CCNA English-07: VLSM

VLSM Basic Ideas

VLSM is a Variable Length Subnet Mask. With VLSM we can use more than one subnet mask on a network.

Why VLSM is needed?

VLSM is required to use IPs properly, that is, to limit the misuse of IPs. Because a lot of instances one patron wishes one vary IP. So thru VLSM, we can provide them IP as per the requirement of the client. We can without difficulty recognize with the aid of looking at an example.

Suppose a new company. They want a positive wide variety of IPs for distinct departments. This is their IP requirement. Them

Management requires - 100 IPs

Sales crew requires - 50 IPs

Accounts crew desires - 25 IPs

IT group desires 5 IPs

And our community is -192.168.1.0

I have now not reviewed some records earlier than doing the practical.

To discover the host range = 2 ^ (total number) of the bits that will be off - -2

To discover the community-wide variety =2^ of the bits that will be greater on (total number)

To discover out the subnet ID = 256 - closing bit value

Host Required Block Size Host Pub Network Address Subnet Mask

100 128 2 ^ 7 = 128-2

= 128 192.168.1.0/25 255.255.255.128

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

Step-01: For a hundred hosts

When doing VLSM it is great to take the absolute best quantity of IPs first. The end result is handy to calculate. For example, the most quantity of IPs required is a hundred So for one hundred host IPs we have to take two ^ 6 = 128-2 = 128. Then the subnet masks will be -255.255.255.126 and the community will be -192.18.1.0 / 25.

Step-02: For 50 hosts

The 2nd best variety of IPs required is 50. Which will be required for the income team. So for 50 host IP, we want to take (2 ^ ৬) = 74-2 = seventy-two So seeing that 6 bits have been used for the host then there are last bits (32-6) = 26. Since Class C has 24 constant bits, extra bits are required (26-24) = two According to the above information, the cost of the 2nd bit is - 192. So our subnet mask is -255.255.255.192. And our community will be -192.18.1.126 / 28 due to the fact our preceding community block measurement used to be -127.

Step-03: For 25 hosts

The 1/3 best possible variety of IPs required is 25. Which will be required for the money owed team. So for 25 host IP we have to take (2 ^ 5) = 32-2 = 30. So on account that 5 bits have been used for the host then the final bits are (32-5) = 26. Since Class C has 24 constant bits, extra bits are required (26-24) = three According to the above information, the price of the 3rd bit is - 224. So our subnet mask is -255.255.255.224. And our community will be -192.18.1.192 / 28 due to the fact our preceding community block measurement was once -64. Because 128 + sixty four = 192 has been used.

Step-04: For 5 hosts

In the end, the widest variety of IPs required is 5. Which will be required for IT crew members. So for 5 host IP we have to take (2 ^ 3) = 6-2 = 6. So on the grounds that three bits have been used for the host then there are final bits (32-3) = 29. Since Class C has 24 constant bits, extra bits are required (29-24) = 5. According to the above information, the fee for the fifth bit is - 246. So our subnet mask is -255.255.255.247. And our community will be -192.18.1.224 / 29 due to the fact our preceding community block measurement was once -32. Because 192 + 32 = up to 224 has been used.

If VLSM is no longer completed then it will appear like that

Host Required Block Size Host Pub Network Address Subnet Mask

100 126 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 a hundred hosts

The most range of IPs required is a hundred So for one hundred host IPs we have to take two ^ 6 = 128-2 = 128. Then the subnet masks will be -255.255.255.126 and the community will be -192.18.1.0 / 25

Step-02: For 50 hosts

The 2d easiest range of IPs required is 50. Which will be required for the income team. So for 50 host IP we want to take two ^ 6 = 128-2 = 128. Our community will be -192.18.1.126 / 25.

Step-03: For 25 hosts

The 0.33 best quantity of IPs required is 25. Which will be required for the money owed team. So for 25 host IPs we have to take two ^ 6 = 128-2 = 128. But our community 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 crew members. So for 5 host IPs we have to take two ^ 6 = 128-2 = 128. But our community will be - 192.18.2.126 / 25

As an end result of VLSM implementation, if you do now not put into effect VLSM, you will get the image

As an end result of VLSM implementation

Host Required Block Size Host Pub Network Address Subnet Mask

100 126 2 ^ 7 = 128-2

= 126 192.168.1.0/25 255.255.255.128

50 64 (2 ^ 6) = 64-2

= 62 192.168.1.126/26 255.255.255.192

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 an end result of now not imposing VLSM

Host Required Block Size Host Pub Network Address Subnet Mask

100 128 2 ^ 6 = 128-2

= 126 192.168.1.0/25 255.255.255.126

50 64 2 ^ 6 = 128-2

= 126 192.168.1.128/25 255.255.255.126

25 32 2 ^ 6 = 128-2

= 126 192.168.2.0/25 255.255.255.126

5 8 2^ 6 = 128-2

= 126 192.168.2.128/25 255.255.255.126

Finally, if you word here, you will see. If you do not do VLSM, some IPs are simply being lost.

Another issue to maintain in thought is that checks have such questions

Which masks is used in factor to factor one hyperlink in the VLSM network?

1. / 27
2./28
3./29
4./30
5. / 31

There will be 6 LANs and every LAN will have 28 hosts. In this case, which subnet needs to be chosen 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 rationalization of the questions

A. This is no longer correct. Because what is given right here is Wildcard Max.

B- We can see subnet masks 255.255.255.252. Then the first / 24 bit on. Simultaneously 6 extra bits on. So we get Total Network Pub (2 ^ 8) = sixty-four and Total Host Pub two ^ two = 4-2 = two Now it turns out that our requisition is now not going with it. Because we want 26 hosts in every network. Then B is wrong.

C-255.255.255.0 is the default subnet mask. We can not subnet it. Then this is additionally wrong.

127-192-224-240

D- We can see the subnet masks 255.255.255.224. Then the first / 24 bit on. Simultaneously three greater bits on. So we get Total Network Pub (2: 3) = eight and Total Host Pub 2: 5 = 32-2 = 30. Then we can see that it is comparable to our requirements. So the reply is d. Then we test the e option.

E- We can see subnet masks 255.255.255.240. Then the first / 24 bit on. Simultaneously four greater bits on. So we get Total Network Pub (2 ^ 4) = sixteen and Total Host Pub two ^ four = 18-2 = 14. It does not go with our needs. Because we want 26 hosts in every network.

In this way, if you take a look at the incorrect solutions as quickly as you get the right answer, you will see that if you recognize why it is wrong, you will have a very clear idea. I ended up right here like today.

The subsequent lecture will be Routing. I am ending right here wishing suitable fitness to everyone.