route summarization - could this be wrong ?

[SJ K]

Hi,

I am reading up about CIDR and came across this diagram as below

If ISP1 router advertise 192.168.0.0/20 to ISP2,  what if ISP3 owned 192.168.16.0/20  ?  Wouldn't end-user of ISP2 go to ISP1 and found nothing ?

It is then assumed that ISP3 have to advertise the route to ISP2 or ISP1 or both then.

I am seeing that 192.168.16.0 still falls under the /20 network and hence my assumption above.

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But somehow I feel that I might be wrong as

.16 is  192.168 | 0001 0000 | 0000 0000  whereas for the earlier networks advertise by ISP1,  all of them have the 1st 4 bits in the 3rd octet as 0000.

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q1) which of my assumption is correct; How should route summarization be calculated ?

Thanks for reading.

Regards,

Confused Noob.

[Jon Marshall · Accepted solution]

I am seeing that 192.168.16.0 still falls under the /20 network and hence my assumption above

It doesn't fall under that summarised range ie -

192.168.0.0/20 = 192.168.0.0 255.255.240.0

so following on from your other post the maths is the same ie.

256 - 240 = 16

so that summary route covers the IP range 192.168.0.0 -> 192.168.15.255.

You have no idea what the actual subnets are within that range ie. you don't know what subnet masks have been used for individual subnets but you don't need to.

With route summarisation it is assumed that the router advertising the summary route has more specific routes for all the subnets within that summary range.

You should not advertise a summary route that includes subnet ranges for which you are not responsible.

Jon

[SJ K]

Hi all,

Got the below from a website, but I feel its wrong

As a final example, consider the following networks:

• 192.168.1.0/25
• 192.168.1.128/25
• 192.168.2.0/24
• 192.168.3.0/24
• 192.168.4.0/26
• 192.168.4.64/26
• 192.168.4.128/26
• 192.168.4.192/26

Try to figure out the summary address that can be used for these networks. If you look carefully the third octet forms a contiguous block of 4 and can be summarized with the address 192.168.1.0 255.255.252.0 or 192.168.1.0/22.

If we use 192.168.1.0/22 ->

1st, we are not sure if 192.168.0.0 network belong to the group of networks

2nd, the network above shown that we must include 192.168.4.0, but using the above summarization only covers up to 192.168.3.255, 192.168.4.0 network isnt included.

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Is my observation right ?

Regards,
Noob

[Jon Marshall]

It's wrong.

192.168.1.0/22 actually works out as 192.168.0.0/22 and as you say covers the IP range -

192.168.0.1 -> 192.168.3.254 with 192.168.3.255 as the broadcast IP address.

This is what I was talking about before ie. you cannot just pick the start address and apply a mask, you have to work out where it actually starts.

You've definitely got the hang of this :-)

Jon

[SJ K]

Hi Jon,

Thanks, some of the teaching material online are just asking us to do the 2 steps (esp on point 2)
 

1. You can only summarize in the block sizes you learned about in VLSM – 128,64,32,16,8,4.
2. The network address used for the summarized address is the first network address in the block.
    

But it wont necessary work as per the example above.

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Btw Jon, i have a missing thread again but its a very simple question, hope you can take a look over there

https://supportforums.cisco.com/discussion/12480701/simple-binary-question-double-confirm

I keep having this issue when working with binary and it confuses me sometimes.

Thanks.

[Jon Marshall]

I can't access that thread.

Don't know where they are ending up but I am not authorised.

Jon

[SJ K]

Hi Jon,

Here you go.

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Hi all,

There are still time to time whereby i can confuse on (8bits or 2^7) and 2^8

It is known that 8 bits give you 256 value.

q1) But in actual fact 8 bits turn on is not 2^8 isnt it ?, it is actually 2^7 + 2^6  all the way to 2^0
 

7	6	5	4	3	2	1	0
2^7	2^6	2^5	2^4	2^3	2^2	2^1	2^0
128	64	32	16	8	4	2	1

q2) And adding all of them up (128+64+32....) only give me 255 which i believe the missing "1" is the digit 0; hence 0-255 = 256 values, am i right ?

Regards,
Noob

[SJ K]

Hi Jon,

1) Not sure if you see this thread above on the binary, am still waiting for your advises ;)

2) Also while viewing some practices for route summaries, I have encounter some difficulties as below, hope you are able to share some insight on this

a) in my ccna textbook ard (8 years back) by todd lammle, there are some examples and method on finding route summaries  -> below is an example

#1

"Summarization is actually somehwat simple because all you really need to have down are the block sizes that we just used in learning subnetting and VLSM design..

For example.... if you want to sumamrize the following network into one network, you just have to find the block size 1st"

192.168.16.0 to 192.168.31.0 = block size = 16
16; subnet mask = 240
 

The network address used to advertise the summary address is always the first network address in the block, in this example 192.168.16.0/20

#2

http://www.freeccnastudyguide.com/study-guides/ccna/ch2/2-5-route-summarization/

Summarization is somewhat simple if you remember the following:

1. You can only summarize in the block sizes you learned about in VLSM – 128,64,32,16,8,4.
2. The network address used for the summarized address is the first network address in the block.

For example, if you want to summarize networks 192.168.8.0 through 192.168.15.0, first find the block size you can use. There are 8 networks so the block size of 8 can be used. The first network address in the block is 192.168.8.0. Now to find the mask of the summarized route, remember the mask used for a block of 8 – 248. You can also deduct the block size from 256 to find the mask. Since we are summarizing the third octet the subnet mask for the summary address will be 255.255.248.0.

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It seems to me , the way to do summarization is to

method 1

a) find difference in block size and determine the mask using that

b) add the mask to the 1st network in the block.

method 2

a) list down all the networks in binary

b) find the common bits and use them as the mask to the 1st network in the block

q1) to me, for either method, step b of adding the mask to the 1st network in the block seem to be wrong; just like my earlier example 192.168.1.0/24 to 192.168.3.0/24 doesnt mean it can be 192.168.1.0/22 (because 192.168.0.0 might not be in the network) -- am i right ?

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q2) In this article, http://www.pearsonitcertification.com/articles/article.aspx?p=2168927&seqNum=7 , all the way to the bottom cram quiz

192.168.1.0/29	192.168.1.128/29
192.168.1.8/29	192.168.1.136/29
192.168.1.16/29	192.168.1.144/29
192.168.1.24/29	192.168.1.152/29
192.168.1.32/29	192.168.1.160/29
192.168.1.40/29	192.168.1.168/29
192.168.1.48/29	192.168.1.176/29
192.168.1.56/29	192.168.1.184/29

I apply the same method of finding the block size from 192.168.1.0 to 192.168.1.56 and was wrong.  Over here, it seems that the block size is 0 to 56, but actual fact, it is 0 to 63.

So the examples above of finding the block size directly by just comparing the network address isn't really accurate

e.g. 192.168.16.0 to 192.168.31.0  = 16 to 31 = block size 16 = correct

e.g 192.168.8.0 through 192.168.15.0 = block size 16 = correct

current = 192.168.1.0 to 192.168.1.56 = 57 = wrong

It only works for network with classful boundaries, if there are fixed subnets or VLSM involved, I better use method 2 to list down all the binaries to find the common bits/subnet mask for summarization.

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Hope you understand what I meant and hope to hear your thoughts on it .

Regards,
Noob

[Jon Marshall]

q1) it doesn't mean add the mask to the first block (network) you are given it means find the first network in the summary range.

So if I gave you 192.168.1.0 and a mask of 255.255.252.0 and assuming each network within that range had a mask of 255.255.255.0 then the first network is 192.168.0.0/24 and the last network is 192.168.3.0/24.

If you notice with the example given the first network 192.168.16.0 and the mask is 255.255.240.0.

The example wouldn't have worked if the first network as 192.168.17.0.

That's the point ie. you need to find the bit boundary and sometimes you may need to use multiple summary ranges to cover a specific range ie. you may not be able to summarise all the networks with just one address.

q2) you don't need to use binary and it does apply but I understand where you confusion is coming from.

What you have to understand is the examples you list from the book are ranges you can summarise which is what I meant above about if he had used a different starting subnet you wouldn't be able to summarise it so neatly.

You can summarise them because they fall on bit boundaries.

And you can summarise your next example but you need to understand that you haven't written it out properly ie. you wrote -  

192.168.1.0/29 -> 192.168.1.56/29 as  192.168.1.0 -> 192.168.1.56 and then worked out 57 which is indeed wrong.

But 192.168.1.0/29 -> 192.168.1.56/29 is actually -

192.168.1.0 -> 192.168.1.63   which is where your 0 - 63 block size (if you want to call it that) is coming from.

And this can be summarised as 192.168.1.0 255.255.255.192

But you cannot summarise 192.168.1.0 -> 192.168.1.56 with one summary address because it does not fall on a bit boundary.

So the method works for all addressing.

You have to find the method that works for you but for me I simply think of it in terms of bit boundaries and there are only so many values you can have and you get used to them very quickly once you use them a lot.

If the above isn't clear or you still have questions please feel free to come back.

Jon

[SJ K]

Hi Jon,

Thanks for replying and the assurance of my confusion.

q1) I am referencing to your previous thread in terms of bits boundaries. To me, bit boundaries are like " the boundaries or range in which bits values might varies"

e.g.

for /24, the bits boundaries are from 0 to 255

for /25, the bit boundaries are 0, 128 whereby anything between 0 and 128 will varies and 128 to the end will varies

for /26, the boundaries are 0,64,128 etc, whereby anything between 0 to 64 will varies and anything from 64-128 will varies

am i right ?
 

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q2) Hence when you mentioned that

But you cannot summarise 192.168.1.0 -> 192.168.1.56 with one summary address because it does not fall on a bit boundary.

I do not understand why 1.56 doe not fall on a bit boundary, it is with a block size of 8 so the boundaries are 0,8,16,24,32,40,48,56,64 etc.. and 56 is on the boundary.

 

q3) in that case, how do I know when should i expand the network to find the actual range if i meet similar examples like 192.168.1.56/29 which is actually up till 63.

Right now, i am seeing it this way to use the upper bound boundary to see the range, and not the lower bound boundary.

Meaning if it is 56, then the the range should go just right before the next boundary of 64 (=63).

Regards,
Noob

[Jon Marshall]

q1) okay I don't know what you mean when you say varies so what I meant was using 256 as the base you can only have so many values ie.

255.255.255.128 = 256 - 128 = 128
255.255.255.192 = 256 - 192 = 64
255.255.255.224 = 256 - 224 = 32
255.255.255.240 = 256 - 240 = 16
255.255.255.248 = 256 - 248 = 8
255.255.255.252 = 256 - 252 = 4
255.255.255.254 = 256 - 254 = 2

the values in on the right hand side are the only  values you can have.

And these values are in the increments in which you go up eg. 255.255.255.192 means your subnets go up in increments of 64.

The above applies to all networks not just class C networks so 255.255.192.0 would still mean your subnets go up in increments of 64 but you would have a much larger host range per subnet.

q2) 192.168.1.56 does fall on a bit boundary if you use a /29 subnet mask because 56 / 8 = 8 but I thought you were trying to summarise 192.168.1.0/29 -> 192.168.1.56/29.

So when I said you cannot summarise 192.168.1.0 -> 192.168.1.56  I meant with one address and you can see you can't because if you look at the values above 56 isn't one of them.

But because 192.168.1.0/29-> 192.168.1.56/29 is actually 192.168.1.0 -> 192.168.1.63 then you can summarise them because that is 0 - 63 (64 values) and that is one of the values in the above.

q3) If you summarise 192.168.0.0 through 192.168.3.0 255.255.252.0 don't forget what you are actually summarising is -

192.168.0.0 -> 192.168.3.255 ie. you go up to the next network.

And that isn't what you did when you worked out 192.168.1.0/29 ->192.168.1.56/29 because instead of going to the end of the 192.168.1.56 subnet you used the start.

The reason it looks different is because the 192.168.0.0/22 example is summarising multiple class C networks and the 192.168.1.0/29 -> 192.168.1.56/29 example is actually subnetting within a class C.

But the principles are the same.

Again, please feel free to come back.

Jon

[SJ K]

Hi Jon,

Thank you for your reply. I understand what you meant. It just that initially i have always started by finding the range required directly by looking at the network. e.g

192.168.0.0 to 192.168.3.0 i have just see it that i need to go from 0 to 3 (4) without caring the last octet and 4 will be the block size which is 256-4 = 252 subnet mask.

so using the same the method, i apply it on 192.168.1.0/29 to 192.168.1.56/29 by directly looking at 0 to 56 which is obviously wrong.

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So i am thinking if there is a straight/direct method without falling into such tricky situation. (e.g. if it is at classful, just directly find the difference, if it is a subnet, then look from the front to the end of the network and see if the differences fits into any bit boundaries

Regards,
Noob

[Jon Marshall]

So i am thinking if there is a straight/direct method without falling into such tricky situation.

The point I was trying to make was to understand the maths and you don't need different shortcuts depending on classful vs classless because in essence it's all the same.

When you summarise 192.168.0.0/22 you are not summarising 192.168.0.0 ->192.168.3.0 you are actually summarising 192.168.0.0 -> 192.168.3.255.

Which is directly equivalent to saying when you summarise 192.168.1.0/29 to 192.168.1.56/29 you are not summarising 192.168.1.0 -> 192.168.1.56 you are summarising 192.168.1.0 -> 192.168.1.63.

So it's exactly the same principle.

Once you see that then it all makes sense.

Jon

[SJ K]

Hi Jon,

Yeap. I do see the underlying principle for both situation.
It just that I have no direct answer for explaining the portion whereby

when we calculate 192.168.1.0/29 -> 192.168.1.56/29 we are summarising 192.168.1.0 -> 192.168.1.63.  (so thats 0 to 63 =64) hence it fall into the 64 blocksize or 26 bit boundary

when we calculate 192.168.0.0 ->192.168.3.0 , we are summarising 192.168.0.0 -> 192.168.3.255 (but we do not see the blocksize as 0.0 to 3.255), we just concentrate ( 0 to 3 =4), without the 0-255.

I know its a full block, but I am not able to explain directly on why there is no need to take note of that range at the last octet. 0 - 255

Regards,
Noob

[Jon Marshall]

I know its a full block, but I am not able to explain directly on why there is no need to take note of that range at the last octet. 0 - 255

I do understand what you are saying but you are taking note of the last range with your summarisation.

Like I said bear in mind that one is subnetting and one is summarising which are different things so maybe you could just use that as a guide but for me I would just concentrate on how it is worked out and I tend not to think of it being one or the other.

In essence whenever I look at either I just see the bit boundaries and work it out from there but that works for me and you can only use something that works for you.

Jon