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If I subnet the IP prefix 172.16. 64.0/18, shouldn't I be able to get 4 subnets? I was only able to get the following subnets:

Network ID - Range - Broadcast ID:
172.16.64.0    -     172.16.127.255
172.16.128.0   -     172.16.191.255
172.16.192.0   -     172.16.255.255

Which subnet, if any, am I missing?

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4 Answers 4

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By splitting the 172.16.64.0/18 network in 4 subnets you obtain:

172.16.64.0/20

172.16.80.0/20

172.16.96.0/20

172.16.112.0/20

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  • Could you explain? I thought the network IDs would increment by 64 in the third octet. Where am I going wrong?
    – user17715
    Sep 4, 2015 at 8:54
  • 172.16.64.0/18 is a network from 172.16.64.0 to 172.16.64.127.255. If you need to subnet it to get 4 networks you have to take 2 more bits (2^2=4). It means /20.
    – jcbermu
    Sep 4, 2015 at 11:52
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You are subnneting in a wrong way. This one if what you should do. Will take 172.16.64.0/18 into binary(.64.) 172.16.0100 0000.0

you need 4 subnets:

0100 0000 /base

0100 0000 0101 0000 0110 0000 0111 0000

All of this subnets are /20. We take 2 bits of host parts in 172.16.64.0/18, that has been used in network.

172.16.64.0/20, 172.16.80.0/20, 172.16.96.0/20, 172.16.112.0/20

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ip address X.X.X.X consisted of two main parts [Network part] , [host part] Network part can be consisted of two parts , fixed part called [network summary address] and tunnable part called [Subnet ID] , will reserved by host IPs . in your example you provide us with the network summary address 172.16. 64.0/18 and you want to subnet this address into 4 subnets (sub 0 ,1 ,2 ,3 ). let us present your summary address in other way enter image description here

as you know in binary if you want to represent up to 3 things you need at lest 2 binary digits (00 ,01,10,11) , simply to represent those 4 subnets we will get the two binary digits we need from the host portion as next

enter image description here

so that the network portion of the IP became 20 digits instead of 18 and the subnets will be as shown

enter image description here

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There are 4 available subnets here

Network ID - Range - Broadcast ID:

172.16.0.0 - 172.16.63.255

172.16.64.0 - 172.16.127.255

172.16.128.0 - 172.16.191.255

172.16.192.0 - 172.16.255.255

But your question states :: If I subnet the IP prefix 172.16.64.0/18

So we have to start @ 172.16.64.0

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