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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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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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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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