VLSM variable length subnet mask partition tips

1、 Determine the number of subnets to be divided , And the internal requirements of each subnet IP The number of

For example, there are four subnets , what is needed IP The addresses are 100 individual ,60 individual ,20 individual ,10 individual ,

2、 As needed IP Determine the number of host bits occupied by each subnet

Then we can see from experience , want 100 individual IP The number of host bits allocated to the subnet of must be 7 position ,60 Yes 6 position ,20 Yes 5 position ,10 Yes 4 position .

3、 Start dividing

We must start with large subnets .

Let's say I have a C Class address 192.168.0.0, Assigned to four subnet A,B,C,D, They need IP The addresses are 100 individual ,60 individual ,20 individual ,10 individual ,

192.168.0.0~（0+127）127 Give me your address A subnet , among 192.168.0.0 It's the network number ,192.168.0.127 It's a broadcast number （ Why 0~127, Because our address distribution must be divided from the beginning , And we give A Subnet assignment 7 A host , So if 7 If all the host bits are 1, Then they are converted into decimal 127）

192.168.0.128~（128+63）191 Give me your address B subnet , among 192.168.0.128 It's the network number ,192.168.0.191 It's a broadcast number .

192.168.0.192~（192+31）223 Give me your address C subnet , among 192.168.0.192 It's the network number ,192.168.0.223 It's a broadcast number .

192.168.0.224~（224+15） Give me your address D subnet , among 192.168.0.224 It's the network number ,192.168.0.239 It's a broadcast number .

If you want to explore the number games inside, there are more things , I'm too lazy to write , But remember that there are just a few tricks ：

1、 First determine the number of host bits

2、 First from IP Subnets with high demand

3、 Tips in the above address allocation Explain

What we allocate first is A subnet , Because we know that only assign it 7 Host bits can meet its IP Address requirements , And because it is the first subnet to allocate addresses , So we simply know A The starting address of the subnet is 192.168.0.0, Also because 7 Bit host bit , therefore A Of the subnet to which the subnet is assigned IP The number of addresses is 128 individual （ from 0 Starting number , Count to 127, The number is 128 individual ）, The scope is 192.168.0.0 To 192.168.0.127.

Then is B subnet , If you want to play a number game here, it doesn't matter to calculate slowly , But here's the trick from the beginning , For example, the object of address allocation is B subnet , It is one of the four subnets IP The second largest subnet is required , Its need is 60 individual , We will soon know that we are going to assign it 6 Host location , because 2^6=64. Let's first make sure B The starting address of the subnet , stay A Add 1 that will do , that B The starting address of the subnet is 192.168.0.128, Then we have to make sure B The end address of the subnet , As long as 128+（64-1） that will do , among 64 yes 2^B The host bits of the subnet are calculated . Finally, we can get B The address distribution of the subnet is ：192.168.0.128~192.168.0.191, among 192.168.0.128 It's the network number ,192.168.0.191 It's a broadcast number .

Then knowing this rule, we can quickly calculate the start address and end address of other subnets .