Increasing DHCP IP addresses to an existing LAN

Posted on 2014-07-30
Last Modified: 2014-07-30

I have a LAN with the IP scheme 192.168.5.x connected via MPLS and VPN to multiple other physical sites with individual subnets 192.168.1.x, 192.168.2.x, 192.168.3.x and 192.168.4.x. Subnet mask is for all.

I need to increase available DHCP addresses of just the 5.x network, and allow the other subnets to continue to communicate with IPs on the 5.x network.

Is it better to:

1) change the subnet mask on 5.x to I believe that will then  increase the available DHCP to include 192.168.6.x range as well. Correct?


2) Set up a separate 6.x network and tie it into the 5.x (and other subnets) using a router?

I would prefer the first option, but will that have an effect on traffic speed across the LAN and the WAN? Will the 6.x addresses be able to communicate with 1.x and 2.x etc...? Or does it require a routing scheme update on the Cisco site to site routers?

Question by:cfgchiran
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Accepted Solution

rauenpc earned 500 total points
ID: 40230154
If you change the subnet mask to (I assume that's what you meant), this would make the subnet contain IP's, not So this would not work.

You will need to setup a separate subnet and do some routing, or change the subnet to which would give you Either choice requires that the other sites be updated with the new subnet on both MPLS and VPN.

Author Comment

ID: 40230243
Thank you for the comment. Why is it not possible to extend 5.1-6.255 with a subnet change? (By making it


Author Comment

ID: 40230261
And yes, I meant :) Thanks.
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LVL 20

Assisted Solution

rauenpc earned 500 total points
ID: 40230353
It all comes down to the binary of it. A subnet mask is simply a bunch of one's followed by zero's. That is matched up with the IP/subnet. Essentially, the 1's of the subnet mask are the "care bits" and the 0's are the "don't care" bits. We would determine the subnet by matching the "care bits" with all the bits of the IP address in question

11000000 10101000 00000101 00000000 (
11111111 11111111 11111110 00000000 (
equals a subnet of
11000000 10101000 00000100 00000000 (

All of the remaining bits would be host bits, which in this case are the last 9 bits
11000000 10101000 00000100 00000000
If you fill out all the possible combinations of the host bits, you end up with the IP range of

If you were to read a book on the topic of subnetting (assuming that the lesson will be binary based and not block based), the author will likely have a much better way of explaining it. It's been a long time since I sat through a subnetting lecture.

Author Comment

ID: 40230513
Thank you again.

I have always had some trouble following this binary, and have tried to read up on it on-line just now as well.

I think the part that's confusing for me is if gives you a range of 4.0-5.255 and 6.0-7.255 respectively, why doesn't the same option work for

If we were to say 192.168.10/23 would it not be

Author Comment

ID: 40230516
I meant

Author Comment

ID: 40230518
I just used a subnet calculator and I actually do understand it better. So thank you for your help.

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