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mikeyb4760
Premium Member
join:2004-09-25
El Cajon, CA

mikeyb4760

Premium Member

Route Summarization

I was working on a CCNA type simulation exam and a route summarization question came up. It's a time killer changing all these routes to binary.. Anybody know of a way to figure these out really fast?

Thanks
meta
join:2004-12-27
00000

meta

Member

if you see 2 /24's you can summarize them into a /23. the same applies for /23's as a pair can be summarized into a /22, meaning there are 2*2=4 /24s in a /22.

That kind of simple way to think about subnetting and summarization is how i do it in my head, not sure if it makes sense to other people though. I have never converted a prefix to binary to try and summarize though, not sure what method you are using.
HELLFIRE
MVM
join:2009-11-25

HELLFIRE to mikeyb4760

MVM

to mikeyb4760
Got some example questions? I'm a brute force kinda guy, so I'll convert
the address into binary, but the trick I use is only convert the octet
that the mask covers -- ie. for a /16 only convert the 2nd octet, for
a /24 only convert the 3rd octet. Also make SURE you don't forget about
common 0's when summarizing... I always found that to be the killer when
quizzed about summarization.

Regards
aryoba
MVM
join:2002-08-22

aryoba to meta

MVM

to meta
said by meta:

if you see 2 /24's you can summarize them into a /23. the same applies for /23's as a pair can be summarized into a /22, meaning there are 2*2=4 /24s in a /22.
Note that this method only apply when the smaller subnets of /24 or /23 are contiguous, such as 192.168.0.0/24 and 192.168.1.0/24. When the smaller subnets are not contiguous such as 192.168.7.0/24 and 192.168.8.0/24, then route summarization (supernetting in this case) need to take deeper consideration of those smaller subnets that you don't have (in this case 192.168.2.0/24 - 192.168.6.0/24 and 192.168.9.0/24 - 192.168.15.0/24).
meta
join:2004-12-27
00000

1 edit

meta

Member

Yep aryoba, i neglected to mention that. They need to be "summarizable" by the larger network. There is also the option of over-summarizing intentionally.

mikeyb4760
Premium Member
join:2004-09-25
El Cajon, CA

mikeyb4760

Premium Member

Here was the sample exam question I was presented with:

10.1.1.0/30
10.1.1.4/30
10.1.1.8/30
10.1.2.0/23
10.1.4.0/24
10.1.5.0/25

And the multiple guess was:

a) 10.0.0.0/21

b) 10.1.0.0/22

c) 10.1.0.0/21

d) 10.1.1.0/21
meta
join:2004-12-27
00000

meta

Member

Obviously that will never summarize cleanly, im guessing they wanted the most specific summary that covered all of them though, so my guess is C.

phantasm11b
Premium Member
join:2007-11-02

phantasm11b

Premium Member

If I remember correctly the answer is B. The reason being is that the first 2 octets are the same so you would have 16 network bits right there. Then in order to reach the number 5 you would need to add an additional 6 bits which would give you a total of 22 network bits and thus resulting in a netmask of 255.255.252.0. Of course I haven't had my coffee yet so we'll see what someone else says.

usa2k
Blessed
MVM
join:2003-01-26
Westland, MI

usa2k

MVM

I as a newbie just know its not A or D
(But I hate to guess, I prefer to know - still studying for CCNA)

TomS_
Git-r-done
MVM
join:2002-07-19
London, UK

TomS_ to mikeyb4760

MVM

to mikeyb4760
Bit wise D is exactly the same as C. If they are marking you based on choosing the correct supernet address, then C for sure.
TomS_

TomS_ to phantasm11b

MVM

to phantasm11b
B will only cover the first 4 subnets in that list, /22 only gives you 4 /24s worth of space. /21 therefore gives you 8 and covers 10.1.0.0 through 10.1.7.255.

phantasm11b
Premium Member
join:2007-11-02

phantasm11b

Premium Member

said by TomS_:

B will only cover the first 4 subnets in that list, /22 only gives you 4 /24s worth of space. /21 therefore gives you 8 and covers 10.1.0.0 through 10.1.7.255.
Well shoot, that's what I get for doing binary math before having coffee. Thanks TomS.

TomS_
Git-r-done
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join:2002-07-19
London, UK

1 edit

TomS_

MVM

Weak. I dont even drink coffee!

edit: but it is bed time...

phantasm11b
Premium Member
join:2007-11-02

phantasm11b

Premium Member

said by TomS_:

Weak. I dont even drink coffee!

edit: but it is bed time...
Now I just feel stupid. lol. Back to reading my CCNP:ROUTE book. lol.

mikeyb4760
Premium Member
join:2004-09-25
El Cajon, CA

1 edit

mikeyb4760

Premium Member

That was my point: If I get this puppy on the CCNA exam, what is the fast way to resolve it.. Heh heh, Im kinda glad this stirred up some interest ;{D I figured it out the long way, convert all to binary

Yeah, the answer is "C"

usa2k
Blessed
MVM
join:2003-01-26
Westland, MI

4 edits

usa2k

MVM

I guess since the biggest VLAN address is 10.1.5.0/25 and the smallest is 10.1.1.0/30, you need that range

0 thru 5 is 000 thru 101 in binary.

That is 3 bits you need to steal from the Network address for the Host address.

/24 is all 24 bits 11111111 11111111 11111111
And you borrow the last three 11111111 11111111 11111000
21 bits mean c) 10.1.0.0/21

typo mania!

phantasm11b
Premium Member
join:2007-11-02

phantasm11b

Premium Member

said by usa2k:

I guess since the biggest VLAN address is 10.1.5.0/25 and the smallest is 10.1.1.0/30, you need that range.
Where do VLAN's come into play? VLAN's are switching, route summarizing pertains to routing.

usa2k
Blessed
MVM
join:2003-01-26
Westland, MI

usa2k

MVM

Well I guess they could be subnets. I guess I am not familiar with all the right jargon. I called them subnets, then thought better. Or should I refer to them as LANs?

If I understand correctly 10.1.0.0/21 at least has the right subnet mask to reach those addresses.
HELLFIRE
MVM
join:2009-11-25

HELLFIRE to mikeyb4760

MVM

to mikeyb4760
Tried my hand at this and I'd also agree with C, here's what
I found while working out this particular question out by hand:

1) discount any octets that have the same number, ie the first
and second as they're identical and you shouldn't have to
convert them to binary to do this problem.

2) /23 was a clue the summary should be somewhere "behind" this
subnet mask as it was the shortest mask of all the routes.

3) the third octets for all 6 routes were values of 1 - 5 which
are easy enough to turn into binary, and from there it was just
a matter of lining them up and seeing where the common bits were
to all of them.

Did this in under five minutes by hand, and truth to tell, I'd
agree there's no real "shortcut" to summarizing. You HAVE to
know binary, you HAVE to know how subnet masks work...
and you plain just know it or you don't, unfortunately.

@usa2k
It's subnets for sure. VLANs are a layer 2 / switch / non-IP
address construct. LAN refers to a network under LOCAL control.

Regards

usa2k
Blessed
MVM
join:2003-01-26
Westland, MI

usa2k

MVM

said by HELLFIRE:

@usa2k
It's subnets for sure. VLANs are a layer 2 / switch / non-IP
address construct. LAN refers to a network under LOCAL control.

Regards
Thanks! I'm CompTIA A+ and Security+ certified and in the 3rd week of a 4 week CCNA training course. I've a ways to go yet!!
meta
join:2004-12-27
00000

meta

Member

edit, the correct answer is:

conf t
 router bgp 32768
  auto-summary
 end
wr
 
;P jkjk
(for those of you that dont know, always turn auto-summarization off when using bgp)

TomS_
Git-r-done
MVM
join:2002-07-19
London, UK

1 edit

TomS_ to HELLFIRE

MVM

to HELLFIRE
said by HELLFIRE:

You HAVE to know binary
I dont think you do, I think a lot of it can come down to simply remembering numbers. Eventually once youve done this enough its just a matter of recognising patterns.

e.g. to work out the answer I didnt use any binary at all, but I am working from years of experience here, so bear with me as I *try* explain this (I hope it makes sense, it is 2am here right now...)

The first set of numbers to remember are the number of hosts you can find in any subnet from /24 to /32, the number of /24's there are in any subnet from /16 to /23, and the number of /16's there are in any subnet from /8 to /15.

So this leaves us with:

256, 128, 64, 32, 16, 8, 4, 2, and 1.

And the second set of numbers are the subnet mask equivalents:

0, 128, 192, 224, 240, 248, 252, 254, and 255.

So we then need to try and turn a list of arbitrary subnets into a supernet address, and the way we do this is by using the this set of numbers to work out the smallest figure that will cover all of our subnet addresses.

In the OPs case, we are trying to work out the smallest number that will cover:

10.1.1.0/30
10.1.1.4/30
10.1.1.8/30
10.1.2.0/23
10.1.4.0/24
10.1.5.0/25

As the first and second octets are not changing, we know that we are working in the /16 to /23 range of subnets (the first octet covers /0 through /7, second octet /8 through /15, and 4th octet /24 through /32). Immediately we can start building a subnet mask, so we'll start with "255.255".

Now work out the smallest figure that will cover our list of subnets using the first set of numbers. We all know that 4 is going to be too small, as 10.1.5.0 falls outside of the 10.1.0.0 to 10.1.3.255 range, so we try 8.

10.1.0.0 through 10.1.7.255 covers all of our subnets just fine, so we add the subnet mask equivalent of 8 to our subnet mask, and we are left with "255.255.248".

Then we pad it out with zeroes till we get "255.255.248.0".

So this is our supernet mask, and were half way there. We now need to find the supernet address, and this only uses a little math. Since we know we are working in blocks of 8, we simply need to find the lowest multiple of 8 that covers all of our subnets.

In our case it happens to be quite simple, but what if we had a set of subnets like the following:

10.1.169.0/30
10.1.169.4/30
10.1.169.8/30
10.1.170.0/23
10.1.172.0/24
10.1.173.0/25

Looks pretty hairy...

But use some basic math like "8 x 20 = 160", well 160 is pretty close to 169. 160 + 8 = 168, and 169-173 falls nicely within 168 + 8, so the 3rd octet of our supernet address would be 168, giving us 10.1.168.0.

Converting the subnet mask above to a CIDR slash is relatively easily, its just counting backwards. Theres 1 x /24 in a /24, 2 x /24s in a /23, 4 x /24's in a /22 ... 8 x /24's in a /21. Bingo

10.1.168.0/21, or as in the OPs case, 10.1.0.0/21.

Granted, the above doesnt use any binary, and in a test situation they may want to see you use binary to work it out, so its probably worth knowing binary anyway.

Im sure absolutely none of the above makes sense (again it is 2am), but whatever I was trying to explain works for me and I can do this pretty quickly in my head - no pen, no paper, no binary.

There are many ways to do this, some things work for some people, others work for the rest. Its just a matter of finding the one that works for you. I use much of the above to also do subnetting in my head. Again its just counting and perhaps a little math.

usa2k
Blessed
MVM
join:2003-01-26
Westland, MI

usa2k

MVM

Re: Route Summarization

Good cheat sheets links here

I've known about Binary, Octal, and Hexadecimal from electronics experience going back 30 years.
Memorizing the cheat sheets is more effective than knowing binary.
Knowing binary though, helps put it all together
HELLFIRE
MVM
join:2009-11-25

HELLFIRE to mikeyb4760

MVM

to mikeyb4760

Re: Route Summarization

// passes a Friday night beer to nosx

Thanks man, put a smile on my face

@TomS_
Okay, let me rephrase "you HAVE to know binary when you're starting off learning this stuff."
Like you said, as you keep doing this you find little tricks and shortcuts you help you out
but if you don't have a grounding in the fundamentals, fuggetabutit.

Thanks for pointing out the whole networks / hosts thing... should've remembered it way back
when but it was one of those things I never bothered to clue into.

Regards