This post is about subnetting part 3, You will learn here final concept of subnetting in this post and this is last post on subnetting. We have already learn two post in previous on subnetting.
If you have not seen that two post, I will suggest you to first go through that post to understand final concept of subnetting well. I ensure you after study of three post, You will never forget the concept of subnetting and subnet mask, All three post written as simple. Refer previous post what is subnetting ? .
Today we will do subnetting of class A IP address network in previous post we have done subnetting on class C where limited subnet ID was possible in class C very less only 4.
Where as in Class A large number of subnet ID is possible.
So today we have taken class A IP address for subnetting.
First you have to design Network. How much IP required in network each and every interface. So lets our network design is given below with IP requirement.
Lets take Class A IP 18.104.22.168, Now we have to provide subnet ID for all five Interfaces as shown in above picture. Hence five new network is required in our design of network.
Now we have take class A IP address try to understand the steps.
Steps for Subnetting questions
1 write down the given class A IP 22.214.171.124
2 We know that this is class A IP hence first octet is reserved for network and three octet are reserved for host see below picture.
Keep Network part as it and convert host part into binary, Then it will be like this.
IP address 12.00000000.00000000.00000000 ( host id converted into binary )
Now check the maximum number of IP required among the five new network interface,
Here in this case 60 PC interface is maximum number required.
Select maximum value interface to design network, small will be adjusted itself.
3 write the equation 2ⁿ -2 ≥ maximum value interface so in this case we can write
2ⁿ -2 ≥ 60 , Put n =1, 2, 3, 4,….. so on such that the value of n which must satisfy this equation.
If we put n =1 which will not satisfy
if we put n= 2 , n =3, n= 4, n= 5, ( all these value of n will not satisfy the above equation ).
But n = 6 then equation will become 2⁶ -2 ≥ 60 or 62≥60 this is correct, Hence value of n =6 satisfy the equation.
4 Now write IP 12.00000000.00000000.00000000
Now you have to reserved 6 bits for host from left side, Rest make the values 1,
Science network bits represented by 1 and host bits represented by 0 . If i am making host bits 1 then its mean that , I am giving host bits to network and this is called subnetting, Then it will look like this.
5 IP address 12.11111111.11111111.11000000 now convert this binary bits into decimal then we will get 126.96.36.199
6 what will be subnet mask for 188.8.131.52 ?
We know that subnet mask for class A by default is 255.0.0.0, Hence subnet mask for
184.108.40.206 will be 255.255.255.192 because you have changed host bits to network bits so it will be not same as by default.
Now if you want to write CDIR value, Hare count how much bits for network is used in IP address, Here in this case total network bits 8+8+8+2 = 26 .
Hence first subnet ID will be 220.127.116.11/26 so its subnet mask will be 255.255.255.192
How to make next subnet mask ?
Now take the first bits of 1 from left to right in IP 12.11111111.11111111.11000000
Find the decimal value of that 1 , So its decimal value is 64.Add this value into first subnet ID 18.104.22.168/26, then it will be 22.214.171.124
Hence Subnet ID -2 will be 126.96.36.199/26.
First subnet ID = 188.8.131.52/26
Second subnet ID = 184.108.40.206/26 ( broadcast ID will be 220.127.116.11 )
Third subnet ID = 18.104.22.168/26 ( broadcast ID will be 22.214.171.124 ) one less than maximum value.
fourth subnet ID = 126.96.36.199/26 (broadcast ID will be 188.8.131.52 ).
Now we have found four subnet ID, But we have five new network, So we required one more subnet ID . So add 64 in 4th subnet ID hence 192 +64 = 256, this value is not possible because maximum value in any octet will be 255.
So for this you have to do like this.
Use watch concept, How time in watch change.
4 : 59 pm here m is minute and s is second, When one minute increase then watch not show like 4 : 60 , It shows 5 : 00
Hence here same watch concept will be use, when reach maximum value so this is main concept. Now we can write fifth subnet ID.
Fifth Subnet ID = 184.108.40.206/26 subnet mask 255.255.1.255
Sixth Subnet ID = 220.127.116.11/26 subnet mask 255.255.1.64
seventh subnet D = 18.104.22.168/26
eighth Subnet ID = 22.214.171.124/26
ninth subnet ID = 126.96.36.199/26 and so on
Last Subnet ID is 188.8.131.52/26
Now suppose you have an IP 184.108.40.206/26 then question is how many maximum host can link to this or how many computer we can add to this subnet ID.
Ans It is simple first you see which class IP is this, So this is class C IP now in class C IP three octet are reserved for network and one for host so total 32 bits.
But as per CIDR value it is clear that here 26 bits are used for network, So 32 – 26 = 6 bits are left for host .
Hence number of computer can be add to this subnet ID will be 2⁶ – 2 = 64- 2 = 62
So we can connect 62 maximum computer to this subnet ID.
We will continue in next post.I hope you have enjoyed to learn subnetting questions if you like comment and share on social media thanks for reading and sharing keep learning more and more
Dated 27th Oct 2018