DHCP ADDRESS POOL

[abarb002]

Question for the community concerning ip address space for a network with 512 addresses. 

I have attempted to add a large dhcp pool of 512 address spaces for one of my vlans. We are a training facility and that large number of addresses were needed at the time. I attempted to use 192.168.135.0 and combine with 192.168.136.0 with the final address being 192.168.135.0/23 with a mask of 255.255.254.0. When I attempted to add this vlan to the switch and or dhcp I got rejected saying that the address space overlapped with another network, thus dhcp wanting to create a super pool. When i used a subnet calculator I noticed when entering 135.0 as a /23, the range covered becomes 192.168.134.0 255.255.254.0. 

Can anyone enlighten me as to why this happens? 

Was able to work around this with 2 other network spaces, but the 135.0 network was the only network space to do this. 

[balaji.bandi]

if you like 192.x.x.x/16 network in to /23 subnet - then you need to use 192.168.134.0/23.

if you like to like use 192.168.135.0  and 192.168.136.0  network you have option only /24

here is the good GUI for subnetting.

http://jodies.de/ipcalc?host=192.168.0.0&mask1=16&mask2=255.255.254.0

[Seb Rupik · Accepted solution]

Hi there,

When using a /23 netmask, 192.168.135.0 is at the end of the subnet boundary. As you point out, the subnet-id for 192.168.135.0/23 is actually 192.168.134.0 .

Due to the mathematics involved with subnet boundaries, if you wanted to combine 192.168.135.0 and 192.168.136.0 into a single subnet you would need to use a /20 mask, giving a subnet-id of 191.168.128.0, and a range of 192.168.128.0-191.168.143.255 .

cheers,

Seb.