05-20-2019 01:08 PM
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?
Solved! Go to Solution.
05-20-2019 01:31 PM
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
05-20-2019 01:21 PM
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
05-20-2019 01:31 PM
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
05-20-2019 01:51 PM - edited 05-20-2019 01:56 PM
why did you make 3 groups for 64 hosts and 2 groups for 32 hosts? what is the rule you followed?
05-20-2019 01:35 PM
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
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