Finding number of valid IPs in subneting.

[ahmadqayyum7]

I am doing sub netting and want to confirm that if my method is correct or not. Suppose we are to design a network that has 2 departments. Department 1 has 32 number of hosts and department 2 has 64 number of hosts. We are given the following IP:

192.168.6.10/24

Now I will give 8 sub nets. For 32 hosts, `2^5` is required. The sub net mask for this network will be 255.255.255.224

Now:

192.168.6.000|01010

These three zeros (the ones before |) will also go with the network bits. Now for the first sub net my IP range will start from

192.168.6.0 for network
.
.
.
192.168.6.31 for Broadcast ID

For second sub net:

192.168.6.32
.
.
.
192.168.6.63

Now for the department having 64 hosts, I will have 4 sub nets and `2^6` will give me 64. The sub net mask will be 255.255.255.192.

Now from where will my IP start for these hosts?

[balaji.bandi]

easy way is make subnettting 192.168.6.0 network in to /26 like below as :

192.168.6.0/26  <--this one one you like to have 2  /27 ( that is 192.168.6.0/27 and 192.168.6.32/27 ) - for your 32 host network

below one you can use for 64 network.

192.168.6.64/26

192.168.6.128/26

192.168.6.192/26

[Jaderson Pessoa · Accepted solution]

I am doing sub netting and want to confirm that if my method is correct or not. Suppose we are to design a network that has 2 departments. Department 1 has 32 number of hosts and department 2 has 64 number of hosts. We are given the following IP:

192.168.6.10/24

NETWORK: 192.168.6.0/24

   NETWORK                FIRST IP                  LAST IP              BROACAST

FOR 64 ADDRESS

192.168.6.0/26      - 192.168.6.1       -   192.168.6.62     192.168.6.63

192.168.6.64/26    - 192.168.6.65     -   192.168.6.126   192.168.6.127

192.168.6.128/26  - 192.168.6.129   -   192.168.6. 190  192.168.6.191

FOR 32 ADDRESS

192.168.6.192/27  - 192.168.6.193   -   192.168.222       192.168.6.223

192.168.6.224/27  - 192.168.6.225   -   192.168.254        192.168.6.255

[ahmadqayyum7]

why did you make 3 groups for 64 hosts and 2 groups for 32 hosts? what is the rule you followed?

[Giuseppe Larosa]

Hello,

be aware that the number of available hosts in a /27 is 2^5 -2 = 30 hosts because the first IP address is the subnet and the last one is the broadcast address.

Similarly to support 64 hosts you would need a /25 because a /26 supports only 62 hosts again because first IP address is the subnet the last is the subnet broadcast address. 2^6 -2 = 62 hosts.

However, if you mean that you want to support 30 hosts and 62 hosts respectively the right method is to use /26 subnets and then take one of them and divide it in two /27 subnets as explained by BB.

Hope to help

Giuseppe