supernetting question

newtechie1988

could somebody please help me with a supernetting question?
What is the mask required to supernet 4 networks starting at 192.168.8.0/24
Could somebody please go through the steps to get the correct answer?
Thank you

chX

Hi,

Hopefully I don't come across as rude - because I'm not intending to be - but didn't you create another thread with the same question? There's replies in there, so in future it might be worth posting in there again if you're still not sure.


So, you're wanting to summarize 4 networks beginning with 192.168.8.0/24, which gives us:

192.168.8.0/24
192.168.9.0/24
192.168.10.0/24
192.168.11.0/24


A lot of people find that the easiest way to figure this out initially is to convert it into binary, like so:

11000000.10101000.00001000.00000000
11000000.10101000.00001001.00000000
11000000.10101000.00001010.00000000
11000000.10101000.00001011.00000000

Now, as they're /24s, the third octet is the one we're interested in. Find the point where the binary matches up.


11000000.10101000.00001000.00000000
11000000.10101000.00001001.00000000
11000000.10101000.00001010.00000000
11000000.10101000.00001011.00000000

Mentally (or physically, if you're writing this down) draw a red line between the bold and normal text. Count the number of bits up until that point, and you have your new subnet mask.

In this case, there's 22 bits - so our subnet mask is /22, or 255.255.252.0.


The answer is thus 192.168.8.0/22.

This encompasses 192.168.8.0 through to 192.168.11.255.


I hope that helps. It's been ages since I've even looked at summarization, so hopefully that's all correct. As always, let me know if something is not quite right.

AvidNetworker

I just look at it like this. You need 4 /24 networks, right? a /24 network is 256 hosts(-2 of course)

so look at it this way

/23 =512 = 2 networks
/22 =1024 = 4 networks
/21 = 2048 = 8 networks

and so on.

So you need 1024/4 = 256, which means you need a /22

This is all stuff you can do in your head if you think of it that way. You just need to think of each octet on its own, and how it relates to the following octet.