newb subnetting question :-) help

chickenlicken09

Guys,

New to subnetting, i understand whats being done on the below question but why does it sit on the 192.168.12.72 subnet?
What am i missing? Its prob something really simple.




What subnet does 192.168.12.78/29 belong to?

You may wonder where to begin. Well to start with let's find the next boundary of this address.

Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc

Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

networker050184

Where else would you expect it to be?

chickenlicken09

192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64

Why .72?

NetworkVeteran

eddo1 wrote: »

Guys,

New to subnetting, i understand whats being done on the below question but why does it sit on the 192.168.12.72 subnet?
What am i missing? Its prob something really simple.

If the explanation you quoted wasn't enough to understand, I'd recommend writing out the address 192.168.12.78 in binary, then the mask 255.255.255.248 in binary just beneath in. The host bits will be the mask bits that equal zero. If you convert all the host bits in the address to zeroes, you'll get the subnet address. Being able to work in binary is a good skill in networking.

(You might also try converting all the host bits in the address to ones, to get the broadcast address!)

I won't do this myself just now, as the learning is in the doing, not the watching.

networker050184

The valid hosts are the numbers between one subnet and the next. The . 72 subnet has host .73-.78. The network and broadcast addresses are .72 and .79. The next network starts at .80 and it keeps going in the same fashion.

Zartanasaurus

32 40 48 56 64

What's the next number in this sequence?

chickenlicken09

Zartanasaurus wrote: »

32 40 48 56 64

What's the next number in this sequence?

Yes i see the increments of 8, thats fine. But why .72

Xyro

eddo1 wrote: »

Why .72?

Because...

The ranges are jumping by 8 in the 4th octet (since this is the changing octet), so the 78 (which is in the 4th octet) will fall into the 192.168.12.72 - 192.168.12.79 range.

For me, it helps to look at the SM.
In this example you have 255.255.255.248 <--- The octet that is "split" here is your changing octet.
256-248 = 8

The rest is just simple division.

Let me know if you're still unclear. This is a bit of a rushed explanation because I'm OMW out.

chickenlicken09

Ah yes, the penny has dropped lol.

The ranges are jumping by 8 in the 4th octet (since this is the changing octet), so the 78 (which is in the 4th octet) will fall into the 192.168.12.72 - 192.168.12.79 range.

Seeing it like this it jumped out straight away. Thanks Xyro, and the rest of ye

boredgamelad

The bold numbers are your subnet IDs:

...
48
56
64
72 < subnet ID
73 < host
74 < host
75 < host
76 < host
77 < host
78 < host
79 < broadcast
80

78 is between 72 and 80. So 78 belongs to the .72 subnet.