Need help with supernetting

This is an area I need work in. (Getting ready for ICND2)

The following networks are present:

10.0.4.0/24
10.0.5.0/24
10.0.6.0/24
10.0.7.0/24

You would like to summarize these
networks, what address/mask should you
use?

A.10.0.4.0/22
B.10.0.4.0/26
C.10.0.4.0/24
D.10.0.4.0/25
E.10.0.4.0/23
F.10.0.0.0/21

Answer: A

I do understand the concept of
auto-summary and why it's sometimes
needed and why sometimes it isn't
needed (classful boundaries), but
can someone please advise on the
process of obtaining the correct answer
in this type of question?

I usually do all the subnetting in
my head, and without binary.

Thanks!

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Comments

oli356

I think this is right, hope it is. I am CCNA now! If anything my explanation might be confusing but I'm sure someone can explain it better

00001010.00000000.00000100.00000000
00001010.00000000.00000101.00000000
00001010.00000000.00000110.00000000
00001010.00000000.00000111.00000000

So those are the address in binary (4.0, 5.0, 6.0, 7.0)

I have put the 22nd bit in bold. If the subnet mask was /22 the increment would be 4. If we are on 10.0.4.0 when the increment is 4, this covers the 4,5,6,7.0 networks.
The subnet mask could be /21 but then we are advertising the 8.0 , 9.0, 10.0.11.0 network as well. So you could summarize with /21 but /22 keeps it more "secure" and only advertises the networks what the question asks you to.

cpartin

I like to think of it this way, in terms of the number of bits you'd have to borrow. When I say "equals", I mean that X consecutive subnets of that size can be "contained" within the larger supernet. They are not exactly equivalent (bonus question: why aren't they?). There are 2 assumptions here.

1) The subnets are contiguous. If they're not you have to factor in the gaps.
2) The subnets fit within the boundaries of your smallest supernet. This is explained further below.

Two /24's = One /23 (2^1=2, borrow 1 bit)
Two /23's = One /22
Four /24's = One /22 (2^2=4, borrow 2 bits)
Eight /22's = One /19 (2^3=8, borrow 3 bits)
Odd numbers - Three /24's can be contained in one /22 (There is room for an extra /24, but a /22 is the smallest container which will hold everything we're trying to summarize)

In your example we have four /24's so you know you need to borrow at least 2 bits (a /22) to cover all of the subnets. Why do I say at least? Well if the subnets you're trying to summarize don't fall completely within the boundaries of your smallest supernet then you'll have to borrow another bit. In your case it does, but say you are trying to summarize the following:

10.0.3.0/24
10.0.4.0/24
10.0.5.0/24
10.0.6.0/24

You still have four /24's, but a /22 in our range can only begin at 10.0.0.0/22 or 10.0.4.0/22, either of which would chop up the networks we're trying to summarize. So we have to borrow an additional bit and we summarize this block as 10.0.0.0/21 (10.0.0.1-10.0.7.254). This is somewhat wasteful, but it's the only way to include all of our addresses.

For more complex scenarios I find the least common denominator. Say you want to summarize:

172.16.0.0/22
172.16.4.0/23
172.16.6.0/24
172.16.7.0/24

A /22 is equal to four /24's, a /23 is equal to two /24's, plus our two /24's total up to 8 /24 networks. 2^3=8 so we need to borrow 3 bits and we would summarize this as 172.16.0.0/21, checking to make sure all subnets are within our bounds, which they are. Once you get faster at this you'll just be able to look at the relevant octet and masks and see you need something that starts at 172.16.0.0 and covers through 172.16.7.0/24.

sratakhin

Come on, why does it have to be so long and complicated?
I never use binary when I do subnetting but to each their own.
You have four extra networks. How many bits do you need to get number 4? Just two. Hence, change your subnet mask from /24 to /22.

oli356

2 bits to get to number 4, hu?

11 = 3 in decimal.
100 = 4 in decimal.

sratakhin

Since when do we start counting from 1 in networking?

Ok, let's do it in binary:
100 - network .4
101 - network .5
110 - network .6
111 - network .7

Only the last 2 bits change, so that's all we need.

atorven

Where can I get decent supernetting practice questions, I find the above examples too neat.