0

Start with 192.168.67.0/24 and split it into 4 equal networks, what is the third usable IP address of the 3rd subnet? how do you solve this?

4

Well let's see if i still remember this. /24 gives you 8 (32-24) bits to play around with. You need 4 networks, so it will take 2 bits to represent them (00 01 10 11) So your network part will be 26 (24+2) and four networks will be :

192.168.67.0/26   (last octet in binary 00'00 0000)
192.168.67.64/26  (last octet in binary 01'00 0000)
192.168.67.128/26 (last octet in binary 10'00 0000)
192.168.67.192/26 (last octet in binary 11'00 0000)

As you can see your third subnet is 192.168.67.128/26 first IP represents subnet so it's not usable, first usable IP is 192.168.67.129, second is 192.168.67.130, and drumroll third 192.168.67.131

Did i just do your homework?

  • Yes you just did his homework, but at least you explained how... – Law29 Nov 16 '16 at 6:50
  • Thank you so much, I know how to solve this now, thank you!!! – Sebs.Alev Nov 16 '16 at 22:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.