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Firstly I'm a beginner so please forgive me.

I can't quite figure out how to do this as I've only ever been taught how to assign IP addresses with the address space of xxx.xxx.xxx.xxx/24 which makes sense to me because you're only dealing with the last octet. But with /23 you're dealing with two octets aren't you?

E.g 192.168.10.0/23 is..

192.168.0000 1010. 0000 0000

Where the last 0 in the 3rd octet is included in the subnet mask.

So to make things more clear in my head, is someone able to show me an example of assigning 256 IP addresses ?

Below is the full question. Please don't fully answer the question, I would just like a push in the right direction.

As part of restructuring at the Cyber University, a new Faculty of Innovation has been established. The faculty encompasses Computer Science and Engineering disciplines. You have been tasked to design the network for the new faculty. The Default_Gateway and Router01 are part of the existing network, and the faculty has been assigned 130.120.10.0/23 address space.

Part 1

Divide the assigned address space between faculty admin, CS and ENG subnets such that the ENG subnet gets 256 IP addresses, Faculty Admin and CS subnets get 128 IP addresses each.

  • I'm not actually physically working with any devices, it's a lab question. The actual question is asking me to use the address space 130.120.10.0/23 and assign 256 address to subnet 1, 128 address to subnet 2 and 128 addresses to subnet 3. But I dont't want to be given the whole answer, I would just like to be shown how to do it so I know for future reference, unfortunately I don't understand how my lecturer is explaining it – oneman Sep 4 '16 at 3:43
  • Look here at this chart diagram and the images: cliffordfeta.tripod.com/supernettingcidr_analysis.htm... Google some terms such as supernetting, IP Addresses in binary, and octets each octet consists of 8 piece on or off 1 or 0... read up on that some to understand. – Pimp Juice IT Sep 4 '16 at 3:44
  • thank you, that sheet has helped already, I hadn't heard of supernetting – oneman Sep 4 '16 at 3:55
  • @PIMP_JUICE_IT so if I have understood correctly. For 130.120.10.0/23. My network range would be 130.120.10.1 - 130.120.11.254 ? – oneman Sep 4 '16 at 4:03
  • For Network: 130.120.10.0/23 you have HostStart: 130.120.10.1 and you have HostEnd: 130.120.11.254 where your Netmask: 255.255.254.0 = 23 11111111.11111111.1111111 0.00000000, and yes 510 hosts. – Pimp Juice IT Sep 4 '16 at 5:44
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For the /24 subnet here is your IP range

192.168.0.1 - 254 (254 host addresses)

192.168.0.0 - network address

192.168.0.255 - broadcast address

For the /23 subnet here is you IP range

Network 1

172.16.0.1 - 172.16.1.254 (510 host addresses per network)

172.16.0.0 - Network Address

172.16.1.255 - Broadcast Address

Network 2

172.16.2.1 - 172.16.3.254 (510 host addresses per network)

172.16.2.0 - Network Address

172.16.3.255 - Broadcast Address

Network 3

172.16.4.0

Network 4

172.16.6.0

And so on...

  • To assign 256 host addresses you need to use 510 host per network (/23) since /24 will only give you 254 hosts per network, missing two hosts. – D007 Sep 4 '16 at 3:52

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