Supernetting Question !!

Hi Guys, i had this question in one of our IJP's, but even after a lot of thought i am not able to get to an answer on this one. Can anyone help me out with the answer.

The question is:

The IP adrress that is provided is 192.168.1.150, the default gateway is 192.168.0.1, the broadcast Address for the current subnet is 192.168.2.255. What would be the subnet mask ?

Hi Tarun,

Looking at this quickly, I would guess it would be a class C subnet which is 255.255.255.0.

But if you have a look at the home page on the site, at the top there is a "tab" called "networking". If you have a look through this, you will find lots of useful stufff on Networking and protocols.

Hope this helps.

Hi Tarun,

I forgot to mention something, so here it is.

The general rule for deciding what subnet an IP address should be on is this.

If the first Octet of the IP address decides the Subnet, I've listed the guide below:

Class A - 1 - 126
Class B - 128 - 191
Class C - 192 ->

But, like I said before, have a look at the networking stuff on this site, it really helps

Thanks Rock,

But i am still looking for an answer because i am not able to figure out the answer to this.

I think the subnet mask should be 255.255.253.0

mask 255.255.255.0 gives a broadcast of 192.168.0.255
mask 255.255.254.0 gives a broadcast of 192.168.1.255
mask 255.255.253.0 gives a broadcast of 192.168.2.255
mask 255.255.252.0 gives a broadcast of 192.168.3.255

but check out the guide that Rockape pointed you at at the top of the page its great

Well, 255.255.253.0, i doubt if that can ever be a subnet mask.

The only valid subnet mask values in the 3rd octet that i know of are:

128, 192, 224, 240, 248, 252, 254

253 is not even a valid subnet value.