VLSM, Subnetting and the CCNA..

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Here is my question. I'm just curious if on the CCNA (the 802) the subnetting questions mainly deal with class C networks.

I feel comfortable with all three but was just curious. I assume that due to time they won't ask you to do a VLSM layout with a class A.

Anyway, thanks in advance. Thinking of taking the test in about 30 days. Can't wait.

Thanks
Bruce

 

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Getting too specific would risk violating the NDA. But, you do need to be just as comfortable working in the second and 3rd octet as you are working in the 4th octet when it comes to subnetting.

 

Fair enough. Thanks.

Bruce

 

class c networks i can handle but subnetting class a and b networks makes me feel .... dumb
so... study study study... we *have to* know this stuff, right?! and relying on a subnet calculator isn't practical as you might not have it available when you need it.

good luck everyone!

md

 

Personally I find working in the 4th octet slightly more difficult.
in the 2nd and third octet you always have:
network address: a.x.0.0 or a.b.x.0
first host address: a.x.0.1 or a.b.x.1
last host address: a.y.255.254 or a.b.y.254
broadcast address: a.y.255.255 or a.b.y.255
where y=x+block size-1

You can see the pattern there. Not only is it slightly less complicated it also means it's easy to check if you've made a mistake.

It's all about the block size. If you know the block size the rest is easy.
calculating block size
If you're given the subnet mask, subtract the value in the significant octet (not 255 or 0) from 256, that's your block size.
If you're given it in CIDR notation, subtract the /x number from 32. This is the number of host bits. Subtract 8 until you have a number less than 8, that's the number of host bits in the significant octet.
Count up, multiplying by 2:
1=2
2=4
3=8
4=16
5=32
6=64
7=128
until you reach the number of host bits in the significant octet.

example:
subnet mask: 255.255.240.0 is the same as /20
block size by subnet mask:
256-240=16
block size 16
block size by CIDR
32-20=12
12>8 so subtract 8 (these 8s are the .0 octets in a subnet mask)
12-8=4
count up
1=2
2=4
3=8
4=16
block size 16

 

Have a click on the jpeg attachment, as mentioned, it's all about the increment size.

The easiest way to work out your increment size though, is simply to find out which bit was reserved for network use last and look at its binary placement value. i.e if it's a /28, then keeping in mind we are going from left to right, the 28th bit has a placement value of 16 so increments of 16. Simples.

If you like using the powers of 2 method then:

If the mask is above a /24, e.g. 172.16.18.16 /29, then deduct your mask from 32 and use the powers of 2 against the result to get the increment.

e.g 32 - 29 = 3.
2^3 = increments of 8 in last octet.

If your mask is less than a /24, then deduct your mask from 24.
e.g. 10.10.0.0 /20

24 - 20=4
2^4= increments of 16 in third octet.

If your mask is less than a /16, then deduct from 16.
e.g. 4.0.0.0.0 /13
16-13 = 3
2^3= increments of 8 in second octet.