subnetting problem

[jaymare33]

I have the following problem. Think I have an inside ip address on my pix 192.168.1.128/25.. now I have to subnet so that I have 1 32 ip subnet, 1 64 ip subnet end 4 8 ip subnets. Is it possible? The reason I'm asking is that I can't subnet 64-32-8-8-8-8 but 32-64-8-8-8-8. I have to add some other insides on my pix to keep traffic phisically in different subnets.

[lgijssel]

IP's are binary oriented. You cannot subnet 32-64-8-8-8-8.

You cannot assign a single block of 64 over a 64-border.

When this does not suit your network you will have to make some re-arrangements on the inside to allow division of IP's.

Regards,

Leo

[Bobby Thekkekandam]

To add to what Leo said,

64-32-8-8-8-8 will work, but 32-64-8-8-8 will not due to the way bit boundaries work.

192.168.1.128 /27 will give you the following subnet:

network: 192.168.1.128 broadcast: 192.168.1.159 for a total of 30 hosts.

192.168.1.160 /26 will give you the following subnet:

network: 192.168.1.128 broadcast: 192.168.1.191 for a total of 62 hosts.

However note that 192.168.1.160 lies in the middle of this subnet, overlapping with the first subnet. Because of the bit boundaries, when you use VLSM subnetting, you have to hierarchily break them down from largest to smallest. This is why you can have

64-32-8-8-8-8 but not 32-64-8-8-8-8

HTH,

Bobby

*Please rate helpful posts.

[kamlesh.sharma]

I hope below ip subetting can give you solution.

192.168.1.128/25

192.168.1.128/26 128-191 62

192.168.1.192/27 192-223 30

192.168.1.224/29 224-231 6

192.168.1.232/29 232-239 6

192.168.1.240/29 240-247 6

192.168.1.248/29 248-255 6

HTH

[jaymare33]

no this way it s not possible because 192.168.1.211 and 192.168.1.168 have to be in the same subnet.. so the only way is subnetting 96 .. 8 .. 8 .. 8 .. 8 .. let me se if it is possible..