This document discusses IP addressing and subnetting. It begins by explaining IPv4 addressing in terms of its 32-bit structure and common address classes. It then covers subnetting, giving an example of how to plan an IP scheme for a network with 3 subnets and varying host requirements using variable length subnet masking. It also provides the network address, start address, end address and broadcast address for each subnet in the example IP plan.

1. Insight IP and Subnet
I – IPv4
II – Subnet
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2. I – IPv4
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4. Application
Presentation
Session
Transport
Network
Data link
Physical
IP – Internet Protocol
OSI
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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
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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
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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)
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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)
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9. Class D:
5 classes of IPv4
 IP range: 224.0.0.0 - 239.255.255.255
1110xxx Network Network Host
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10. Class E:
5 classes of IPv4
 IP range: 240.0.0.0 - 255.255.255.255
 http://tools.ietf.org/html/rfc6890
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11. Write down start and end IP
addresses which use to assign
to PC for class A, B and C.
Challenge 1
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12. - 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
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13. 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
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14. II - SUBNET
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15. How to plan IP for this network???
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16.  Has 3 networks  need 2 more bits for Network part
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
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17.  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 2
VLMS: Variable Length Subnet Mask
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18.  Write down:
• Network IP,
• Start IP,
• End IP,
• Broadcast IP,
for each network.
Challenge 2
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19. 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
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