How to subnet in a detailed way without any type of introduction that makes you feel bored.
I know I have promised you not to read too much, but let's be familiar with some concept before move on.
What is a subnet?
Subnetting is basically just a way of splitting a TCP/IP network into smaller, more manageable pieces.
What its purpose?
The main purpose of subnetting is to help relieve network congestion such as traffic.
What's the benefit?
Prevents Unnecessary Broadcasts, Increases Security Options, Simplifies Administration and last but not least Controls Growth.
What is binary?
Binary describes a numbering scheme in which there are only two possible values for each digit: 0 and 1
In computer electrical terms 1 means ON 0 mean OFF.
IPv4 Address Classes?
Class A addresses are used for very large networks and always start with a leftmost bit being a zero.
A class A network can hold as many as 16,777,214 hosts.
Class | Private IP Address Range | Subnet Mask |
---|---|---|
A | 10.0.0.0 to 10.255.255.255 | 255.0.0.0 |
Class B These addresses are used for medium sized networks and always start with the leftmost 10 bits. lass B network can hold as many as 65,534 hosts.
Class | Private IP Address Range | Subnet Mask |
---|---|---|
B | 172.16.0.0 to 172.31.255.255 | 255.240.0.0 |
Class C These addresses are used for smaller networks, the one you have at home and always start with the leftmost bits 110. Each class C network can only hold up to 254 hosts.
Class | Private IP Address Range | Subnet Mask |
---|---|---|
C | 192.168.0.0 to 192.168.255.255 | 255.255.0.0 |
What's VLMS?
Variable Length Subnet Masking "VLSM" is a technique that allows network administrators to divide an IP address space into subnets of different sizes, unlike simple same-size Subnetting. In a way, means subnetting a subnet
**End of the introduction**
Now, looking at the diagram, we have three LANs connected to each other with two WAN links. The first thing to look out for, is the number of subnets and number of hosts. In this case, an ISP allocated 192.168.1.0/24. Class C
R1 = 50 host
R2 = 20 host
R3 = 90 host
2 WAN links.
We will try and subnet 192.168.1.0 /24 to sooth this network which allows a total number of 254 hosts I recommend you get familiar with this table below. Or at least learn the binary values.
NOTE: As a first rule we need to start using the network that require the more hosts until we reach the lowest.
Let's begin with R3 with 90 hosts.
We are borrowing 1 bit to the value 128 on the table above, why? Remember, we need 90 hosts therefore 128 fits in the range. Now, our subnet is 192.168.1.0 /24 but after we've found an extra Bit we can do the follows:
1- Convert to binary our old subnet mask adding the new borrowed bit.
(Bits) 11111111.11111111.11111111.1000000 = 255.255.255.128
Our new subnet is 192.168.1.0 /25, why 25? If you count the Bits you will notice that you have a total of 25.
So here is the outcome of what we did.
R3 subnet: 192.168.1.0 /25
Network: 192.168.1.0
First IP: 192.168.1.1
Last IP: 192.168.1.126
Broadcast: 192.168.1.127
Easy right? now keep in mind that 128 it's our next network for the following subnet below.
R3 has been completed.
Let's work with R1 50 hosts:
As we were working on R3 we know for fact that our next network for R1 is 128 and therefore we have our network ID and first IP.
Network: 192.168.1.128
First IP: 192.168.1.129
Now, we need to find where this network ends. R1 require 50 hosts. If we look at the table above we know that 64 will give us the right range for the mentioned hosts, therefore we need to borrow 2 bits. so let's add more bits to our old subnet mask.
1- Convert to binary our old subnet mask adding the new borrowed bit.
(Bits) 11111111.11111111.11111111.1100000 = 255.255.255.192
Our new subnet is 192.168.1.128 /26, why 26? If you count the Bits you will notice that you have a total of 26.
So what's next? we need to find our incremental.
"Always remember that our incremental is the last Bit you have taken borrowed". In this case 64.
So let's do the math 128+64=192
R1 subnet: 192.168.1.128 /26
Network: 192.168.1.128
First IP: 192.168.1.129
Last IP: 192.168.1.190
Broadcast: 192.168.1.191
Getting in there right? now, keep in mind that 192 it's our next network for the following subnet below.
R1 has been completed.
Let's work with R2 20 hosts:
Again, as we were working on R1 we know for fact that our next network for R2 is 192 and therefore we have our network ID and first IP.
Network: 192.168.1.192
First IP: 192.168.1.193
Now, we need to find where this network ends. R2 require 20 hosts. If we look at the table above we know that 32 will give us the right range for the mentioned hosts, therefore we need to borrow 3 bits. So let's add more bits to our old subnet mask.
1- Convert to binary our old subnet mask adding the new borrowed bit.
(Bits) 11111111.11111111.11111111.1110000 = 255.255.255.224
Our new subnet is 192.168.1.192 /27, why 27? If you count the Bits you will notice that you have a total of 27.
So what's next? we need to find our incremental.
"Always remember that our incremental is the last Bit you have taken borrowed". In this case 32.
So let's do the math 192+32=224
R1 subnet: 192.168.1.192 /27
Network: 192.168.1.192
First IP: 192.168.1.193
Last IP: 192.168.1.1.222
Broadcast: 192.168.1.223
Even easier right? now keep in mind that 224 it's our next network for the following subnet below.
R2 has been completed.
Let's move onto WAN 1 (2 hosts)
Same methodology, my old subnet is 192.168.1.0 /24, we need 2 host for each wan link right?. So our magic number will be 4 and we need to borrow 6 bits.
1- Convert to binary our old subnet mask adding the new borrowed bit.
(Bits) 11111111.11111111.11111111.11111100 = 255.255.255.252
Our new subnet is 192.168.1.224 /30, why 30? If you count the Bits you will notice that you have a total of 30.
So what's next? we need to find our incremental.
"Always remember that our incremental is the last Bit you have taken borrowed". In this case 4.
So let's do the math 224+4=228
WAN link subnet: 192.168.1.224 /30
Network: 192.168.1.224
First IP: 192.168.1.1225
Last IP: 192.168.1.1.226
Broadcast: 192.168.1.227
Let's move onto WAN 2 (2 hosts)
Same methodology, my old subnet is 192.168.1.0 /24, we need 2 host for each wan link right?. So our magic number will be 4 and we need to borrow 6 bits.
1- Convert to binary our old subnet mask adding the new borrowed bit.
(Bits) 11111111.11111111.11111111.11111100 = 255.255.255.252
Our new subnet is 192.168.1.228 /30, why 30? If you count the Bits you will notice that you have a total of 30.
So what's next? we need to find our incremental.
"Always remember that our incremental is the last Bit you have taken borrowed". In this case 4.
So let's do the math 228+4= 232
WAN link subnet: 192.168.1.228 /30
Network: 192.168.1.228
First IP: 192.168.1.1229
Last IP: 192.168.1.1.230
Broadcast: 192.168.1.231
Phew! That was all. Now we know what subnet need each router we can start using it's respective IP addresses.
R3: 192.168.1.0 /25
R1: 192.168.1.128 /26
R2: 192.168.1.192 /27
WAN1: 192.168.1.224 /30
WAN2: 192.168.1.228 /30
I hope this is been informative for all who wants to know how to subnet.
Later in the future I will be adding more post with class A and B, stay tuned
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This website will help you to practice subnet until you learn. Also they can teach you CCNA and CCNP for a minimum cost of 5 bucks a month what a deal right?
I learned a lot with this website and if you have any question they'd help you with your question to reach your goals.
Thank you for reading and if you enjoyed my article please endorse!
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