Route Summarization. How?

Hi,

Just learning route summarization by Todd Lammle and he says that all we need is the block size.

He gives a example of 192.168.16.0 - 192.168.31.0 and says the block size is 16.

The only way I can think he got to this figure is by the difference which is not the exact figure.

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joneno

lon21 wrote: »

hi,

just learning route summarization by todd lammle and he says that all we need is the block size.

He gives a example of 192.168.16.0 - 192.168.31.0 and says the block size is 16.

The only way i can think he got to this figure is by the difference which is not the exact figure.

192.168.16.0 - 31.0
192.168.32.0 - 47.0
192.168.48.0 - 63.0

Forsaken_GA

lon21 wrote: »

Hi,

Just learning route summarization by Todd Lammle and he says that all we need is the block size.

He gives a example of 192.168.16.0 - 192.168.31.0 and says the block size is 16.

The only way I can think he got to this figure is by the difference which is not the exact figure.

You have to remember 0 is a valid number. 0 to 15 is 16 distinct digits. 0 to 31 is 32 distinct digits.

And he's correct.

If I have 192.168.16.0/28 and 192.168.16.31/28, those are contiguous blocks that I can summarize as 192.168.16.0/27. That represents the exact same # of IP addresses, but takes one less route to do it, thereby decreasing routing table sizes.

Now, if I have 192.168.16.0/28 and 192.168.16.240/28, the first summarization available for them is 192.168.16.0/24, but that may not be desirable because there may be other routes in that range that I don't want summarized. If that were the case, I'd continue to advertise the individual /28's (and I'd kick myself or a predecessor for putting a bad IP addressing scheme into play)

lon21

Ok, now I'm completely lost.

How did he get the block size 16 tho?

Thanks

Forsaken_GA

I don't have a copy of the book, so I'd need to see the specific text cite to be able to answer that. I'm assuming that since it's a discussion on route summarization and you mention a block of 32 IP addresses with a range of 16, that there's something in the context that gives it away

jwashington1981

Forsaken_GA wrote: »

If I have 192.168.16.0/28 and 192.168.16.31.0/28, those are contiguous blocks that I can summarize as 192.168.16.0/27. That represents the exact same # of IP addresses, but takes one less route to do it, thereby decreasing routing table sizes.

There's a huge typo. In that second address, you have 5 octets.

This is the example in the Lammle book that he's referring to.

Lammle says networks 192.168.16.0 through 192.168.31.0. He mentions these as 16 class C networks. That would have had to been known in order to properly answer the question.

Forsaken_GA

jwashington1981 wrote: »

There's a huge typo. In that second address, you have 5 octets.

This is the example in the Lammle book that he's referring to.

Lammle says networks 192.168.16.0 through 192.168.31.0. He mentions these as 16 class C networks. That would have had to been known in order to properly answer the question.

Ah crap, you're right, I do.

And yeah, I just misread the original list of subnets, if I'd been paying attention, I would have realized that was aggregating /24's and not summarizing a deaggregated single /24, so my bitmasking is wrong as well

Given that, it would be 192.168.16.0/20 for the summary instead, as that's the first prefix that can be used to cover just those ranges.

Forsaken_GA

lon21 wrote: »

Ok, now I'm completely lost.

How did he get the block size 16 tho?

Thanks

Ok, so given what the other guy said, he told you it was a range of 16 class c's, and gave the range. Since those are a contiguous block that falls on a binary boundary, you can represent all 16 subnets with one route, that's what summarization is for. So instead of propagating 16 individual /24's, you can simply advertise one /20

lon21

Forsaken_GA wrote: »

Ok, so given what the other guy said, he told you it was a range of 16 class c's, and gave the range. Since those are a contiguous block that falls on a binary boundary, you can represent all 16 subnets with one route, that's what summarization is for. So instead of propagating 16 individual /24's, you can simply advertise one /20

Hi Forsaken,

Sorry mate I'm still lost.

The books asks to work out the range, all I'm given is the two ip address.

But I can't seem to figure out how the number 16 was achieved?

jwashington1981

lon21 wrote: »

Hi Forsaken,

Sorry mate I'm still lost.

The books asks to work out the range, all I'm given is the two ip address.

But I can't seem to figure out how the number 16 was achieved?

Ok, knowing that the range of networks is 16 class C networks, we can determine this.

192.168.16.0
192.168.17.0
192.168.18.0
192.168.19.0
.......
192.168.31.0

Notice what's in common and what's different. The first two octets are the same for each network. The difference comes in the third octet so this is our octet of interest. If you were to write each network in binary format, you will notice that each of the networks has the first 20 bits in common.

11000000.10101000.00010000.00000000
.....
11000000.10101000.00011111.00000000

192.168.16.0 just happens to fall on a boundary. Using a /20, that means that the subnet mask would be 255.255.240.0 and would give us blocks of 16.

256-240 = 16 (block size)

Always subtract the portion of the subnet mask that has the octet of interest from 256 to get your block size.

256 - 128 = 128
256 - 192 = 64
256 - 224 = 32
256 - 240 = 16
256 - 248 = 8
256 - 252 = 4
256 - 254 = 2

lon21

jwashington1981 wrote: »

Ok, knowing that the range of networks is 16 class C networks, we can determine this.

192.168.16.0
192.168.17.0
192.168.18.0
192.168.19.0
.......
192.168.31.0

Notice what's in common and what's different. The first two octets are the same for each network. The difference comes in the third octet so this is our octet of interest. If you were to write each network in binary format, you will notice that each of the networks has the first 20 bits in common.

11000000.10101000.00010000.00000000
.....
11000000.10101000.00011111.00000000

192.168.16.0 just happens to fall on a boundary. Using a /20, that means that the subnet mask would be 255.255.240.0 and would give us blocks of 16.

256-240 = 16 (block size)

Always subtract the portion of the subnet mask that has the octet of interest from 256 to get your block size.

256 - 128 = 128
256 - 192 = 64
256 - 224 = 32
256 - 240 = 16
256 - 248 = 8
256 - 252 = 4
256 - 254 = 2

Hi Thank you very much.

Just to clarify

The 16 would be because there are 3 bits which are not used and the 3rd octet starts from the 4th bit?

Did you choose the /20 because the block size of the 3rd octet was 16 and a block size of 16 give a 240 mask i.e. 256-16 = 240 ?

Sorry just to add, is there ANY way possible to work out the block size without writing the ip address to be summarize in bit format?

Nevins

YouTube - ‪Quick Summary Route‬‏

I made this for summary route earlier this year. Feel free to use it.

lon21

Nevins wrote: »

YouTube - ‪Quick Summary Route‬‏

I made this for summary route earlier this year. Feel free to use it.

Thanks mate that helped me on a few questions which had given the ip address.

lon21

Guys,

I'm really sorry and I don't want to seem rude.

There are two types of questions which I wish to answer with summarization.

The question which has the IP address listed (single) you guys have answered excellent. Really appreciated.

The second question is where I'm only given 'Network through 172.16.32.0 - 172.16.50.0'

I don't understand how we can get a summary route with only two ip address when we don't have the ip address in-between as we don't know the block sizes.

If I knew the block size I could start to write down the network between the two ip address.

Thanks

jwashington1981

lon21 wrote: »

Guys,

I'm really sorry and I don't want to seem rude.

There are two types of questions which I wish to answer with summarization.

The question which has the IP address listed (single) you guys have answered excellent. Really appreciated.

The second question is where I'm only given 'Network through 172.16.32.0 - 172.16.50.0'

I don't understand how we can get a summary route with only two ip address when we don't have the ip address in-between as we don't know the block sizes.

If I knew the block size I could start to write down the network between the two ip address.

Thanks

Read the situation again. Lammle says networks 172.16.32.0 through 172.16.50.0. That through means everything in between.

Now you have to approach this one differently than you did the last one because the amount of networks you have to summarize does not equal that of a perfect block size. You have a total of 19 networks.

First thing you need to do, is look for your octet of interest. In both of those addresses, you'll notice that the first two octets are the same and it's the third octet where they differ. The third octet will be our octet of interest.

After that, there's two ways to approach this. Considering that there are 19 networks, a block of 16 will not summarize them all. If you did this, then you would leave the other three networks as separate entries in the routing table. So one answer will look like this.

172.16.32.0 /20
172.16.48.0 /24
172.16.49.0 /24
172.16.50.0 /24

The other thing you can do is that since you have 19 networks to summarize, go the next block size higher which would be a block size of 32. So what mask would give us a block size of 32? 256-32 = 224. So your mask will be 255.255.224.0 which is equivalent to a /19. So you can summarize those networks with 172.16.32.0 /19. Now if you go this route, know that you're summarizing more networks that what's in the question and that's perfectly fine as long as you don't need those additional networks for use later on. If you did, then you'd just simply change that to reflect whatever it is you're doing.

lon21

jwashington1981 wrote: »

Read the situation again. Lammle says networks 172.16.32.0 through 172.16.50.0. That through means everything in between.

Now you have to approach this one differently than you did the last one because the amount of networks you have to summarize does not equal that of a perfect block size. You have a total of 19 networks.

First thing you need to do, is look for your octet of interest. In both of those addresses, you'll notice that the first two octets are the same and it's the third octet where they differ. The third octet will be our octet of interest.

After that, there's two ways to approach this. Considering that there are 19 networks, a block of 16 will not summarize them all. If you did this, then you would leave the other three networks as separate entries in the routing table. So one answer will look like this.

172.16.32.0 /20
172.16.48.0 /24
172.16.49.0 /24
172.16.50.0 /24

The other thing you can do is that since you have 19 networks to summarize, go the next block size higher which would be a block size of 32. So what mask would give us a block size of 32? 256-32 = 224. So your mask will be 255.255.224.0 which is equivalent to a /19. So you can summarize those networks with 172.16.32.0 /19. Now if you go this route, know that you're summarizing more networks that what's in the question and that's perfectly fine as long as you don't need those additional networks for use later on. If you did, then you'd just simply change that to reflect whatever it is you're doing.

Where did the block size 16 come from?

jwashington1981

lon21 wrote: »

Where did the block size 16 come from?

Here's where memorizing that chart from Lammle's book that you posted in another thread would come in handy.

Let's take a look at what I posted earlier from your first question with the bits for the range of subnet masks for 192.168.16.0 to 192.168.31.0

11000000.10101000.00010000.00000000
.....
11000000.10101000.00011111.00000000

You will notice that the first 20 bits are all common. That's where the /20 comes from. They don't differ until you get to the 21st bit. Now if we just concentrate on the third octet, you'll notice that the first 4 bits are the same throughout the entire range. So if we were to turn on those first 4 bits on in the subnet mask, the mask would have 11110000 which is equal to 240. Since everything before that matches, all those bits are going to be turned as well so your mask would be 11111111.11111111.11110000.00000000, or 255.255.240.0

1's equal to being part of the network and 0's being part of the hosts.

If you take that 240 and subtract it from 256, that's where the 16 comes from.

If we take the example from your second question, networks 172.16.32.0 to 172.16.50.0, it would look something like this

10101100.00010000.00100000.00000000 (172.16.32.0)
.......
10101100.00010000.00101111.00000000 (172.16.47.0)
10101100.00010000.00110000.00000000 (172.16.48.0)
10101100.00010000.00110001.00000000 (172.16.49.0)
10101100.00010000.00110010.00000000 (172.16.50.0)

Remember there were two ways to approach this. If we take approach one, notice that from 172.16.32.0 to 172.16.47.0, the first 20 bits are all common and they don't change until you get to the 21st bit. Once you get to 172.16.48.0, that 20th bit is not common with the networks before it.

If we take approach two, notice that for all of those networks listed, the first 19 bits are all common and they don't change until you get to the 20th bit, hence why you could use a larger block size to summarize them all, but you'll be summarizing more than what was required.

lon21

OK, thanks.

I'll give the theory a go with some other questions and let you know.

Really appreciate your help.

jwashington1981

You're welcome. Glad I could help out.

lon21

Just to add... is there not a easier way of working out the block sizes instead of writing the binary values?

If not... in the exam would I need to write the ip address in-between or could I just get away with the first and last ip address

First ip address = 172.16.32.0 last ip address = 172.16.50.0

10101100.00010000.00100000.00000000 (172.16.32.0)

10101100.00010000.00110010.00000000 (172.16.50.0)

From here the subnet would be /19 ?

I guess I would have to write all the ip address in-between as otherwise I will summarize more than what is needed.

Is there not a shortly method instead of writing all the binary of ALL the ip address?

jwashington1981

lon21 wrote: »

Just to add... is there not a easier way of working out the block sizes instead of writing the binary values?

If not... in the exam would I need to write the ip address in-between or could I just get away with the first and last ip address

First ip address = 172.16.32.0 last ip address = 172.16.50.0

10101100.00010000.00100000.00000000 (172.16.32.0)

10101100.00010000.00110010.00000000 (172.16.50.0)

From here the subnet would be /19 ?

The easiest way is to memorize the block values that go with each value in the subnet mask. Using the bits is just a descriptive way to show you how those values actually came about. Go to Lammle's chart that you posted and memorize that chart. If you have a question about how he came up with those values, we'll be glad to explain.

Also, I moved the bold tag over. In the second address listed, you only bolded the first two numbers in the third octet when the first three numbers match the first three numbers of the first address in that octet. And you would use the /19 for this problem if you're going to use the second method of the two that I described.

lon21

Sorry what I meant was... is there a easier way of working out the block sizes without writing ALL the binary values of the ip address, without having the subnet?

As Lammle and you/others were able to work out the block sizes without writing the binary values and with no subnet?

p.s. thanks for the correction.

Thanks

jwashington1981

lon21 wrote: »

Sorry what I meant was... is there a easier way of working out the block sizes without writing ALL the binary values of the ip address, without having the subnet?

As Lammle and you/others were able to work out the block sizes without writing the binary values and with no subnet?

p.s. thanks for the correction.

Thanks

It wasn't really a matter of us not knowing the subnet values. Mr. Lammle forgot to add that information and I'm going to chalk that up as a he made an assumption that we would think the same way we did about that. He should have let us know that in those examples, the subnet masks for those networks were all /24s.

lon21

jwashington1981 wrote: »

It wasn't really a matter of us not knowing the subnet values. Mr. Lammle forgot to add that information and I'm going to chalk that up as a he made an assumption that we would think the same way we did about that. He should have let us know that in those examples, the subnet masks for those networks were all /24s.

A /24 is class C subnet but the 172 ip address is class B.

I'm looking at the chart and tbh I can't seem to find any way of working out the block sizes. I can only see the block sizes relating to the subnet masks. But your saying that we don't really need to know the subnet.

All I can think is that 172 is a class B which is a 255.255.0.0 and this gives a /16 but there is no /16 on that table.

If I'm correct we don't really need the block sizes, because what we do when given two ip address is find out the ip address in-between and on the table look for host which cover that number e.g. 19 network would be a 14 host, 16 block size as its the one closest one before the next host. Is this correct?

Forsaken_GA

lon21 wrote: »

A /24 is class C subnet but the 172 ip address is class B.

I'm looking at the chart and tbh I can't seem to find any way of working out the block sizes. I can only see the block sizes relating to the subnet masks. But your saying that we don't really need to know the subnet.

All I can think is that 172 is a class B which is a 255.255.0.0 and this gives a /16 but there is no /16 on that table.

If I'm correct we don't really need the block sizes, because what we do when given two ip address is find out the ip address in-between and on the table look for host which cover that number e.g. 19 network would be a 14 host, 16 block size as its the one closest one before the next host. Is this correct?

Ok, for starters, don't take anything I'm about to say in a harsh light.

You should not be studying route summarization yet. You need to go back and learn how to properly subnet. This will all make alot more sense once you get subnetting.

There are a few key concepts you need to understand -

The router doesn't comprehend IP addresses. The only thing the router cares about are two things - the binary value of all 32 bits of the IP address, and the binary value of the 32 bit mask. The 32-bit IP address is what the router uses to identify individual nodes on a network, and the 32-bit mask is what the router uses to identify nodes that are grouped together as a common network. From those two values, the router derives the route, and uses that to build the routing table

The dotted decimal notation we work with is done that way to make the representation easier for humans to understand. If we were just doing simple binary to decimal conversion, we'd be dealing with numbers the size of 2^32. This is too large for the human brain to process in a meaningful way, so instead of dealing with 2^32, it's broken up so that we're dealing with 4 sets of 2^8 instead. This doesn't change the fact that at the basic level, it's all just binary, and all the little tricks we learn are just ways to manipulate the binary, and comprehend what we're doing. If you don't understand this now, this will REALLY screw with you when you get to ipv6, because ipv6 is *not* simply dealing with ip addresses that are 4 times bigger (32 bit addressing as opposed to 128 bit addressing).

Now, with all that being said - You need to be able to do this kind of thing in your head, because unless you write it out yourself beforehand, you're not going to have a table to look at on the exam, and if you're going to memorize the table, you're already memorizing the info you need in order to do it in your head.

For starters, the concept of classes went out the window with CIDR. They're nothing more than a convenient (if confusing) way of referring to subnetting. When folks say a Class C network these days, what they really mean is a /24, they are *not* the same thing. Classful routing means the subnet mask is implied, classless routing means the subnet mask is explicitly defined as part of the prefix.

There are only 8 possible block sizes when it comes to subnetting. 128, 64, 32, 16, 8, 4, 2, 1

Fortunately, these are also the first 8 powers of 2. That makes them fairly easy to remember (that's why 256 is a magic number, it's 2^8 ).

In order to do route summarization, you absolutely do need to know the subnet. For example, if I asked you to summarize 172.16.12.0 and 172.16.127.128, you wouldn't have a clue where to begin, you'd have to guess, and you'd probably summarize them as 172.16.0.0/16, which would certainly be a correct summarization, but it's not smaller summary, in the same way that you could also summarize back to 172.0.0.0/8, and it'd be valid.

If I told you to summarize 172.16.12.0/27 and 172.16.127.128/25, you'd have somewhat of a clue where to begin, and once you get practiced at it, you'd be able to spot that 172.16.0.0/17 is the first possible summary that contains both of them.

Route summarization is basically just subnetting in reverse. If you can take a block of IP's and break it up into smaller blocks of IP's, you already know how to do summarization, you just need to train your brain to recognize it. Summarization is sometimes called supernetting because the concepts for subnetting transfer, it's just a different application. In subnetting, you're manipulating the bit boundaries in order to take a block of IP's and break them into smaller blocks of IP's to represent more networks. Summarization functions exactly the same way, except instead of taking blocks of IP's and making them smaller, you're combining multiple blocks of IP's to represent them as a single bigger network.

So for example, let's say I gave you 192.168.12.0/24. And I asked you to create 4 networks out of that space. Well, if you learned your subnetting properly, you'd know that in order to create 4 networks, you have to borrow two bits from the fourth octet, because that's how many bits it takes to represent 4, 2^2 = 4.

So your subnets would be /26's.

192.168.12.0/26, 192.168.12.64/26, 192.168.12.128/26 and 192.168.192.0/26

Now, what if I flipped the question and gave you those 4 /26's, and asked you to create a summary route that represented all 4 of those networks?

You already have the answer, it's the original 192.168.12.0/24. But how do you get back to that? In order to get back to the summary route, I basically have to give the bits I borrowed back. So what happens if I give one bit back? Ok, that creates me 2 /25's. That won't represent all 4 of the created subnets. What happens if I give 2 bits back? That creates me a /24 that covers all 4 subnets, which makes it the first valid summary route.

lon21

Firstly of all I would like to thank each and everyone for taking the time to help me out. I really, really appreciate it.

I've taken what you guys said, nothing taken to heart. Really appreciate all you input.

I've been working on subneting with Route Summarization ALL day today since I've woken up.

With all the questions I have. I have come up with a method which I believe works for me.

I use this table to obtain the answers.

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

e.g. 1

192.168.16.0 - 192.168.31.0

1. I calculate the network in between the IP address = 16.

2. From the table above I see that a Block 16 sits in the fourth column.

3. This means that, I will take 4 bits from the network. As the first two octets are the same 8 + 8 = 16 the third octet will only have 4 bits taken 16 + 4 = 20.

4. Therefore my / notification would be a /20

5. To calculate the summary IP address I write the IP address in binary format.

192.168. 0001|0000.00000000
192.168. 0001|1111.00000000

The line | marks all the network bits on left and host on right.

6. I look at the first IP address and work out the summary IP address based on every bit value from left of the line. Also at this point I can verify my / notification to make sure that the bits on the left of the line are common.

Therefore my summary route is 192.168.16.0 /20 (/20 = 255.255.(256-16=240).0

e.g. 2

172.16.32.0 - 172.16.50.0

1. I calculate the network in between the IP address = 19.

2. From the table above I see that a Block 19 sits in-between the fourth and third column. I will choose the third column (round up to next block size).

3. This means that, I will take 3 bits from the network. As the first two octets are the same 8 + 8 = 16 the third octet will only have 3 bits taken 16 + 3 = 19.

4. Therefore my / notification would be a /19

5. To calculate the summary IP address I write the IP address in binary format.

172.16. 001|00000.00000000
172.16. 001|10100.00000000

The line | marks all the network bits on left and host on right.

6. I look at the first IP address and work out the summary IP address based on every bit value from left of the line. Also at this point I verify my / notification to make sure the the bits on the left of the line are common.

Therefore my summary route is 172.16.32.0 /19 (/19 = 255.255.(256-32=224).0

My only concern is in example 2 the block 32 will summarise more ip address than required, and a block 16 will summarise only 32.0 - 47.255. Leaving out 48 - 50.

In the exam which one should I use as the correct method do I summarise more than I need or only the best possible answer and leaving the rest in the routing table of the router?

Thanks

jwashington1981

lon21 wrote: »

My only concern is in example 2 the block 32 will summarise more ip address than required, and a block 16 will summarise only 32.0 - 47.255. Leaving out 48 - 50.

In the exam which one should I use as the correct method do I summarise more than I need or only the best possible answer and leaving the rest in the routing table of the router?

Thanks

I mentioned this in one of my previous posts. Both methods are correct.

dougefresh

PLEASE HELP.;.. i have been working on subnetting for about 2 days now straight and using various "magic number charts". For the most part i am getting my answers correct. But I cannot for the life of me get this one. Can anyone please help? possible with the the magic number chart showing how.
172.16.87.56 /22 or 192.168.45.101/25. Need network,first host, last host, BC. and how many hosts and subnets? thanks a bunch. doug

EdTheLad

172.16.87.56/22; /22 = 255.255.252.0

256 - 252 = 4

Subnets ranges go as follows:

172.16.0.0 -> 172.16.3.255
172.16.4.0 -> 172.16.7.255
etc

Now i know the 3rd octet has an increment of 4, looking at 87 i can see it would fall into subnet 172.16.84.0

172.16.86.0 subnet
172.16.84.1 first host
172.16.87.254 last host
172.16.87.255 broadcast

You should be able to work the other one out if you follow the same logic!

jjmuriel

Please your help with this exercise:

192.168.1.0/24 through 192.168.12.0/24

I can understand how to get the mask: The third octet of second IP address less

the third octet of first IP address: 12-1 = 11. So there is 11+1 = 12 networks (because I count starting from 0). The best block size is = 16, then the mask is 255.240.0.0 or /20. At this point everything is ok, but how can I get the network address with the "Lammle method". In the book, the answer is 192.168.0.0, but

mr. Lammle

said that always is the first address of the range. For me, in this case is 192.168.1.0.

PD: Sorry for my english.