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Using more then 254 IP addresses on same network (multiple subnets)

Posted on 2009-04-07
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Last Modified: 2012-05-06
I have a client who has 90 network drops. They go into multiple offices. The idea is that they will be leasing office space to small companies/non-profit organizations.

 We will be using a NETGEAR FVX538 router.

My problem is this. Some of those office areas may have 3-4 computers, and then maybe 1 printer and maybe a vonage device, so thats about 5-6 IP's for one office. As you can imagine we'd run out of useable IP addresses on our standard 192.168.1.xxx network.

Is there a way I can span across multiple subnets to assign more ip's say a range of:

192.168.1.xxx - 192.168.2.xxx which will give me roughly 500+ ip's I can use?

I know it can be done, my main question is: How do I do this, as i've never really used more then 1 subnet before. What settings should I use. Its a brand new network, nothing installed on it yet.
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Question by:leemandelintralogic
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by:mfhorizon
ID: 24094450
This article will be a good help for you, just try:
http://www.tomshardware.co.uk/forum/page-21918_17_0.html
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by:mfhorizon
ID: 24094453
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mfhorizon earned 500 total points
ID: 24094469
HOW TO CREATE A SUBNET is not that easy to comlete in one sitting but this link can give you VERY GOOD OVERVIEW:
http://articles.techrepublic.com.com/5100-10878_11-6089187.html
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by:vertsyeux
ID: 24094474
I would say to configure the router for class B operation (by setting its LAN-side mask to 255.255.0.0 instead of 255.255.255.0) and use a class B IP address like 10.0.100.100 - this would give you 65533 addresses instead of 253. However, I took a quick look at your router spec, and it does say it supports only 253 users, so it may not support class B operation on the LAN side.. In our office we can have up to 4 local subnets (eg 192.168.1.xxx, 192.168.2.xxx etc) on our Draytek 2900 router, but I didn't see any facility for virtual LAN's on your router's spec.. I think you should phone Netgear and avail yourself of their technical support..
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by:andy_deru
ID: 24095541
In the beginning all IP addresses are grouped in classes A (mask8), B (mask16), C (mask24). These days most network devices, including computers NIC, are capable to subnet or supernet addresses.
What you want is supernets 2 class C. Your 192.168.1.x-192.168.2.x  happens not to be in the same subnet in mask 23 (which contain double as much hosts of mask24).
If you google on IPCALC, you will find some online tools to calculate.
http://jodies.de/ipcalc?host=192.168.1.0&mask1=23&mask2=
result:
Address:   192.168.1.0           11000000.10101000.0000000 1.00000000
Netmask:   255.255.254.0 = 23    11111111.11111111.1111111 0.00000000
Wildcard:  0.0.1.255             00000000.00000000.0000000 1.11111111
=>
Network:   192.168.0.0/23        11000000.10101000.0000000 0.00000000 (Class C)
Broadcast: 192.168.1.255         11000000.10101000.0000000 1.11111111
HostMin:   192.168.0.1           11000000.10101000.0000000 0.00000001
HostMax:   192.168.1.254         11000000.10101000.0000000 1.11111110
Hosts/Net: 510                   (Private Internet)
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