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CCNA-Notes/14. Subnetting (Part 2).md

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Subnetting (Part 2)

Problem

You have a network topology where each segment requires 45 hosts:

45 hosts -> SW1 -> R1 <- SW3 <- 45 hosts
45 hosts -> SW2 -> R1 <- SW4 <- 45 hosts

You are given the network:

  • 192.168.1.0/24

Your task is to divide it into 4 subnets, each capable of supporting at least 45 hosts.


Step 1: Choose the Right Subnet Size

To support 45 hosts, we need:

  • ( 2^n - 2 \geq 45 )
  • ( n = 6 ) → ( 2^6 - 2 = 62 ) hosts

So each subnet must be a /26.


Step 2: Subnet Breakdown (/26)

Each subnet increases by 64 addresses.


Subnet 1

  • Network: 192.168.1.0/26
  • Range: 192.168.1.0 192.168.1.63
  • Broadcast: 192.168.1.63

Binary:

11000000.10101000.00000001.00 000000

Subnet 2

  • Network: 192.168.1.64/26
  • Range: 192.168.1.64 192.168.1.127
  • Broadcast: 192.168.1.127

Subnet 3

  • Network: 192.168.1.128/26
  • Range: 192.168.1.128 192.168.1.191
  • Broadcast: 192.168.1.191

Subnet 4

  • Network: 192.168.1.192/26
  • Range: 192.168.1.192 192.168.1.255
  • Broadcast: 192.168.1.255

Quick Trick 💡

Each subnet jumps by 64:

  • 0 → 64 → 128 → 192

Think of it like stepping stones across a river. Same stride, different landing spots.


Subnetting Basics

  • Number of subnets: ( 2^x ) (x = borrowed bits)

  • Number of hosts per subnet: ( 2^n - 2 ) (n = host bits)


Identify the Subnet

Example 1

Host: 192.168.5.57/27

  • /27 → block size = 32
  • Subnets: 0, 32, 64, ...

57 falls between 32 and 63

Network: 192.168.5.32/27


Example 2

Host: 192.168.29.219/29

  • /29 → block size = 8
  • Subnets: 0, 8, 16, ..., 216, 224

219 falls between 216 and 223

Network: 192.168.29.216/29


Class C Reference Table

Prefix Subnets Hosts
/25 2 126
/26 4 62
/27 8 30
/28 16 14
/29 32 6
/30 64 2

Subnetting Class B Networks

The method is exactly the same. Only the starting mask changes.


Example 1

Network: 172.16.0.0/16 Required subnets: 80

  • ( 2^7 = 128 ) → enough
  • New prefix: /23

Example 2

Network: 172.22.0.0/16 Required subnets: 500

  • ( 2^9 = 512 )
  • New prefix: /25

Example 3

Network: 172.18.0.0/16 Required subnets: 250

  • ( 2^8 = 256 )
  • New prefix: /24

Class B Reference Table

Prefix Subnets Hosts
/17 2 32766
/18 4 16382
/19 8 8190
/20 16 4094
/21 32 2046
/22 64 1022
/23 128 510
/24 256 254

Final Review

  • Subnetting is about borrowing bits
  • Larger prefix = more subnets, fewer hosts
  • Smaller prefix = fewer subnets, more hosts