Subnetting: Why am I always 2 off???

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Hey guys,

Just got to the subnetting chapter of the AIO and have been doing practice questions online re: subnetting, not having too much problems with it but there is one thing that i am consistently doing wrong and i don't know why, for example:

Which subnet does host 172.30.99.75/21 belong to?

The way i would work it out would be to put the subnet mask into binary so:

11111111.11111111.11111000.00000000

from the 3rd octet i would work out that the intervals for the subnets would be of 7 so the closest divisor of 7 to 99 would be 98, so 172.30.99.75/21 sits on 172.30.98.0..... in my crazy mind at least!

The answer according to the site is .30.96.0 and the other similar questions i have done also resulted in me being 2 off so i must be doing something incredibly simple wrong somewhere!

BTW forgive me if my working out is a bit messy, the AIO chapter and methods confused the hell out of me so my subnetting knowledge comes from various online sources

 

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Subnet intervals are always powers of 2. In this case, it'd be 8...

0 to 7
8 to 15
16 to 23...

...etc.

I think you're seeing the 7 and assuming that there's 7 digits from 1 to 7... but you're forgetting 0.

In this scenario, the closest multiple of 8 (without going over) is 96.

 

use your binary chart

3rd oct
The numbers line up in my text box but not when i publish it so sorry for the binary chart not displaying correctly
128 64 32 16 8 4 2 1
1 1 1 1 1 0 0 0

look whats above your last 1 of the subnet mask

or 256-248=8

not 255 as boson 0 is a value

Jumps would be
0-7
8-15
16-23
24-31
32-39
40-47
48-55
56-63
64-71
72-79
80-87
88-95
96-103
104-111
and so on

 

If you make a chart like this:-

128 64 32 16 8 4 2 1 0

This can help you when working out the whole subnet

so say the IP and subnetmask is 192.168.1.0/24 or 255.255.255.0

You have 1 IP address and you have 4 networks with the subnet mask given we can only have 1 network so you need to steal some bits and make the subnetmask into 255.255.255.192 which /26 this will give us the 4 networks.

so to work out the subnet (if you do this a few times it will start to sink in). You need to steal some bits from one side and give to the other so if you use your binary chart you have the IP

192.168.1.0 = 11000000.10101000.00000001.00000000
255.255.255.192 = 11111111.11111111.11111111.11000000 (the two ones in the final octect are the subnet leaving 6 bits which is 62

so using the chart that gives us 62 hosts and 4 networks.

Hopefully that makes a bit of sense subnetting is a bit confusing when you first start to do, I'm still not good at it yet.

 

This

Looks to me as well that you're looking at the 3 zeros at the end of 11111000 and seeing 7 different combinations rather than eight.

000
001
010
011
100
101
110
111

I suppose you could always take the subnet mask binary number 11111000 convert it to decimal 248, then subtract 248 from 256, this will give you the correct number of intervals in the octect.