CCNA English-06: Subnetting in class
Subnetting in class
A friend of mine got a job in a multinational company before finishing his studies. Then of course no one else among our friends joined the job. And this friend of mine would get paid on the 25th or 26th. So at the end of the month, when we were in a tug of war, this friend was shopping and shopping. Even then we used to have a lot of fun with him because he talked very little and always tried to be alone. One day I asked this friend if you have joined IT then you can do some IT work! Oto got angry and said if he can't work, did he see my face and give me a job! I thought to myself that something is going to be known now. So I immediately said sorry, so if you don't know the job, how did the job happen and I immediately asked what class IP in your office using the class immediately said. I asked again, why in class? He said with a little bit of wisdom, "Hey, you don't know. The purpose of using it in our class is to have less network but more hosts." From this, we understand that if the number of network IDs is less and the number of host IDs is more then we will select its IP in class. Also,
The first 8 bits are used in NetAir in Class A and the next 24 bits are used as host bits. So it is understood that when selected in the class, almost three times the network address can be used as the host address. And the default mask of the class is 255.0.0.0.
Let's subnet in class A with a (1) bit on.
10.0.0.0/9
255.126.0.0
Network number = 2 1 = 2
Number of hosts = 2: 23-2 = 8338606
Subnet ID = 256-128 = 128
Network-1 = 10.0.0.0
First host = 10.0.0.1
10.0.0.1
10.0.0.3
Last host = 10.127.255.254
Broadcast address = 10.127.255.255
Network-2 = 10.128.0.0
First host = 10.128.0.1
10.128.0.2
10.128.0.3
Last host = 10.255.255.254
Broadcast address = 10.255.255.255
===================================
In the same way, in the case of class B, we subnet by turning on two (2) bits.
10.0.0.0/10
255.192.0.0
Network number = 2 ^ 2 = 4
Number of hosts = 2: 22-2 = 4194302
Subnet ID = 256-192 = 64
Network-1 = 10.0.0.0
First host = 10.0.0.1
10.0.0.1
10.0.0.3
Last host = 10.63.255.254
Broadcast address = 10.63.255.255
Network-2 = 10.64.0.0
First host = 10.64.0.1
10.64.0.2
10.64.0.3
Last host = 10.127.255.254
Broadcast address = 10.127.255.255
Network-3 = 10.128.0.0
First host = 10.128.0.1
10.128.0.2
10.128.0.3
Last host = 10.255.255.254
Broadcast address = 10.255.255.255
==================================
Network-4- = 10.192.0.0
First Host = 10.192.0.1
10.192.0.2
10.192.0.3
Last host = 10.255.255.254
Broadcast address = 10.255.255.255
===================================
The type of questions that the text contains.
10.0.0.0/8 Each subnet will have at least 5000 hosts.
Q: How many bits do you need to turn on to create 5,000 hosts?
Answer:
2 ^ 12 = 4096-2 = 4094
2 ^ 13 = 8192-2 = 8190
In other words, if we turn on 13 bits, our need will be fulfilled.
Q: So what will be the new subnet?
Answer:
Since the default subnet of the class has been turned on 255.0.0.0 and 13 more bits then (32-13) = 19 bits. If you turn on 19 bits, then 255 for the first 6 bits. 255 for the next 8 bits, then the rest is 3 bits. The value of 3 bits is 224. So the subnet mask
255.255.224.0
We got this information from the section we learned earlier
128-192-224-240-248-252-254-255
1-------2------3-------4------5------6-------7-------8
Answer: What is the block size?
256-224
= 32
Q: What is the 5th network?
10.0.0.0
10.0.32.0
10.0.64.0
10.0.96.0
10.0.128.0
I am ending here like today. Our next tutorial is VLSM. And to see the previous lectures, go to the main page of the course by following the link below.
CCNA English-06: VLSM
VLSM is a Variable Length Subnet Mask. With VLSM we can use multiple subnet masks on a network.
Why VLSM is needed?
VLSM is required to use IPs properly, that is, to reduce the misuse of IPs. Because a lot of times one client needs one range IP. So through VLSM, we can give them IP as per the requirement of the client. We can easily understand by looking at an example.
Suppose a new company. They need a certain number of IPs for different departments. This is their IP requirement. Them
Management requires - 100 IPs
The sales team requires - 50 IPs
Accounts team needs - 25 IPs
IT team needs 5 IPs
And our network is -192.168.1.0
I have not reviewed some information before doing the practical.
To find the host number = 2 ^ (total number) of the bits that will be off - -2
To find the network number = 2 গুলোর of the bits that will be extra on (total number)
To find out the subnet ID = 256 - last bit value
Host Required Block Size Host Pub Network Address Subnet Mask
100 126 2 ^ 6 = 128-2
= 126 192.168.1.0/25 255.255.255.128
50 64 (2 ^ 6) = 64-2
= 62 192.168.1.128/26 255.255.255.192
25 32 (2 ^ 5) = 32-2
= 30 192.168.1.192/27 255.255.255.224
5 8 (2 ^ 3) = 8-2
= 6 192.168.1.224/29 255.255.255.247
Let's see how we are doing the above work
When doing VLSM it is best to take the highest number of IPs first. The result is easy to calculate. For example, the maximum number of IPs required is 100. So for 100 host IPs we have to take 2 ^ 6 = 128-2 = 128. Then the subnet mask will be -255.255.255.126 and the network will be -192.18.1.0 / 25.
The second-highest number of IPs required is 50. Which will be required for the sales team. So for 50 host IP we need to take (2 ^ ৬) = 74-2 = 72. So since 6 bits have been used for the host then there are remaining bits (32-6) = 26. Since Class C has 24 fixed bits, additional bits are required (26-24) = 2. According to the above information, the value of the 2nd bit is - 192. So our subnet mask is -255.255.255.192. And our network will be -192.18.1.126 / 28 because our previous network block size was -127.
Step-03: For 25 hosts
The third-highest number of IPs required is 25. Which will be required for the accounts team. So for 25 host IP we have to take (2 ^ 5) = 32-2 = 30. So since 5 bits have been used for the host then there are remaining bits (32-5) = 26. Since Class C has 24 fixed bits, additional bits are required (26-24) = 3. According to the above information, the value of the 3rd bit is - 224. So our subnet mask is -255.255.255.224. And our network will be -192.18.1.192 / 28 because our previous network block size was -64. Because 128 + 64 = 192 has been used.
In the end, the maximum number of IPs required is 5. Which will be required for IT team members. So for 5 host IP we have to take (2 ^ 3) = 6-2 = 6. So since 3 bits have been used for the host then there are remaining bits (32-3) = 29. Since Class C has 24 fixed bits, additional bits are required (29-24) = 5. According to the above information, the value of the 5th bit is - 246. So our subnet mask is -255.255.255.247. And our network will be -192.18.1.224 / 29 because our previous network block size was -32. Because 192 + 32 = up to 224 has been used.
If VLSM is not done then it will look like that
Host Required Block Size Host Pub Network Address Subnet Mask
100 128 2 ^ 7 = 128-2
= 126 192.168.1.0/25 255.255.255.128
50 64 2 ^ 7 = 128-2
= 126 192.168.1.128/25 255.255.255.128
25 32 2 ^ 7 = 128-2
= 126 192.168.2.0/25 255.255.255.128
5 8 2 ^ 7 = 128-2
= 126 192.168.2.128/25 255.255.255.128
Step-01: For 100 hosts
The maximum number of IPs required is 100. So for 100 host IPs we have to take 2 ^ 6 = 128-2 = 128. Then the subnet mask will be -255.255.255.126 and the network will be -192.18.1.0 / 25
Step-02: For 50 hosts
The second-highest number of IPs required is 50. Which will be required for the sales team. So for 50 host IP we need to take 2 ^ 6 = 128-2 = 128. Our network will be -192.18.1.126 / 25.
Step-03: For 25 hosts
The third-highest number of IPs required is 25. Which will be required for the accounts team. So for 25 host IPs we have to take 2 ^ 6 = 128-2 = 128. But our network will be - 192.18.2.0 / 25
Step-04: For 5 hosts
In the end, 5 IPs are needed. Which will be required for IT team members. So for 5 host IPs we have to take 2 ^ 6 = 128-2 = 128. But our network will be - 192.18.2.126 / 25
As a result of VLSM implementation, if you do not implement VLSM, you will get the image
As a result of VLSM implementation
Host Required Block Size Host Pub Network Address Subnet Mask
100 128 2 ^ 7 = 128-2
= 126 192.168.1.0/25 255.255.255.128
50 64 (2 ^ 6) = 64-2
= 62 192.168.1.128/26 255.255.255.192
25 32 (2 ^ 5) = 32-2
= 30 192.168.1.192/27 255.255.255.224
5 8 (2 ^ 3) = 8-2
= 6 192.168.1.224/29 255.255.255.248
As a result of not implementing VLSM
Host Required Block Size Host Pub Network Address Subnet Mask
100 128 2 ^ 7 = 128-2
= 126 192.168.1.0/25 255.255.255.126
50 64 2 ^ 7 = 128-2
= 126 192.168.1.128/25 255.255.255.126
25 32 2 ^ 7 = 128-2
= 126 192.168.2.0/25 255.255.255.126
5 8 2 ^ 7 = 128-2
= 126 192.168.2.128/25 255.255.255.126
Finally, if you notice here, you will see. If you don't do VLSM, some IPs are just being lost.
Another thing to keep in mind is that exams have such questions
Which mask is used in point-to-point one link in the VLSM network?
1. / 26
2./26
3./29
4./30
5. / 31
- There will be 6 LANs and each LAN will have 28 hosts. In this case, which subnet should be selected from here?
A -0.0.0.240
B- 255.255.255.252
C-255.255.255.0
D-255.255.255.224
E-255.255.255.255.240
Let's see the explanation of the questions
A. This is not correct. Because what is given here is Wildcard Max.
B- We can see subnet mask 255.255.255.252. Then the first / 24 bit on. Simultaneously 6 more bits on. So we get Total Network Pub (2 ^ 8) = 64 and Total Host Pub 2 ^ 2 = 4-2 = 2. Now it turns out that our requisition is not going well. Because we need 26 hosts in each network. Then B is wrong.
C-255.255.255.0 is the default subnet mask. We can't subnet it. Then this is also wrong.
127-192-224-240
D- We can see the subnet mask 255.255.255.224. Then the first / 24 bit on. Simultaneously 3 more bits on. So we get Total Network Pub (2: 3) = 8 and Total Host Pub 2: 5 = 32-2 = 30. Then we can see that it is similar to our requirement. So the answer is d. Then we check the e option.
E- We can see subnet mask 255.255.255.240. Then the first / 24 bit on. Simultaneously 4 more bits on. So we get Total Network Pub (2 ^ 4) = 16 and Total Host Pub 2 ^ 4 = 18-2 = 14. It doesn't go with our needs. Because we need 26 hosts in each network.
In this way, if you check the wrong answers as soon as you get the correct answer, you will see that if you know why it is wrong, you will have a very clear idea. I ended up here like today.
The next lecture will be Routing. I am ending here wishing good health to everyone.
0 মন্তব্যসমূহ