CODEWITHSUNDEEP
network-programmingipip-addresssubnet
user1770049
user1770049

Reputation: 29

How to calculate subnets

It's been a while and haven't been able to figure out how I can subnet the below ranges. Could someone help me to subnet these ranges and provide some insight?

Subnet the 10.0.0.0/8 address space into 30 subnets.
Subnet the 172.16.0.0 address space into 13 subnets.
Subnet the 192.168.1.0/24 address space into 4 subnets.
Subnet the 192.168.0.0/24 address space into 7 subnets.

Upvotes: 1

Views: 4909

Answers (3)

Rahul Goyal
Rahul Goyal

Reputation: 774

I love to solve problems on IP sub-netting. I will teach for the 1st problem and you can do the rest by yourself. Following are the steps:

  1. Write the given networks address with its subnet mask. 10.0.0.0 / 8 => we can not change the first 8 bits, N=8.

  2. Write it in binary form with each octet separated by . 000001010 . 00000000 . 00000000. 00000000 /8

  3. Now comes the question, how many subnets do you need. in this case : 30 subnets

  4. Find the next nearest power of 2, 2^5 = 32 (nearest to 30)

  5. The number for bits required to subnet, M=5

  6. Leave out the first N bits (step 1), and for the next M bits, write all the binary combination. In our example, first 8 bits remain same, then we write all 5 bit combination and append it. 1st: 000001010 . 00000000 . 00000000. 00000000 /13 N+M 2nd: 000001010 . 00001000 . 00000000. 00000000 /13 N+M 3rd: 000001010 . 00010000 . 00000000. 00000000 /13 . 30th subnet: 000001010 . 11101000 . 00000000. 00000000 /13 31th subnet: 000001010 . 11110000 . 00000000. 00000000 /13 32nd sunbet: 000001010 . 11111000 . 00000000. 00000000 /13

  7. Convert each thing to the decimal equivalent and you will get 32 different subnets. I hope its clear.

Upvotes: 5

CustomX
CustomX

Reputation: 10113

I have done 2, the rest is upto you! Also have a look here, it will help you refresh your memory :)

10.0.0.0/8 - 30 subnets

N = network - S = subnet - H = host

10.0.0.0 = NNNNNNNN.HHHHHHHH.HHHHHHHH.HHHHHHHH /8

We need 30 subnets, so 2^5 = 32 subnets (provides 524286 usable hosts [2^19 - 2])

10.0.0.0 = NNNNNNNN.SSSSSHHH.HHHHHHHH.HHHHHHHH /13

1: 10.0.0.0/13 
2: 10.8.0.0/13
3: 10.16.0.0/13
...
30th: 10.232.0.0/13

192.168.1.0/24 - 4 subnets

192.168.1.0 = NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH /24

We need 4 subnets, so 2^2 = 4 subnets (provides 62 usable hosts [2^6 - 2])

192.168.1.0 = NNNNNNNN.NNNNNNNN.NNNNNNNN.SSHHHHHH /26

1: 192.168.1.0/26
2: 192.168.1.64/26
3: 192.168.1.128/26
4: 192.168.1.192/26

Upvotes: 3

Andrew Ring
Andrew Ring

Reputation: 3343

You're creating a simple binary mask (i.e. perform AND between the make and the number to resolve the network's address, remaining is host id). There is more detailed information here and here.

Upvotes: 0

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