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Increasing DHCP IP addresses to an existing LAN

Hi,

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 255.255.255.0 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 255.255.255.254? I believe that will then  increase the available DHCP to include 192.168.6.x range as well. Correct?

or

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?

Thanks,
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cfgchiran
Asked:
cfgchiran
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2 Solutions
 
rauenpcCommented:
If you change the subnet mask to 255.255.254.0 (I assume that's what you meant), this would make the subnet contain IP's 192.168.4.0-192.168.5.255, not 192.168.5.0-192.168.6.255. So this would not work.

You will need to setup a separate subnet and do some routing, or change the subnet to 192.168.6.0/23 which would give you 192.168.6.0-192.168.7.255. Either choice requires that the other sites be updated with the new subnet on both MPLS and VPN.
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cfgchiranAuthor Commented:
Thank you for the comment. Why is it not possible to extend 5.1-6.255 with a subnet change? (By making it 192.168.5.0/23)?

Thanks,
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cfgchiranAuthor Commented:
And yes, I meant 255.255.254.0. :) Thanks.
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rauenpcCommented:
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 (192.168.5.0)
11111111 11111111 11111110 00000000 (255.255.253.0)
equals a subnet of
11000000 10101000 00000100 00000000 (192.168.4.0)

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 192.168.4.0-192.168.5.255.

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.
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cfgchiranAuthor Commented:
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 192.168.4.0/23 192.168.6.0/23 gives you a range of 4.0-5.255 and 6.0-7.255 respectively, why doesn't the same option work for 192.168.5.0/23?

If we were to say 192.168.10/23 would it not be 192.168.1.0-192.168.2.255?
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cfgchiranAuthor Commented:
I meant 192.168.1.0/23
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cfgchiranAuthor Commented:
I just used a subnet calculator and I actually do understand it better. So thank you for your help.
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