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My question is different from the one being marked as duplicate of. The problem with my question arises when the octet in which I'm working in is used up but there are still networks left to be assigned suitable addresses.

I have 4 networks connected to 1 router. Number of users on each network are; 99, 77, 13 and 5. The starting or working network address is 192.168.10.0/24. Using VLSM, I'm supposed to get each network suitable range of host addresses. I started from the network with the highest users and realized by the time i got to network with 13 users, all the host addresses in the 4th octet were used up. So i decided to enter the 3rd octet since all the hosts there are not used up and i get network address of 192.168.11.0/28. Am i correct? Or in this case, i should have started from network with the smallest users?

Richardson Ansong
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  • "_The problem with my question arises when the octet in which I'm working in is used up but there are still networks left to be assigned suitable addresses._" Octets play no part in this. IP addresses are 32-bit binary numbers, and you must do this in binary. The dotted decimal notation is simply to make it easier for humans to read, but you will fail if you try to do this in decimal. The answer to the previous question explains how to do it correctly. – Ron Maupin Apr 06 '17 at 20:19
  • @RonMaupin calculating prefix, network, subnet, and host numbers was not the problem. Which network to start from was the problem. I just wanted to know if it's best to start from the smallest network and build up or vice versa! – Richardson Ansong Apr 06 '17 at 23:34
  • Per my calculations, I arrived at the same solution as @SujeetSingh in his comment under his answer and i modeled the whole thing in packet tracer which worked perfectly. I set up the RIP of the router to the common network address which is 192.168.10.0/24 and I was able to ping from a PC in 192.168.10.0/25 network (99 users) to a PC in 192.168.10.128/25 network (77 users) and also in 192.168.11.0/28 network (13 users). Earlier, I had not tested my proposed solution out in packet tracer. – Richardson Ansong Apr 06 '17 at 23:36
  • Then, it is a duplicate of http://networkengineering.stackexchange.com/q/28486/8499. You question about the octets lead one to believe you are not doing this in binary because octets play no part in subnetting, and trying to use them when subnetting leads to errors. IPv4 addresses are 32-bit binary numbers, and subnetting should use them as such; forget the dotted decimal notation when subnetting. – Ron Maupin Apr 06 '17 at 23:38
  • I use a different method when calculating which is as accurate as the one you directed me at and it involves working with the dotted decimal notations. – Richardson Ansong Apr 06 '17 at 23:43
  • Then you are making a foolish mistake. I have seen everyone who does it that way get into trouble. Using binary numbers in decimal leads to errors, and you will get in trouble at some point. – Ron Maupin Apr 06 '17 at 23:45
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/56690/discussion-between-nana-osaberima-and-ron-maupin). – Richardson Ansong Apr 06 '17 at 23:46

2 Answers2

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  • A /25 can normally support 125 end users.
  • A /26 can normally support 61 end users.
  • A /27 can normally support 29 end users.
  • A /28 can normally support 13 end users
  • A /29 can normally support 5 end users
  • A /30 can normally support 1 end users

So to do this with traditional subnetting (even with variable subnet masks) assigning one subnet to each network you would need two /25's, a /28 and a /29. That isn't going to fit in your /24.

So how can we solve this problem? I see a couple of options.

  1. Assign two subnets to the network with 77 users. If we assign it a /26 and a /27 we cover it's users while leaving a /26 free for further subdivision. Downside is that some formerly local traffic will end up going via the router and local firewalls may make incorrect assumptions about what is on the local network.
  2. Use proxy arp on the subnet with 77 users. We can then assign smaller subnets which overlap with unused parts of that subnet and proxy arp will pick up the traffic. Again local firewalls may make incorrect assumptions about what is on the local network. Proxy arp can also magnify the problems experienced in the event of a misconfiguration.
Peter Green
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  • Shouldn't the router also be assigned an address using vlsm? – Richardson Ansong Apr 06 '17 at 19:30
  • The router will normally need an address on each subnet it serves, I have allowed for this in the number of client addresses I have listed for each subnet size. If the router has an upstream link that may need addressing too but we have been given no information on that so I assumed that it is not intended to be addressed out of this address pool. – Peter Green Apr 06 '17 at 20:08
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You can use following network/subnet mask

192.168.10.0/25 for 96 and 77 users

192.168.11.0/28 for 13 users 192.168.11.0/29 for 5 users

Sujeet Singh
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    You can't use a /25 network for 96 users and the same /25 network for 77 other users. – JFL Apr 06 '17 at 18:52
  • The network/ subnet mask will be like this. – Sujeet Singh Apr 06 '17 at 19:02
  • For Network 1 for 96 users:- Network: 192.168.10.0/25 Broadcast: 192.168.10.127 HostMin: 192.168.10.1 HostMax: 192.168.10.126 Hosts/Net: 126 For Network 2 for 77 users:- Network: 192.168.10.128/25 Broadcast: 192.168.10.255 HostMin: 192.168.10.129 HostMax: 192.168.10.254 Hosts/Net: 126 – Sujeet Singh Apr 06 '17 at 19:04
  • @JFL Actually, I think you can. With 99 users, usable IP addresses would be 192.168.10.1/25 to 192.168.10.127/25 and 77 users, usable IP addresses would be; 192.168.10.129/25 to 192.168.10.254/25 – Richardson Ansong Apr 06 '17 at 19:32
  • please reread my previous comment. The key word is "same". – JFL Apr 06 '17 at 19:44