subneting

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Guest

hello out there.
i have to seperate one network 192.168.15.0/28 into two of differten size
without changing IPs! I need the hosts from 192.168.15.1 to 192.168.15.190 to
be in one network and the hosts from 192.168.15.194 to 192.168.15.254 to be
in the other network. Now I´m not really sure if that can be done. I thought
of the following subneting:

Network 1 (small):
192.168.15.192/255.255.255.192
Hosts: 192.168.15.194 -192.168.15.254
GW: 192.168.15.193
BC: 192.168.15.255

Network 2 (big):
192.168.15.0/255.255.255.64
Hosts: 192.168.15.2 -192.168.15.190
GW: 192.168.15.1
BC: 192.168.15.191

I´d appreciate if someone could just tell me if this subneting is correct
and in case it is not some hints how i could achive this separation (.. or if
it is just impossible environment..)

thanks a lot in advance, Mike
 
You would need a router between the subnets. Shame you don't have managed
vlan capable swithes. No subnetting required with vlans.

Have you used one of the online subnet calculators?

255.255.255.64 is not a class c subnet. Class c goes 0, 128, 192...
 
freytshy said:
hello out there.
i have to seperate one network 192.168.15.0/28 into two of differten size
without changing IPs! I need the hosts from 192.168.15.1 to 192.168.15.190 to
be in one network and the hosts from 192.168.15.194 to 192.168.15.254 to be
in the other network. Now I´m not really sure if that can be done. I thought
of the following subneting:

Network 1 (small):
192.168.15.192/255.255.255.192
Hosts: 192.168.15.194 -192.168.15.254
GW: 192.168.15.193
BC: 192.168.15.255

Network 2 (big):
192.168.15.0/255.255.255.64
Hosts: 192.168.15.2 -192.168.15.190
GW: 192.168.15.1
BC: 192.168.15.191

I´d appreciate if someone could just tell me if this subneting is correct
and in case it is not some hints how i could achive this separation (.. or if
it is just impossible environment..)

thanks a lot in advance, Mike
Click to expand...

I hate to do other people's homework, but...

A subnet mask of 255.255.255.192 splits the last octet into 4 parts
(subnets). The subnets would be (given the IP addresses you provided)
192.168.15.0
192.168.15.64
192.168.15.128
192.168.15.192

So you can't actually achieve what you are trying to do. Using the
192.168.15.192 subnet gets you .193 - 254 in one subnet, but the closest
you can come to getting the remaining addresses into a single subnet is
with a mask of 255.255.255.128.

192.168.15.0 mask 255.255.255.128 gives you .1 - .126.

BTW, a /28 is mask 255.255.255.240 and is a subnet of 16 IP addresses
(14 usable), so the original premise of your question is incorrect.
Better read the homework question again!


....kurt
 
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