5.2 KiB
id, aliases, tags
| id | aliases | tags | |
|---|---|---|---|
| 1777871422-VTLQ |
|
Subnetting (Part 2)
Question
45 hosts -> SW1->+--+<-SW3<-45 hosts |R1| 45 hosts -> SW2->+--+<-SW4<-45 hosts
Divide the 192.168.1.0/24 network into four subnets that can accommodate the number of hosts required.
The first subnet (Subnet 1) is 192.168.1.0/26 What are the remaining subnet
Hint: Find the broadcast address of Subnet 1. The next address is the network address of Subnet 2. Repeat the process for Subnets 3 and 4
Subnet 1
Subnet 1: 192.168.1.0/26 11000000.10101000.00000001.00000000 192 . 168 . 1 . 0 broadcast address: 11000000.10101000.00000001.00111111 192 . 168 . 1 . 63
192.168.1.0 to 192.168.1.63
Subnet 2
Subnet 2: 192.168.1.64/26 11000000.10101000.00000001.01000000 192 . 168 . 1 . 64 11000000.10101000.00000001.01111111 192 . 168 . 1 . 127
192.168.1.63 to 192.168.1.127
Subnet 3
Subnet 3: 192.168.1.128/26 11000000.10101000.00000001.10000000 192 . 168 . 1 . 128 11000000.10101000.00000001.10111111 192 . 168 . 1 . 191
192.168.1.128 to 192.168.1.191
Subnet 4
Subnet 4: 192.168.1.192/26 11000000.10101000.00000001.11000000 192 . 168 . 1 . 192 11000000.10101000.00000001.11111111 192 . 168 . 1 . 255
192.168.1.192 to 192.168.1.255
Trick
By adding 64 the last bit of the network portion you can find the network address
Subnetting
11000000.10101000.00000001.00000000 192.168.255.0
11000000.10101000.11111111.10000000 192.168.255.0 Borrowing 1 bit = can make 2 subnets
Formulas
2^x = number of subnets (x=number of 'borrowed' bits)
2^n-2 = number of hosts (x=number of hosts bits)
Identify the subnet
What subnet does host 192.168.5.57/27 belong to?
11000000.10101000.00000101.00111001 192.168.5.57
The borrowed bits are 11000000.10101000.00000101.00111001
to find the network address we need to change the host bits by zero
11000000.10101000.00000101.00100000 192.168.5.32
What subnet does host 192.168.29.219/29 belong to?
11000000.10101000.00011101.11011011 192.168.29.219
to find the network address we need to change the host bits by zero 11000000.10101000.00011101.11011000 192.168.29.216
Subnets / hosts (class C)
| Prefix Length | Number of subnets | Number of Hosts |
|---|---|---|
| /25 | 2 | 126 |
| /26 | 4 | 62 |
| /27 | 8 | 30 |
| /28 | 16 | 14 |
| /29 | 32 | 6 |
| /30 | 64 | 2 |
| /31 | 128 | 0 (2) |
| /32 | 156 | 0 (1) |
Subenetting Class B
The process of subnetting Class A, Class B and Class C networks is : Exactly the same !
Example 1
You have been given the 172.16.0.0/16 network. You are asked to create 80 subnets for your company's various LANs, What prefix length should you use ?
172.16.0.0/16
10101100.00010000.00000000.00000000 172 . 16 . 0 . 0
We have to borrow 7 bits = 128 subnets
10101100.00010000.00000000.00000000 172 . 16 . 0 . 0
Example 2
You have been given the 172.22.0.0/16 network. You are asked to create 500 subnets for your company's various LANs, What prefix length should you use ?
172.22.0.0/16
10101100.00010000.00000000.00000000 172 . 22 . 0 . 0
We have to borrow 9 bits = 512 subnets = /25
10101100.00010000.00000000.00000000 172 . 22 . 0 . 0
Example 3
You have given the 172.18.0.0/16 network. Your company requires 250 subnets with the same number of hosts per subnet. What prefix length should you use
172.18.0.0
10101100.00010000.00000000.00000000 172 . 18 . 0 . 0
We have to borrow 8 bits = 256 subnets = /24
10101100.00010000.00000000.00000000 172 . 18 . 0 . 0
Subnets / hosts (class B)
| Prefix Length | Number of subnets | Number of Hosts |
|---|---|---|
| /17 | 2 | 32766 |
| /18 | 4 | 16382 |
| /19 | 8 | 8190 |
| /20 | 16 | 4094 |
| /21 | 32 | 2044 |
| /22 | 64 | 1022 |
| /23 | 128 | 510 |
| /24 | 256 | 254 |
| /25 | 2 | 126 |
| /26 | 4 | 62 |
| /27 | 8 | 30 |
| /28 | 16 | 14 |
| /29 | 32 | 6 |
| /30 | 64 | 2 |
| /31 | 128 | 0 (2) |
| /32 | 156 | 0 (1) |
Review
- Subnetting practice questions (class C networks)
- Subnetting Class B networks