The best way of learning subnetting is to do it. Here are a selection of worked examples to help you get started.
At the end are some links to online quizes so you can do it yourself.
255.255.0.0 and or 255.255.255.0
Yes both would work.
Mask 255.255.0.0 has 8 bits for the subnet and 16 bits for the host
8 bits would accommodate 28=256 subnets
16 bits would accommodate 216= over 64000 hosts
Mask 255.255.255.0 has 16 bits for the subnet and 8 bits of the host.
Have possible 28 -2 hosts =254 which is enough
Easiest is to sub net on a byte boundary which would mean a subnet mask of 255.255.255.0
This would allocate 8 bits for the subnet and 8 bits for the host.
We need to accommodate around 200 hosts which requires 8 bits which we have.
We need 4 subnets which require 4 bits and we have 8 bits. So we have more than enough.
Decimal 192 =11000000 binary which means that 2 bits of this octet are used for the subnet. Now add the 24 bits 255.255.255 and we have 26 bits. So we write:
222.1.1.20/26
Decimal 248 =11111000 binary which means that 5 bits of this octet are used for the subnet. Now add the 16 bits 255.255. and we have 21 bits. So we write:
135..1.1.25/21
A class C address has 8 bits of the host which will give 28 -2 =254 hosts
Current mask= 255.255.255.0
Bits needs for 10 subnets =4 =24 =16 possible subnets
Bits needs for 12 hosts = 4 = 24 = 16-2=14 possible hosts.
So our mask in binary =11110000= 240 decimal
Final Mask =255.255.255.240
What is the subnet mask for the maximum number of hosts?
How many hosts can each subnet have?
What is the IP address of host 3 on subnet 2 ?
Current mask= 255.255.255.0
Bits needs for 30 subnets =5 =25 =32 possible subnets
Bits left for hosts = 3 = 23 = 8-2=6 possible hosts.
So our mask in binary =11111000= 248 decimal
Final Mask =255.255.255.248
Address of host 3 on subnet 2 is
subnet 2 =00010000 host 3 =000000011
Add the two together =00010011=19
therefore IP address of host 3 on subnet 2 =205.11.2.19
What are the two possible subnet masks?
Current mask= 255.255.255.0
Bits needs for 48 hosts = 6 = 26 = 64-2=62 possible hosts.
Bits needs for 2 subnets =1 =21 =2 possible subnets
Total of 7 bits needed so therefore we can use either 1 bit or 2 bits for the subnet. So we could have
1 bit subnet 7 bits hosts or 2 bits subnet 6 bit host
masks are 10000000 and 11000000 =128 decimal and 192 decimal.
Final possible masks are:
255.255.255.128 and 255.255.255.192
192 in binary =11000000 gives 4 possible subnets of (showing 2 most significant bits):
00,01,10,11
67 in binary =01000011
So Applying Mask:
Videos
Related Tutorials and Resources:
- binary numbers for beginners tutorial
- IPv4 Addresses and Classes
- Subnets and subnet Masking explained.
- IPv6 Guide
Subnet Quizzes
Test you knowledge using these online quizzes
Hello Steve, your question are very helpful ..but can you guide me on the below ones please
1.You currently use the default mask for your IP network 192.168.1.0. You need to subnet
your network so that you get the following.
a. 30 additional networks, and 4 hosts per network.
b. 20 hosts per subnet
c. Only 4 additional networks
Hi
4 hosts means you need 3 bits for the host so your mask looks like this
11111000
20 hosts means that you need 5 bits for the host so your mask looks like this
11100000
4 additional networks means you need 3 bits for the Network so your mask looks like this
11100000
I’ll let you convert them to decimal
Rgds
Steve
Great job steve!
All your answers are 100% correct and I found the exercises very helpful. Thanks again!
what is the last valid host on the subnetwork 172.21.80.128/26?
pls help me in this
thanks a lot
Hi
you have 2 bits for the host so values are
=00,01,10,11
00 and 11 are not usable so you have hosts 01 and 10
by last I assume the mean 10 so the address is 172.21.80.130
CIDR is /26
172.21.80.128
Binary: 10101100 00010101 01010000 10000000
Let’s seperate the subnet and hosts
10101100 00010101 01010000 10 —> 000000
Lets make all the 0’s 1’s where the seperation is to get Broadcast
10101100 00010101 01010000 10 —> 111111
Broadcast : 172.21.80.191 (191 is 10 111 111)
Last usuable host IP will therefore be 10 111110
Last usuable IP: 172.21.80.190
Q.6 . Subnet 10 host 1 ip address must be 195.1.1.151 as per my calculation
Hi
Just checked it again and 161 is correct
rgds
steve
1. Calculate IP addresses and subnet mask for 3 nodes (indexes are listed into the 3rd line) for three subnets (indexes are listed in the 2nd line). Network can contain some amount of subnets and no less than it is specified (the 1st line, S). Each subnet can contain no less than specific numbers of nodes (the 1st line, H).
Result: final nodes IP and masks with calculaations
300, 3, 140
1, 2, 3
11, 12, 13 / 1, 22, 23
The 1st line: N, S, H; N – total nodes amount, S – subnet amount, H – subnet nodesamount.
The 2nd line: A, B, C; A & B & C – subnet indexes
The 3rd line: a1, a2, a3 / b1, b2, b3; ai и bi – node indexes to calculate. ci is equal to bi
Help please if someone knows
Sorry find the question confusing were did it come from
Rgds
Steve
Question one should be a MINIMUM of 160 hosts, not a MAXIMUM!
No Maximum is correct.But using Minimum would also work in this case.
Rgds
Steve
give the following subnets ,represnt them with a single network address
212.56.120,0
212.56.121.0
212.56.122.0
212.56.123.0
Not sure about the question where is it from
Rgds
Steve
Amazing, amazing work! You have no idea how much these exercises have helped, and the explanations are all very clear as well. Thank you very much!
p.s. I love the font
Hi Steve, thanks for producing this tutorial.
In question 9 – is this assuming a Class C network? E.g. if it were a Class B network with subnet mask of 255.255.255.192 then there would be 10 bits for the subnets rather than 2.
Yes it is a class C address
Four valid subnets are to be created in a class B IP address 129.22.0.0.
Calculate the required Subnet mask and fill in the table below.
The answer “therefore IP address of host 3 on subnet 2 =205.11.2.19” of question 7 is wrong and by the way the question is wrong, because if you calculate the ip address of 2nd host pf 3rd subnet then you will get the answer as 205.11.2.19…….
And if we go according to question the real ans should be 204.11.2.11……
Correct me if I’m wrong!
The answer is correct.
The first sub network address is 00001xxx =decimal 8 and the second is 00010xxx =decimal 16
rgds
steve
is there any one who can explain the 9th example? i am not getting that ans .
What answer do you get
rgds
steve
Very useful exercises 😉 but I have a question about exercise 6: Shouldn’t it be 255.255.255.248 instead of 255.255.255.240? Because if we choose the last one we will get more than the 12 hosts.
Hi
the answer is correct as 248 would only give us 8 hosts (3 bits) and we need at least 12.
Yes but since it is typed “10 subnets each with a maximum 12 hosts on each subnet.” and 8 is smaller than 12 I assume 255.255.255.248 is the correct answer.
Maximum of 12 means that you can have 12. But with 8 bits you can only have 8.
Does that make sense
rgds
steve
it is very good for beggineer but i need know more focus on the type of the class
Not sure what you mean by type of the class
Sir its a very very help full for beginners and also for experience but i need ur more help to learn total subneting as class a b and c,
Have you watch the videos
IPv4 Addresses and Address Classes Explained
https://youtu.be/nVKZs-_-7U4
———–
IPv4 Subnetting Worked Examples
https://youtu.be/eet6SumgW5A
Were are you struggling? Would it help if I sent you a quiz to test your current understanding?