broadcast IP,for each network.

Challenge 2


For network: 192.168.1.0/26:Last block: 00XX XXXX• Network IP: 192.168.1.0• Start IP: 192.168.1.1• End IP: 192.168.1.62• Broadcast IP: 192.168.1.63

Challenge 2 Solution

ipand subnet

Technology

broadcast ip

xxxx network ip

insight ip

xxxxxx host host host

bits ipv4 network host

classes of ipv4 ip range

class b

ip planning solution