Update 14. Subnetting (Part 2).md
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id: 1777871422-VTLQ
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aliases:
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- Subnetting (Part 2)
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tags: []
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---
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# Subnetting (Part 2)
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# Subnetting (Part 2)
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## Question
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## Problem
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45 hosts -> SW1->+--+<-SW3<-45 hosts
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You have a network topology where each segment requires **45 hosts**:
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|R1|
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45 hosts -> SW2->+--+<-SW4<-45 hosts
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Divide the 192.168.1.0/24 network into four subnets that can accommodate
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```
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the number of hosts required.
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45 hosts -> SW1 -> R1 <- SW3 <- 45 hosts
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45 hosts -> SW2 -> R1 <- SW4 <- 45 hosts
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```
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The first subnet (Subnet 1) is 192.168.1.0/26 What are the remaining subnet
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You are given the network:
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Hint: Find the broadcast address of Subnet 1. The next address is the network address of
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* **192.168.1.0/24**
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Subnet 2. Repeat the process for Subnets 3 and 4
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Your task is to divide it into **4 subnets**, each capable of supporting at least 45 hosts.
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---
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## Step 1: Choose the Right Subnet Size
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To support **45 hosts**, we need:
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* ( 2^n - 2 \geq 45 )
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* ( n = 6 ) → ( 2^6 - 2 = 62 ) hosts
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So each subnet must be a **/26**.
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---
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## Step 2: Subnet Breakdown (/26)
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Each subnet increases by **64 addresses**.
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---
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### Subnet 1
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### Subnet 1
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Subnet 1: 192.168.1.0/26
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* **Network:** 192.168.1.0/26
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11000000.10101000.00000001.00**000000**
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* **Range:** 192.168.1.0 – 192.168.1.63
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192 . 168 . 1 . 0
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* **Broadcast:** 192.168.1.63
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broadcast address:
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11000000.10101000.00000001.00**111111**
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192 . 168 . 1 . 63
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192.168.1.0 to 192.168.1.63
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Binary:
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```
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11000000.10101000.00000001.00 000000
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```
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---
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### Subnet 2
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### Subnet 2
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Subnet 2: 192.168.1.64/26
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* **Network:** 192.168.1.64/26
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11000000.10101000.00000001.01**000000**
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* **Range:** 192.168.1.64 – 192.168.1.127
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192 . 168 . 1 . 64
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* **Broadcast:** 192.168.1.127
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11000000.10101000.00000001.01**111111**
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192 . 168 . 1 . 127
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192.168.1.63 to 192.168.1.127
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---
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### Subnet 3
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### Subnet 3
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Subnet 3: 192.168.1.128/26
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* **Network:** 192.168.1.128/26
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11000000.10101000.00000001.10**000000**
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* **Range:** 192.168.1.128 – 192.168.1.191
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192 . 168 . 1 . 128
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* **Broadcast:** 192.168.1.191
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11000000.10101000.00000001.10**111111**
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192 . 168 . 1 . 191
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192.168.1.128 to 192.168.1.191
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---
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### Subnet 4
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### Subnet 4
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Subnet 4: 192.168.1.192/26
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* **Network:** 192.168.1.192/26
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11000000.10101000.00000001.11**000000**
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* **Range:** 192.168.1.192 – 192.168.1.255
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192 . 168 . 1 . 192
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* **Broadcast:** 192.168.1.255
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11000000.10101000.00000001.11**111111**
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192 . 168 . 1 . 255
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192.168.1.192 to 192.168.1.255
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---
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### Trick
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## Quick Trick 💡
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By adding 64 the last bit of the network portion you can find the network address
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Each subnet jumps by **64**:
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## Subnetting
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* 0 → 64 → 128 → 192
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11000000.10101000.00000001.0**0000000**
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Think of it like stepping stones across a river. Same stride, different landing spots.
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192.168.255.0
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11000000.10101000.11111111.1**0000000**
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---
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192.168.255.0
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Borrowing 1 bit = can make 2 subnets
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### Formulas
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## Subnetting Basics
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2^x = number of subnets
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* **Number of subnets:**
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(x=number of 'borrowed' bits)
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( 2^x ) (x = borrowed bits)
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2^n-2 = number of hosts
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* **Number of hosts per subnet:**
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(x=number of hosts bits)
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( 2^n - 2 ) (n = host bits)
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## Identify the subnet
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---
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What subnet does host 192.168.5.57/27 belong to?
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## Identify the Subnet
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11000000.10101000.00000101.00111001
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192.168.5.57
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The borrowed bits are
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11000000.10101000.00000101.**001**11001
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to find the network address we need to change the host bits by zero
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11000000.10101000.00000101.**001**00000
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192.168.5.32
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What subnet does host 192.168.29.219/29 belong to?
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11000000.10101000.00011101.**11011**011
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192.168.29.219
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to find the network address we need to change the host bits by zero
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11000000.10101000.00011101.**11011**000
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192.168.29.216
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## Subnets / hosts (class C)
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| Prefix Length | Number of subnets | Number of Hosts |
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| ------------- | ----------------- | --------------- |
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| /25 | 2 | 126 |
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| /26 | 4 | 62 |
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| /27 | 8 | 30 |
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| /28 | 16 | 14 |
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| /29 | 32 | 6 |
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| /30 | 64 | 2 |
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| /31 | 128 | 0 (2) |
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| /32 | 156 | 0 (1) |
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## Subenetting Class B
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The process of subnetting Class A, Class B and Class C networks is :
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Exactly the same !
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### Example 1
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### Example 1
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You have been given the 172.16.0.0/16 network. You are asked to create 80
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**Host:** 192.168.5.57/27
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subnets for your company's various LANs, What prefix length should you use ?
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172.16.0.0/16
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* /27 → block size = 32
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* Subnets: 0, 32, 64, ...
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10101100.00010000.00000000.00000000
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57 falls between **32 and 63**
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172 . 16 . 0 . 0
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We have to borrow 7 bits = 128 subnets
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→ **Network:** 192.168.5.32/27
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10101100.00010000.**0000000**0.00000000
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---
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172 . 16 . 0 . 0
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### Example 2
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### Example 2
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You have been given the 172.22.0.0/16 network. You are asked to create 500
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**Host:** 192.168.29.219/29
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subnets for your company's various LANs, What prefix length should you use ?
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172.22.0.0/16
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* /29 → block size = 8
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* Subnets: 0, 8, 16, ..., 216, 224
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10101100.00010000.00000000.00000000
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219 falls between **216 and 223**
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172 . 22 . 0 . 0
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We have to borrow 9 bits = 512 subnets = /25
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→ **Network:** 192.168.29.216/29
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10101100.00010000.**00000000.0**0000000
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---
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172 . 22 . 0 . 0
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## Class C Reference Table
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| Prefix | Subnets | Hosts |
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| ------ | ------- | ----- |
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| /25 | 2 | 126 |
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| /26 | 4 | 62 |
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| /27 | 8 | 30 |
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| /28 | 16 | 14 |
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| /29 | 32 | 6 |
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| /30 | 64 | 2 |
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---
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## Subnetting Class B Networks
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The method is exactly the same. Only the starting mask changes.
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---
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### Example 1
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**Network:** 172.16.0.0/16
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**Required subnets:** 80
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* ( 2^7 = 128 ) → enough
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* New prefix: **/23**
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---
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### Example 2
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**Network:** 172.22.0.0/16
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**Required subnets:** 500
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* ( 2^9 = 512 )
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* New prefix: **/25**
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---
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### Example 3
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### Example 3
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You have given the 172.18.0.0/16 network. Your company requires 250 subnets with the
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**Network:** 172.18.0.0/16
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same number of hosts per subnet. What prefix length should you use
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**Required subnets:** 250
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172.18.0.0
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* ( 2^8 = 256 )
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* New prefix: **/24**
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10101100.00010000.00000000.00000000
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---
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172 . 18 . 0 . 0
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We have to borrow 8 bits = 256 subnets = /24
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## Class B Reference Table
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10101100.00010000.**00000000**.00000000
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| Prefix | Subnets | Hosts |
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172 . 18 . 0 . 0
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| ------ | ------- | ----- |
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| /17 | 2 | 32766 |
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| /18 | 4 | 16382 |
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| /19 | 8 | 8190 |
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| /20 | 16 | 4094 |
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| /21 | 32 | 2046 |
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| /22 | 64 | 1022 |
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| /23 | 128 | 510 |
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| /24 | 256 | 254 |
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## Subnets / hosts (class B)
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---
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## Final Review
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| Prefix Length | Number of subnets | Number of Hosts |
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* Subnetting is about **borrowing bits**
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| ------------- | ----------------- | --------------- |
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* Larger prefix = more subnets, fewer hosts
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| /17 | 2 | 32766 |
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* Smaller prefix = fewer subnets, more hosts
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| /18 | 4 | 16382 |
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| /19 | 8 | 8190 |
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| /20 | 16 | 4094 |
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| /21 | 32 | 2044 |
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| /22 | 64 | 1022 |
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| /23 | 128 | 510 |
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| /24 | 256 | 254 |
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| /25 | 2 | 126 |
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| /26 | 4 | 62 |
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| /27 | 8 | 30 |
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| /28 | 16 | 14 |
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| /29 | 32 | 6 |
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| /30 | 64 | 2 |
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| /31 | 128 | 0 (2) |
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| /32 | 156 | 0 (1) |
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## Review
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- Subnetting practice questions (class C networks)
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- Subnetting Class B networks
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