This tutorial explains Subnetting tricks in detail with 50+ Subnetting examples. Subnetting tricks explained in this tutorial not only make Subnetting easier but also allow you to answer any Subnetting related question in less than a minute. Learn Subnetting with examples in easy language.
Subnetting is one of the most complex and highly tested topics in any Cisco entry level exam and interview. In other situation, you may use any regular method of Subnetting. But in exam and interview, where time and accuracy matter, you should always use the easiest and the fastest methods of Subnetting such as explained in this tutorial.
This tutorial is the fourth part of the article “IP Subnetting in Computer Network Step by Step Explained with Examples”. Other parts of this article are following.
Network Address Broadcast Address and IP Address Explained
This tutorial is the first part of the article. It explains IP addressing and network addressing such as IP address, subnet mask, IP address types and IP classes in detail.
Basic Subnetting in Computer Networks Explained
This tutorial is the second part of the article. It explains what Subnetting is and why it is necessary in computer network along with the advantages of Subnetting.
Subnetting Tutorial - Subnetting Explained with Examples
This tutorial is the third part of the article. It explains the Subnetting concepts and terms such as network id, broadcast id, total hosts, valid hosts, power of 2, block size and CIDR in detail.
VLSM Subnetting Explained with Examples
This tutorial is the fifth part of the article. It explains what VLSM Subnetting is and how it is done step by step including differences between FLSM Subnetting and VLSM Subnetting.
VLSM Subnetting Examples and Calculation Explained
This tutorial is the sixth part of the article. It explains VLSM Subnetting examples for Cisco exams and interviews.
Supernetting Tutorial: - Supernetting Explained with Examples
This tutorial is the last part of the article. It explains Supernetting in detail with examples.
Key points of Subnetting
Before we start working with Subnetting examples, let’s have a quick recap of important Subnetting concepts and terms from previous parts of this tutorial.
An IP address is the combination of two addresses; network address and host address. Network address is always written first in sequence. While reading an IP address how much portion should be treated as network address and how much portion should be treated as host address is decided by an another address known as Subnet mask.
An IP address is always used with Subnet mask. Without subnet mask an IP address is an ambiguous address and vice versa.
Both IP address and Subnet mask consists 32 bits. These bits are divided in four octets. Octets are separated by periods and written in a sequence.
A Subnet mask can be written in two ways; in full form and in abbreviated form. In full form a decimal value of each octet is written along with IP address. In abbreviated form only the number of network bits are written along with IP address. Following table lists some examples of both types.
IP address with full subnet mask | IP address with abbreviated subnet mask |
10.0.0.0 255.0.0.0 | 10.0.0.0/8 |
172.168.1.0 255.255.0.0 | 172.168.1.0/16 |
192.168.1.0 255.255.255.0 | 192.198.1.0/24 |
In Subnetting questions, mostly second type is used.
There are five IP classes A, B, C, D and E. From these, only first three classes A, B and C are used in Subnetting. Subnetting cannot be done in class D and E.
In class A, B and C first 8, 16 and 24 bits are reserved for network address respectively. In all three classes last 2 bits are reserved for host addresses.
If we exclude reserved network bits and host bits from total IP bits, we will get Subnetting eligible host bits.
Subnetting can be done only in Subnetting eligible host bits.
To know how many Subnetting bits are required to create how many networks, we use power of 2.
For example if we want to know many networks can be created from 3 Subnetting bits, we will use power 2 three times.
23 = 8
We can create 8 networks from 3 Subnetting host bits.
Power of 2
2X | Value | 2X | Value | 2X | Value | 2X | Value |
1 | 2 | 9 | 512 | 17 | 131072 | 25 | 33554432 |
2 | 4 | 10 | 1024 | 18 | 262144 | 26 | 67108864 |
3 | 8 | 11 | 2048 | 19 | 524288 | 27 | 134217728 |
4 | 16 | 12 | 4096 | 20 | 1048576 | 28 | 268435456 |
5 | 32 | 13 | 8192 | 21 | 2097152 | 29 | 536870912 |
6 | 64 | 14 | 16384 | 22 | 4194304 | 30 | 1073741824 |
7 | 128 | 15 | 32768 | 23 | 8388608 | 31 | 2147483648 |
8 | 256 | 16 | 65536 | 24 | 16777216 | 32 | 4294967296 |
In each network first address and last address are always reserved for network address and broadcast address respectively. Besides these two addresses, all remaining addresses are considered as valid host addresses.
Subnetting Questions
A Subnetting question can be asked in three ways.
- Find the number of networks and number of host addresses in each network
- Find the type of given address, network ID and broadcast ID
- Build the Subnet mask
Let’s understand the each type of Subnetting question in detail with examples.
Finding the number of networks and host addresses in each network
In this type of questions, we are asked to find the number of networks and host addresses in each network from a given address space. Let’s take some examples.
Following table lists 15 imaginary networks. Find the number of networks, total hosts and valid hosts in each network.
Example Subnetting Questions
10.0.0.0/10 | 130.0.0.0/18 | 192.168.1.0/26 |
20.12.0.0/13 | 140.50.60.0/20 | 200.0.0.0/27 |
78.59.12.0/16 | 172.168.1.0/24 | 210.200.0.0/28 |
112.15.0.0/24 | 180.10.20.0/28 | 215.0.0.0/29 |
122.14.25.0/28 | 185.0.0.0/30 | 220.220.10.0/30 |
Step 1
Determine the class of the given address space.
If the value of first octet is in range 0-127, 128-191, 192-223 then it belongs to class A, B and C respectively.
Subtract the given network bits from the reserved network bits. This will give us the number of host bits used as network bits in Subnetting.
Number of host bits used in Subnetting = Total bits used in network portion – reserved bits for network portion
Subtract the given network bits from 32 (total IP bits). This will give us the number of the host bits remaining in host portion.
Remaining host bits in host portion = 32 – Total bits used in network portion
Given Address space | Total bits used in network portion (Value after slash) | IP Class | Reserved network bits | Number of host bits used in Subnetting | Available bits in host portion |
10.0.0.0/10 | 10 | A | 8 | 10 - 8 = 2 | 32 – 10 = 22 |
20.12.0.0/13 | 13 | A | 8 | 13 – 8 = 5 | 32 – 13 = 19 |
78.59.12.0/16 | 16 | A | 8 | 16 – 8 = 8 | 32 – 16 = 16 |
112.15.0.0/24 | 24 | A | 8 | 24 – 8 = 16 | 32 – 24 = 8 |
122.14.25.0/28 | 28 | A | 8 | 28 – 8 = 20 | 32 – 28 = 4 |
130.0.0.0/18 | 18 | B | 16 | 18 – 16 = 2 | 32 – 18 = 14 |
140.50.60.0/20 | 20 | B | 16 | 20 – 16 = 4 | 32 – 20 = 12 |
172.168.1.0/24 | 24 | B | 16 | 24 – 16 = 8 | 32 – 24 = 8 |
180.10.20.0/28 | 28 | B | 16 | 28 – 16 = 12 | 32 – 28 = 4 |
185.0.0.0/30 | 30 | B | 16 | 30 – 16 = 14 | 32 – 30 = 2 |
192.168.1.0/26 | 26 | C | 24 | 26 - 24 = 2 | 32 – 26 = 6 |
200.0.0.0/27 | 27 | C | 24 | 27 – 24 = 3 | 32 – 27 = 5 |
210.200.0.0/28 | 28 | C | 24 | 28 – 24 = 4 | 32 – 28 = 4 |
215.0.0.0/29 | 29 | C | 24 | 29 – 24 = 5 | 32 – 29 = 3 |
220.220.10.0/30 | 30 | C | 24 | 30 – 24 = 6 | 32 – 30 = 2 |
Once we know how many bits are used in Subnetting to create the additional networks, we can use following formulas to calculate number of networks, total hosts and valid hosts.
Number of networks (subnet) = 2N Number of total host addresses in each network = 2H Number of valid host addresses in each network = 2H – 2
Here N is the number of host bits used in Subnetting and H is the available host bits.
Given Address | Host bits used in Subnetting | Available host bits | Networks (Subnets) | Total hosts | Valid hosts |
10.0.0.0/10 | 2 | 22 | 4 (22) | 4194304 (222) | 4194302 (4194304 – 2) |
20.12.0.0/13 | 5 | 19 | 32 (25) | 524288 (219) | 524286 (524288 - 2) |
78.59.12.0/16 | 8 | 16 | 256 (28) | 65536 (216) | 65534 (65536 - 2) |
112.15.0.0/24 | 16 | 8 | 65536 (216) | 256 (28) | 254 (256 - 2) |
122.14.25.0/28 | 20 | 4 | 1048576 (220) | 16 (24) | 14 (16 - 2) |
130.0.0.0/18 | 2 | 14 | 4 (22) | 16384 (214) | 16382 (16384 -2) |
140.50.60.0/20 | 4 | 12 | 16 (24) | 4096 (212) | 4094 (4096 -2) |
172.168.1.0/24 | 8 | 8 | 256 (28) | 256 (28) | 254 (256 -2) |
180.10.20.0/28 | 12 | 4 | 4096 (212) | 16 (24) | 14 (16 -2) |
185.0.0.0/30 | 14 | 2 | 16384 (214) | 4 (22) | 2 (4 -2) |
192.168.1.0/26 | 2 | 6 | 4 (22) | 64 (26) | 62 (64 -2) |
200.0.0.0/27 | 3 | 5 | 8 (23) | 32 (25) | 30 (32 -2) |
210.200.0.0/28 | 4 | 4 | 16 (24) | 16 (24) | 14 (16 - 2) |
215.0.0.0/29 | 5 | 3 | 32 (25) | 8 (23) | 6 (8 -2) |
220.220.10.0/30 | 6 | 2 | 64 (26) | 4 (22) | 2 (4 -2) |
Finding address type, network ID and broadcast ID
In this type of questions, we are asked to figure out the address type, network ID and broadcast ID from a given address. Let’s take some examples.
Find the address type, network ID and broadcast ID of following addresses.
Example Subnetting Questions
10.0.0.0/8 | 50.100.255.255/20 | 150.60.180.0/19 | 192.168.1.240/25 |
10.48.0.0/12 | 1.1.8.255/23 | 160.0.39.255/21 | 200.20.10.191/26 |
20.42.255.255/13 | 100.100.100.110/25 | 172.168.8.0/24 | 210.200.20.100/27 |
30.6.1.0/16 | 110.80.20.128/27 | 180.78.0.64/28 | 215.0.0.47/28 |
40.60.240.0/17 | 120.20.30.15/30 | 185.0.0.26/29 | 220.10.10.50/29 |
Step 1
Find the interesting octet and make two new addresses from the given address in rough area of worksheet as explained following: -
- If octet comes before the interesting octet, write it as it is.
- If octet comes after the interesting octet, in first address write 0 and in second address write 255.
- In interesting octet write a placeholder character “X”.
Interesting octet is the octet in which given network bits separate from host bits. To find it, see the value after slash. If value is in range 1-8, 9-16, 17-24 and 25-32 then the interesting octet is first octet, second octet, third octet and fourth octet respectively.
Given Address | Value after slash | Interesting octet | First address | Second Address |
10.0.0.0/8 | /8 | First (1-8) | X.0.0.0 | X.255.255.255 |
10.48.0.0/12 | /12 | Second (9-16) | 10.X.0.0 | 10.X.255.255 |
20.42.255.255/13 | /13 | Second (9-16) | 20.X.0.0 | 20.X.255.255 |
30.6.1.0/16 | /16 | Second (9-16) | 30.X.0.0 | 30.X.255.255 |
40.60.240.0/17 | /17 | Third (17-24) | 40.60.X.0 | 40.60.X.255 |
50.100.255.255/20 | /20 | Third (17 -24) | 50.100.X.0 | 50.100.X.255 |
1.1.8.255/23 | /23 | Third (17-24) | 1.1.X.0 | 1.1.X.255 |
100.100.100.110/25 | /25 | Fourth (25 -32) | 100.100.100.X | 100.100.100.X |
110.80.20.128/27 | /27 | Fourth (25-32) | 110.80.20.X | 110.80.20.X |
120.20.30.15/30 | /30 | Fourth (25-32) | 120.20.30.X | 120.20.30.X |
150.60.180.0/19 | /19 | Third (17-24) | 150.60.X.0 | 150.60.X.255 |
160.0.39.255/21 | /21 | Third (17-24) | 160.0.X.0 | 160.0.X.255 |
172.168.8.0/24 | /24 | Third (17-24) | 172.168.X.0 | 172.168.X.255 |
180.78.0.64/28 | /28 | Fourth (25-32) | 180.78.0.X | 180.78.0.X |
185.0.0.26/29 | /29 | Fourth (25-32) | 185.0.0.X | 185.0.0.X |
192.168.1.240/25 | /25 | Fourth (25-32) | 192.168.1.X | 192.168.1.X |
200.20.10.191/26 | /26 | Fourth (25-32) | 200.20.10.X | 200.20.10.X |
210.200.20.100/27 | /27 | Fourth (25-32) | 210.200.20.X | 210.200.20.X |
215.0.0.47/28 | /28 | Fourth (25-32) | 215.0.0.X | 215.0.0.X |
220.10.10.50/29 | /29 | Fourth (25-32) | 220.10.10.X | 220.10.10.X |
Step 2
Based on interesting octet, subtract after slash value form the upper value of octet range.
- Subtract it from 8 if interesting octet is the first octet.
- Subtract it from 16 if interesting octet is the second octet.
- Subtract it from 24 if interesting octet is the third octet.
- Subtract it from 32 if interesting octet is the fourth octet.
Given Address | Value after slash | Interesting octet | Upper value of range | Subtraction |
10.0.0.0/8 | /8 | First(1-8) | 8 | 8 – 8 = 0 |
10.48.0.0/12 | /12 | Second (9-16) | 16 | 16 – 12 = 4 |
20.42.255.255/13 | /13 | Second (9-16) | 16 | 16 – 13 = 3 |
30.6.1.0/16 | /16 | Second (9-16) | 16 | 16 – 16 = 0 |
40.60.240.0/17 | /17 | Third (17-24) | 24 | 24 – 17 = 7 |
50.100.255.255/20 | /20 | Third (17 -24) | 24 | 24 - 20 = 4 |
1.1.8.255/23 | /23 | Third (17-24) | 24 | 24 – 23 = 1 |
100.100.100.110/25 | /25 | Fourth (25 -32) | 32 | 32 – 25 = 7 |
110.80.20.128/27 | /27 | Fourth (25-32) | 32 | 32 – 27= 5 |
120.20.30.15/30 | /30 | Fourth (25-32) | 32 | 32 – 30 = 2 |
150.60.180.0/19 | /19 | Third (17-24) | 24 | 24 – 19 = 5 |
160.0.39.255/21 | /21 | Third (17-24) | 24 | 24 – 21 = 3 |
172.168.8.0/24 | /24 | Third (17-24) | 24 | 24 – 24 = 0 |
180.78.0.64/28 | /28 | Fourth (25-32) | 32 | 32 – 28 = 4 |
185.0.0.26/29 | /29 | Fourth (25-32) | 32 | 32 – 29 = 3 |
192.168.1.240/25 | /25 | Fourth (25-32) | 32 | 32 – 25 = 7 |
200.20.10.191/26 | /26 | Fourth (25-32) | 32 | 32 – 26 = 6 |
210.200.20.100/27 | /27 | Fourth (25-32) | 32 | 32 – 27 = 5 |
215.0.0.47/28 | /28 | Fourth (25-32) | 32 | 32 – 28 = 4 |
220.10.10.50/29 | /29 | Fourth (25-32) | 32 | 32 – 29 = 3 |
Step 3
Given Address | Interesting octet | Result of subtraction | Block Size |
10.0.0.0/8 | First | 0 | (20) = 1 |
10.48.0.0/12 | Second | 4 | (24) = 16 |
20.42.255.255/13 | Second | 3 | (23) = 8 |
30.6.1.0/16 | Second | 0 | (20) = 1 |
40.60.240.0/17 | Third | 7 | (27) = 128 |
50.100.255.255/20 | Third | 6 | (24) = 16 |
1.1.8.255/23 | Third | 1 | (21) = 2 |
100.100.100.110/25 | Fourth | 7 | (27) = 128 |
110.80.20.128/27 | Fourth | 5 | (25) = 32 |
120.20.30.15/30 | Fourth | 2 | (22) = 4 |
150.60.180.0/19 | Third | 5 | (25) = 32 |
160.0.39.255/21 | Third | 3 | (23) = 8 |
172.168.8.0/24 | Third | 0 | (20) = 1 |
180.78.0.64/28 | Fourth | 4 | (24 ) = 16 |
185.0.0.26/29 | Fourth | 3 | (23) = 8 |
192.168.1.240/25 | Fourth | 7 | (27) = 128 |
200.20.10.191/26 | Fourth | 6 | (26) = 64 |
210.200.20.100/27 | Fourth | 5 | (25) = 32 |
215.0.0.47/28 | Fourth | 4 | (24) = 16 |
220.10.10.50/29 | Fourth | 3 | (23) = 8 |
Step 4
Starting from 0, calculate the block size until the value given in interesting octet does not come in the range.
In calculation the number 0 is used as the first number. For example if we are calculating block size 4 then it would be calculated as 0,1,2,3 instead of 1,2,3,4.
Once you got the range in block size which cover the given value in interesting octet, stop the calculation. For example value in interesting octet is 27 and block size is 8 then the range will be 24-31 (0-7, 8-15, 16-23, 24-31).
Given Address | Interesting octet | Interesting octet value | Block Size | Range in block size which cover the value of interesting octet |
10.0.0.0/8 | First | 10 | 1 | 0-0, 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8, 9-9, 10-10 |
10.48.0.0/12 | Second | 48 | 16 | 0-15, 16-31, 32-47, 48-63 |
20.42.255.255/13 | Second | 42 | 8 | 0-7, 8-15, 16-23, 24-31, 32-39, 40-47 |
30.6.1.0/16 | Second | 6 | 1 | 0-0, 1-1, 2-2, 3-3, 4-4, 5-5, 6-6 |
40.60.240.0/17 | Third | 240 | 128 | 0-127, 128-255 |
50.100.255.255/20 | Third | 255 | 16 | 0-15, 16-31, 32-47,..... 224-239, 240-255 |
1.1.8.255/23 | Third | 8 | 2 | 0-1, 2-3, 4-5, 6-7, 8-9 |
100.100.100.110/25 | Fourth | 110 | 128 | 0-127,128-255 |
110.80.20.128/27 | Fourth | 128 | 32 | 0-31, 32-63, 64-95, 96 -127, 128-159 |
120.20.30.15/30 | Fourth | 15 | 4 | 0-3, 4-7, 8-11, 12-15 |
150.60.180.0/19 | Third | 180 | 32 | 0-31, 32-63, 64-95, 96 -127, 128-159, 160 - 191 |
160.0.39.255/21 | Third | 39 | 8 | 0-7, 8-15, 16-23, 24-31, 32-39 |
172.168.8.0/24 | Third | 8 | 1 | 0-0, 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8 |
180.78.0.64/28 | Fourth | 64 | 16 | 0-15, 16-31, 32-47, 48-63, 64-79 |
185.0.0.26/29 | Fourth | 26 | 8 | 0-7, 8-15, 16-23, 24-31 |
192.168.1.240/25 | Fourth | 240 | 128 | 0-127, 128-255 |
200.20.10.191/26 | Fourth | 191 | 64 | 0-63, 64-127, 128- 191 |
210.200.20.100/27 | Fourth | 100 | 32 | 0-31, 32-63, 64-95, 96 -127 |
215.0.0.47/28 | Fourth | 47 | 16 | 0-15, 16-31, 32-47 |
220.10.10.50/29 | Fourth | 50 | 8 | 0-7, 8-15, 16-23, 24-31, 32-39, 40-47, 48- 55 |
Step 5
Update the temporary addresses written in first step as following: -
- In first address, replace the placeholder character “X” with the starting value of the range.
- In second address, replace the placeholder character “X” with the ending value of the range.
After this update first address and second address will become the network address and broadcast address of the given address respectively.
Once it’s done, you can easily determine the type of the given address.
- If given address exactly matches with the network address then it’s a network address.
- If given address exactly matches with the broadcast address then it’s a broadcast address.
- If given address is neither a network address nor a broadcast address then it’s a valid host address.
Given Address | Interesting octet | Interesting octet value | Block range | Network address | Broadcast address | Type |
10.0.0.0/8 | First | 10 | 10-10 | 10.0.0.0 | 10.255.255.255 | Network |
10.48.0.0/12 | Second | 48 | 48-63 | 10.48.0.0 | 10.63.255.255 | Network |
20.42.255.255/13 | Second | 42 | 40-47 | 20.40.0.0 | 20.47.255.255 | Valid host |
30.6.1.0/16 | Second | 6 | 6-6 | 30.6.0.0 | 30.6.255.255 | Valid host |
40.60.240.0/17 | Third | 240 | 128-255 | 40.60.128.0 | 40.60.255.255 | Valid Host |
50.100.255.255/20 | Third | 255 | 240-255 | 50.100.240.0 | 50.100.255.255 | Broadcast |
1.1.8.255/23 | Third | 8 | 8-9 | 1.1.8.0 | 1.1.9.255 | Valid host |
100.100.100.110/25 | Fourth | 110 | 0-127 | 100.100.100.0 | 100.100.100.127 | Valid Host |
110.80.20.128/27 | Fourth | 128 | 128-159 | 110.80.20.128 | 110.80.20.159 | Network |
120.20.30.15/30 | Fourth | 15 | 12-15 | 120.20.30.12 | 120.20.30.15 | Broadcast |
150.60.180.0/19 | Third | 180 | 160 - 191 | 150.60.160.0 | 150.60.191.255 | Valid host |
160.0.39.255/21 | Third | 39 | 32-39 | 160.0.32.0 | 160.0.39.255 | Broadcast |
172.168.8.0/24 | Third | 8 | 8-8 | 172.168.8.0 | 172.168.8.255 | Network |
180.78.0.64/28 | Fourth | 64 | 64-79 | 180.78.0.64 | 180.78.0.79 | Network |
185.0.0.26/29 | Fourth | 26 | 24-31 | 185.0.0.24 | 185.0.0.31 | Valid host |
192.168.1.240/25 | Fourth | 240 | 128-255 | 192.168.1.128 | 192.168.1.255 | Valid host |
200.20.10.191/26 | Fourth | 191 | 128- 191 | 200.20.10.128 | 200.20.10.191 | Broadcast |
210.200.20.100/27 | Fourth | 100 | 96 -127 | 210.200.20.96 | 210.200.20.127 | Valid host |
215.0.0.47/28 | Fourth | 47 | 32-47 | 215.0.0.32 | 215.0.0.47 | Broadcast |
220.10.10.50/29 | Fourth | 50 | 48- 55 | 220.10.10.48 | 220.10.10.55 | valid host |
Building the Subnet mask value
In this type of questions, we are asked to build the full subnet mask from an abbreviated subnet mask. Let’s take some examples.
Find the full subnet mask of following networks.
20.10.30.0/8, 111.187.45.34/14, 162.160.46.24/20, 202.100.20.50/27
Step 1
Find the interesting octet as explained above.
Given network | Value after slash | Interesting octet |
20.10.30.0/8 | 8 | First |
111.187.45.34/14 | 14 | Second |
162.160.46.24/20 | 20 | Third |
202.100.20.50/27 | 27 | Fourth |
Step2
In rough area of worksheet, write a temporary subnet mask as following: -
- If octet comes before the interesting octet, write 255.
- In interesting octet, write “X”.
- If octet comes after the interesting octet, write 0.
Given network | Value after slash | Interesting octet | Temporary mask |
20.10.30.0/8 | 8 | First | X.0.0.0 |
111.187.45.34/14 | 14 | Second | 255.X.0.0 |
162.160.46.24/20 | 20 | Third | 255.255.X.0 |
202.100.20.50/27 | 27 | Fourth | 255.255.255.X |
Step 3
Just like we did above, based on interesting octet, subtract the given value from the upper value of interesting octet range.
Given network | Value after slash | Interesting octet | Interesting octet range | Subtraction |
20.10.30.0/8 | 8 | First | 1 – 8 | 8 – 8 = 0 |
111.187.45.34/14 | 14 | Second | 9 – 16 | 16 – 14 = 2 |
162.160.46.24/20 | 20 | Third | 17 – 24 | 24 -20 = 4 |
202.100.20.50/27 | 27 | Fourth | 25 – 32 | 32 – 27 = 5 |
Step 4
Use the result of subtraction in following formula to get the octet value
Octet value = 256 - 2 Remainder
Given network | Value after slash | Subtraction result | Calculation Step 1 | Calculation step 2 |
20.10.30.0/8 | 8 | 0 | 20 = 1 | 256 – 1 = 255 |
111.187.45.34/14 | 14 | 2 | 22 = 4 | 256 – 4 = 252 |
162.160.46.24/20 | 20 | 4 | 24 = 16 | 256 – 16 = 240 |
202.100.20.50/27 | 27 | 5 | 25 = 32 | 256 – 32 = 224 |
Step 5
Replace the “X” with octet value in interesting octet to build the full subnet mask
Given network | Value after slash | Temporary mask | Octet value | Subnet mask |
20.10.30.0/8 | 8 | X.0.0.0 | 255 | 255.0.0.0 |
111.187.45.34/14 | 14 | 255.X.0.0 | 252 | 255.252.0.0 |
162.160.46.24/20 | 20 | 255.255.X.0 | 240 | 255.255.240.0 |
202.100.20.50/27 | 27 | 255.255.255.X | 224 | 255.255.255.224 |
That’s all for this for this tutorial. If you need any assistance regarding this tutorial or have any feedback or comment about this tutorial, please mail me. If you like this tutorial, please don’t forget to share it through your favorite social network.