Welcome

First of all, may I welcome you to my site. My name is Chris and I'm from the UK and work as a Systems Engineer for Cisco. This blog was initially created to post up my subnetting technique but has now got more stuff to do with attaining Cisco certifications. Either way I really hope that the content is sufficent for your needs and I look forward to hearing your feedback. If you find that the content really helps you please feel free to donate using the PayPal link on the right.

To view the index of all my articles please click here.

Route Summarization

Some people get really confused when it comes to route summarization, probably more so with the different names given to it:

Route Summarization
Supernetting
Aggregate Address
Summary Address

Actually, I get more confused typing out this thing as being English I want to replace the z in "summarization" with an s, but as it's Cisco I best keep the z in, lol. I also get this with "neighbor" versus "neighbour".

Anyway, I digress. The above terms all mean the same thing.

What is summarization?

The process of taking a range of IP addresses and advertising them in one address block. The most well-known summarization/supernet is the RFC1918 Class B Range. More commonly we know the range to be 172.16.0.0 - 172.31.255.255, however the supernet is 172.16.0.0/12. You see what has happened there? We've taken a range of addresses and squashed it into one advertisement.

What are the benefits of summarization?

Quite clearly, if we have just one address instead of lots of individual addresses then the routing table is going to be smaller. This in turn means that memory requirements are reduced.

The less obvious benefit is that summarization means you're tracking whether or not you're connected to some subnets of a summary, not the up/down state of every link. Thus when the link goes up or down, you don't have a flurry of traffic announcing the state change.*

How to summarize/supernet?

This is what you really want to know isn't it? Well it's dead simple.

The first method shows you the long way.

1. Starting from the left of the IP address, identify the first octet that has a change of address in it. For example, the Class B RFC1918 range, the IP address first changes in the second octet (i.e. 172.16.x.x - 172.31.x.x).

2. Write out the binary equivalent of the address up to and including the changing octet. So for example the range above is:

10101100.00010000 = 172.16
10101100.00010001 = 172.17
10101100.00010010 = 172.18
10101100.00010011 = 172.19
10101100.00010100 = 172.20
10101100.00010101 = 172.21
10101100.00010110 = 172.22
10101100.00010111 = 172.23
10101100.00011000 = 172.24
10101100.00011001 = 172.25
10101100.00011010 = 172.26
10101100.00011011 = 172.27
10101100.00011100 = 172.28
10101100.00011101 = 172.29
10101100.00011110 = 172.30
10101100.00011111 = 172.31

3. From this list, count from the left how many bits are the same in each address. If we look at it we see that the first 12 bits for each address is the same so that gives us our mask in slash notation. We therefore start at our first address 172.16.0.0 and append our mask so the summary address is 172.16.0.0/12.

Well that was quite simple. But can we be quicker? Yes we can is the good news, after all, you don't want to eat up time in the exam by writing out addresses in binary. So here goes:

1. How many subnets are in the range? The RFC1918 Class B range is 16 subnets.

2. What power of 2 equals our range? 16 subnets = 24 so the answer is four.

3. Subtract the figure from step 2 from the default mask of our address range. In this example our default mask is 16 so the mask after subtracting 4 is /12.

4. Add this mask to the first address in the range - 172.16.0.0/12 in this example

Quick, eh? This is how it works in my head, "mmmm, 16 addresses, 2 to the 4 is 16, mask is 16, minus 4 is 12, so summary address must be first address with /12 mask."

One last example:

Summarise the following:

192.168.0.0/24
192.168.1.0/24
192.168.2.0/24
192.168.3.0/24

There are 4 subnets. 2 to the power of 2 gives us 4 so default mask of 24, minus 2, gives us /22. Address is therefore 192.168.0.0/22.

More difficult supernetting question

There are, however, a few pitfalls with supernetting/summarization. Sometimes you may not be able to get all of the addresses into a supernet without wasting addresses. Experience with supernetting questions will help you to identify this. An example below shows you how this could happen:

You need to summarize the following range:

10.16.31.0/24
10.16.32.0/24
10.16.33.0/24
10.16.34.0/24
10.16.35.0/24
10.16.36.0/24
10.16.37.0/24
10.16.38.0/24
10.16.39.0/24
10.16.40.0/24

What summary address should you use? Well there's 10 subnets. 2 to the power of 3 only gives us 8 subnets which is too small so we need to look at 2 to the power of 4 which is 16. The problem is that if we count in 16s our addresses straddle two subnets: 10.16.16.0 to 10.16.31.255 and 10.16.32.0 to 10.16.47.255. We have the same problem if we use the power of 5 giving us a range of 32 addresses (10.16.31.0 is in a different subnet than the other addresses in the range). We therefore have to go out to 2 to the power of 6 = 64 in order to get all of our addresses in the range (i.e. 10.16.0.0 to 10.16.63.255). What a waste of address space!!

The best answer is the following:

Break the space down into three ranges. We can summarize 10.16.32.0 to 10.16.39.255 with 3 bits (i.e. there are 8 subnets and 2 to the power of 3 is 8). We simply leave the other two address as they are. We have therefore avoided any wasted address space.

If you are unsure that you have the correct answers why not download a subnet calculator to double-check your answers? There is a great one by 3Com and can be downloaded from here.

I hope this has helped you with route summarization.

Chris


*http://www.netcraftsmen.net/welcher/papers/ospf2.htm

Posted byChris Bloomfield at 09:43  

81 comments:

Andres said... 9 August 2008 at 17:50  

Thank you, this has been most useful. However I am having problems trying to apply your concept to the following question:

"Given the addresses 10.1.138.0/27, 10.1.138.64/26, and 10.1.138.32/27, which of the following is the best summary?"

What throws me off is that there are three subnets but they have different prefixes (/26, /27). How do you find the summaries?

The answer is (according to the book) 10.1.138.0/25, but have no clue how they got that.

Any help is much appreciated.

Chris Bloomfield said... 12 August 2008 at 18:34  

Hi,

10.1.138.0/27 covers 10.1.138.0 to 10.1.138.31

10.1.138.32/27 covers 10.1.138.32 to 10.1.138.63

10.1.138.64/26 covers 10.1.138.64 to 10.1.138.127

Therefore you have to find out which summary covers 10.1.138.0 to 10.128.38.127. This is 128 addresses (1 network address, 1 broadcast address, and 126 host addresses). We therefore need to reserve enough bits to cover 128 addresses. We know that 2^7 = 128 so we must not subnet the last 7 bits of the address. As there are 32 bits in an IP address we subtract 7 from 32 to give us 25. Our summary address is therefore 10.128.38.0/25

HTH

michael said... 10 December 2008 at 19:52  

Chris your awareness to observe these math associations are terrific. I will share this with as many of my fellow Cisco Academy students as possible.(or i'll keep it a secret and dazzle them with my 'ahem' our skill)

Ok on to my question:
Can you relate how your method may be used in the question:
A router has a summary route to network 192.168.32.0/20 installed in its routing table. What range of networks are summarized by this route?

In England would you use an "s" or a "z" in Summarized?

Chris Bloomfield said... 10 December 2008 at 20:33  

Hi Michael.

Please feel free to share this technique although I can understand the one-upmanship you may want on your fellow students ;-)

The maths and working out of the question is very simple.

You have a /20 which is a block size of 16 in the third octet (i.e. 2^(24-20) = 16). Count in block sizes of 16 in the third octet so you would have 192.168.0.0, 192.168.16.0, 192.168.32.0, 192.168.48.0 etc

Your summary will therefore cover 192.168.32.0 to 192.168.47.255

To prove this you could always use binary. If we just concentrate on the third octet and pick 32 (the subnet address, in binary 00100000) and the mask for that octet is 11110000 (i.e. the full mask for a /20 is 111111111.11111111.11110000.00000000)

Perform an AND on that octet:

00100000
11110000
=
00100000
=
32

What if we use 47 in the 3rd octet (binary 00101111)?

Perform an AND:

00101111
11110000
=
00100000
=
32

They both resolve to the same value in the third octet so we know it is in the same summary.

Let's go to 48 in the third octet (binary 00110000) and perform the same AND with a /20 mask:

00110000
11110000
=
00110000
=
48

This value is different and therefore is not in the summary.

The binary way is the longest method but can be useful for you to see what is happening with the bits. However, as I trust myself (thousands wouldn't lol) I stick with my technique and it works all of the time!

Good luck with your studies.

Chris

Chris Bloomfield said... 10 December 2008 at 20:35  

BTW, we would use an S and not a Z

Cristi Badea said... 18 December 2008 at 11:13  

Hi,
Can you help me with that ?

Witch address can be used to summarize networks 172.16.0.0/24 thru 172.16.7.0/24 ?

The third block is different so I do not know how to do it

Thank you

joel said... 7 January 2009 at 08:56  

Hi Cristi Badea,

My answer to your question would be:
172.16.0.0/21
This would be enough to summarize the given 8 routes.

Chris Bloomfield said... 7 January 2009 at 18:43  

Yes this is correct. Maybe my tutorial is slightly misleading as you have a Class B address that has been subnetted to a /24.

In this case you still follow the same principals as in the tutorial but subtract the number of bits required to cover the subnet range from the mask given in the question.

Your questions has 8 subnets in all. 3 bits will give us 8 subnets as 2 ^ 3 = 8. Subtract 3 from the mask in the question (24 - 3) which equals 21. Take the first address in your range and change the mask to the new mask and that gives you the summarized address (i.e. 172.16.0.0/21)

HTH

Jatin....Destroyed in Love said... 9 January 2009 at 09:47  

Amazing!...

Never thought it is so easy. Searched a lot on the net but never got something like this.

Thanks mate.

Jatin

Cristi Badea said... 22 January 2009 at 11:24  

A routter has a summary rute to network 192.168.32.0/20 installed in its routing table. What range of networks are summarized by this route ?

- 192.168.0.0 - 192.168.32.0/24
- 192.168.0.0 - 192.168.47.0/24
- 192.168.32.0 - 192.168.47.0/24
- 192.168.32.0 - 192.168.48.0/24
- 192.168.32.0 - 192.168.63.0/24

Mask 20. Next boundary is 24. 24-20=4 2^4=16 block size
So what is tha answer ? - 192.168.32.0 - 192.168.48.0/24

I do not know becaute I do not have the whole range of subnets.

Please clarify me
Thank You

Yusif said... 30 March 2009 at 21:42  

Hi Chris,
Can you please help me the following?
What is the most effiecent summarization that R4 can use to advertise its networks to R5.
172.1.4.0/25
172.1.128.0/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
They came up with 172.1.4.0/22 for the summary route, but how did they get mask. I have no clue.
Please help.
Yusif

matteovinci said... 23 April 2009 at 09:00  

Yusif,

You mentioned the below networks:172.1.4.0/25
172.1.128.0/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

Forgetting about 172.1.128.0 for a moment...

172.1.4.0 in binary is:
10101100.00000001.000001|00.00000000

notice the division in the 3rd octet at the 6th bit

7=4+3
so 1.4 can make 1.5,.6,.7 with just the last 2 bits changing in the 3rd octet

so hence the summary route of 172.1.4.0/22

as 22 bits cover 172.1.4.0 & the 23rd & 24th bit are used to allow 5.0, 6.0 & 7.0

not quite sure why 172.1.128.0/25 was included in the question

because if you included that in your summery, you would end up having 172.1.0.0/24

as the first bit of the 3rd octect has to be allowed to change to 1 to allow 128

for anyone else who may have other suggestions as to the way i interpreted this, please do tell

:)

Sailesh said... 3 May 2009 at 13:55  

Hi Chris,

can ya please help me with this one:

the question is:

Given these subnets:
192.168.10.0 /24
192.168.11.0 /24
192.168.12.0 /24
192.168.13.0 /24
192.168.14.0 /24
192.168.15.0 /24
... what's the summary address / mask?

Now, my answer is 192.168.8.0/21

is that correct?

Please let me know

liveandlearn said... 11 May 2009 at 04:31  

Sailesh,

(First of all, thanks to Chris B's excellent technique.)

Your answer of 192.168.8.0/21 is correct if you can live with over-summarization. Because this block includes address blocks 192.168.8.0/24 and 192.168.9.0/24. When you configure your router to advertise the route summary above, the other routers will forward packets destine to those two extra networks to you.

To get the summarization exactly without extras, you'll need two route summaries:

192.168.10.0/23
192.168.12.0/22

Chris T

Chris Bloomfield said... 27 May 2009 at 23:17  

Hi,

Apologies for my absence. It seems that some of these replies go to my junk mail :-|

It is good to see you guys helping each other and using my technique. Please keep the questions coming so we can help each other.

Best of luck in your studies!

Chris

Tim said... 12 June 2009 at 11:09  

Thoughts on how to summarize two different subnets? Like 10.1.0.0/24 and 10.1.7.0/24 without wasting address space and only summarizing the two networks in one statement?

Thanks for the post, it's very helpful

kaushal said... 14 June 2009 at 09:01  

You and Your trick are awesome...

It been years I was struggling with this but you have made me count it on fingers and in less than a min...

Amazing....

Thanks a ton

Sidney said... 10 August 2009 at 16:28  

I'm having a hard time understanding route summarization in this this example:
172.16.24.0
172.16.16.0
172.16.55.0
172.16.3.0
172.16.2.0

the summarized route was 172.16.2.0/18. How did they came with the answer. Can you please me understand this topic better.

Chris Bloomfield said... 18 August 2009 at 18:40  

Hi Sidney,

You have to summarize from 172.16.2.0 to 172.16.55.0.

Now ask yourself what power of 2 would cover all of those addresses? It would take 6 bits (i.e. 2 ^ 6 = 64).

As your subnets are in the third octet, the default subnet mask for that octet is /24. If you take your 6 bits that you need and subtract it from 24 you get a /18 mask.

Append the /18 to your lowest address to get 172.16.2.0/18. However, to be technically correct it ought to be 172.16.0.0/18 as the /18 covers everything from 172.16.0.0 to 172.16.63.255.

HTH

Agris said... 18 September 2009 at 03:31  

Hi,

What about this case:

10.2.1.0, 10.2.2.0, 10.2.3.0 and 10.2.4.0 , all with mask 255.255.255.0

Using the decimal way it should be simply 10.2.1.0 mask 255.255.252.0

However ICND2 book states that answer is 10.2.0.0 mask 255.255.248.0.
Summary route calculator gives the same result, contradicting with result gained from this magic decimal method.

Not sure what's wrong :(

thanks
John

Chris Bloomfield said... 18 September 2009 at 10:04  

This is a slightly more complicated question. Look at the answer you've given. Your mask is 255.255.252.0 which gives a block size of 4 in the 3rd octet. Therefore the subnets start at 0, 4, 8, 12 etc in the third octet. In your question you have addresses in the 10.2.0.0/22 subnet and one address, 10.2.4.0 which is part of the 10.2.4.0/22 subnet.

You can't simply summarize 4 subnets and expect 2 bits to cover them as they could straddle a boundary. You therefore have to use 3 bits so your subnet mask is 255.255.248.0 and your first address is 10.2.0.0.

In essence ask yourself what is the smallest block size that all addresses you are trying to summarize can fit in?

HTH

ash said... 7 October 2009 at 21:13  

Great Place,, some great mind live here...
can u please further go into details

ash said... 27 October 2009 at 11:31  

Given below address, could you please summarrize without over summarizaon--

172.30.31.0/24
172.30.32.0/24
172.30.33.0/24

172.30.34.0/23
172.30.36.0/23
172.30.38.0/24

This is the answer, but i just dont get it.


172.30.31.0/24

172.30.32.0/22
172.30.36.0/23

172.30.38.0/24


Could you also post few example on how to find network range and host range from given route summarized addresss..

thanks...

Chris Bloomfield said... 30 October 2009 at 20:09  

Hi there. So you have the following:

172.30.31.0/24
172.30.32.0/24
172.30.33.0/24
172.30.34.0/23
172.30.36.0/23
172.30.38.0/24

This is the same as:

172.30.31.0/24
172.30.32.0/24
172.30.33.0/24
172.30.34.0/24
172.30.35.0/24
172.30.36.0/24
172.30.37.0/24
172.30.38.0/24

This is because /23 covers two numbers in the third octet. There is nothing complicated with this and is the same as any subnetting question. If you was given the address 172.30.36.0/23 and was asked to find the valid host range you would work out that the block size is 2^(24-23)=2 in the third octet. Your subnet starts at 172.30.26.0 and ends with 172.30.27.255 with a valid host range between those addresses.

We are asked to summarise those addresses without oversummarisation. Let's think about how many addresses we have there. We have eight addresses which we could say needs 3 bits to cover as 2^3=8. If you count up in 8's from 0 you will have 0,8,16,24,32,40......

So remember we have 31 in the third octet in our range of addreses so that can't be in the same summary as the other addresses.

If we reserve 4 bits we then have 2^4=16 addresses which for one is too many addresses as we cannot oversummarise and two, the numbers go 0,16,32,48 so yet again 31 is outside of the summary.

We therefore have to reserve fewer bits. If we reserve two bits giving us a block size of 2^2=4 we can summarise a few of the addresses. Counting from 0 we have 0,4,8,12,16,20,24,28,32,36,40.

We can therefore summarise 32,33,34,35 without oversumarisation given us an address of 172.30.32.0/22. Remember that we reserved 2 bits so we subtract that from the /24 mask giving us the /22. This leaves us with:

172.30.31.0/24
172.30.36.0/24
172.30.37.0/24
172.30.38.0/24

If we reserve 1 bit we would have a block size of 2^1=2. Countng up in 2's from 0 we have 0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40......

36 and 37 can therefore be summarised together givng us an address of 172.30.36.0/23. Remember we have reserved 1 bit so we subtract that from the /24 to give us the /23. This leaves us with:

172.30.31.0/24
172.30.38.0/24

We have proved that these cannot be summarised any further and hence the answer is

172.30.31.0/24
172.30.32.0/22
172.30.36.0/23
172.30.38.0/24

HTH

Mundul said... 8 January 2010 at 01:56  

Hi

Bunch of geniuses here huh.


I have very siple question .for most of can help me

192.168.100.17 – 192.168.100.30.

range wildcard mask is 0.0.0.15


How this become ?

Thank you

Chris Bloomfield said... 8 January 2010 at 15:10  

Hi

Bunch of geniuses here huh.


I have very siple question .for most of can help me

192.168.100.17 – 192.168.100.30.

range wildcard mask is 0.0.0.15


How this become ?

Thank you

Not quite sure what you're on about but think about what your host range is there:

192.168.100.17 to 192.168.100.30

This is 14 host addresses so the mask of this address is 255.255.255.240 or /28. The subnet is therefore 192.168.100.16/28.

The wildcard version of your subnet mask is 0.0.0.15.

HTH,

Chris

Mundul said... 9 January 2010 at 00:29  

Thankh you ,thats exatly what i needed

Mundul said... 13 January 2010 at 04:36  

Question:

What is the most efficient summarization that can be used by R1 to advertise its network to R2???

R1 connected to R2 via serial..
R1 has 5 fa interfaces as following subnets..

172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

Possible answers are

a.172.1.0.0/21
b.172.1.0.0/22
c.172.1.4.0/22

I choose a. which happened to be d wrong answer..

Correct answer was actually c !!!!!according to the test answer


Can you kindly confirm that the test answer is corect or mine is correct?


Thank you

Chris Bloomfield said... 13 January 2010 at 20:21  

172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

Possible answers are

a.172.1.0.0/21
b.172.1.0.0/22
c.172.1.4.0/22

Your answer of 172.1.0.0/21 gives you a summary of 172.1.0.0 through to 172.7.255.255. Can this be done more efficeintly? Yes it can. You have four /24 subnets (don't get confused with the two /25's as they are the same as 172.1.4.0/24). Will our addresses fit into a block size of 4? Yes they will. Counting in 4's in our 3rd octet gives us:

172.1.0.0
172.1.4.0
172.1.8.0
etc......

To get a block size of 4 we need 2 bits (i.e. 2 ^ 2 = 4). Subtract this from /24 and we get /22. Therefore our address is 172.1.4.0/22.

HTH

NVV2009 said... 24 February 2010 at 09:36  

Wow this is such a greatttt blog Chris could youu continue posting new contents? can i also ask you for a link exchange to my site?
I learned from this one
http://ccna-in-cebu.blogspot.com/
thnx...

Devin said... 8 March 2010 at 21:18  

Hi Chris

I love your technique for summarization. My question is this:

Which of the following summarized subnets is not a valid summary of the 10.1.55.0, 10.1.56.0 and 10.1.57.0, with /24 mask.

a. 10.1.48.0 255.255.240.0
b. 10.1.55.0 255.255.255.0
c. 10.0.0.0 255.0.0.0
d. 10.1.32.0 255.255.224.0

After using your technique, I found that with 3 subnets you would have to use 2^2=4, which gives a mask of /20 on the subnet 10.1.0.0. I concluded that the summary route would be 10.1.0.0 255.255.240
but that doesn't fit with these answers. Please help

I picked answer d because the others seemed to summarize one or more of the subnets in question.

Chris Bloomfield said... 8 March 2010 at 21:37  

Hi Devin,

I think the question is asking which one of the answers does not summarise all 3 addresses given in the question.

Answer A is 10.1.48.0 255.255.240.0 which summarises 10.1.48.0 to 10.1.63.255 so that answer summarises all 3 addresses. Block size of 16 in the 3rd octet.

Answer B is 10.1.55.0 255.255.255.0 which summarises 10.1.55.0 to 10.1.55.255. This does not summarise all 3 addresses therefore this is the correct answer to the question. Block size of 0 in the 3rd octet.

Just to prove a point Answer C is 10.0.0.0 255.0.0.0 which summarises 10.0.0.0 to 10.255.255.255 so that answer summarises all 3 addresses.

Answer D is 10.1.32.0 255.255.224.0 which summarises 10.1.32.0 to 10.1.63.255 so that summarises all 3 addresses. Block size of 32 in the 3rd octet.

HTH and good luck with your studies!

Chris

paul.dye said... 8 August 2010 at 18:08  

Hi Chris

I have seen this on this site concerning cidr

http://www.opennet.ru/soft/cisco-configuration.html#configuring-routes


Classless Inter-Domain Routing.

With the recent explosion of the internet, Dividing address into class A, B, C, and D networks is no longer adequate.
For example, the following example routes 8 class Cs at the specified router.

ip route 221.243.242.0 255.255.248.0 128.230.3.1

Note that the only change from the above examples is the different mask. This command uses subnet style netmasks to split off 8 class C networks beginning at 221.243.242.0 through 221.243.250.0 and lists 128.230.3.1 as the next-hop router.

Is the starting network correct 221.243.242.0 ??

I worked it out as

How many supernets 221.243.242.0 255.255.248.0

take 3rd octet

xxxxxxx.xxxxxxxx.11110010.xxxxxxxx .242 /20
xxxxxxx.xxxxxxxx.11111000.xxxxxxxx .248 /21

do AND on bits

xxxxxxx.xxxxxxxx.11110000.xxxxxxxx .240 starting subnet


find blocksize

2/(boundary bits used - mask bits)

2/(24-21) = 2/3 = 8 = blocksize (Max Supernets)

221.243.240.0 221.243.240.1 - 221.243.247.254

221.243.248.0

Supernet Address Range = 221.243.240.0 - 221.243.247.255

I cant see how it can start @ 221.243.242.0 through to 221.243.250.0

paul.dye said... 8 August 2010 at 18:24  

Hi Chris

I was looking at a site concerning cidr - supernetting and this was suggested

For example, the following example routes 8 class Cs at the specified router.

ip route 221.243.242.0 255.255.248.0 128.230.3.1

This command uses subnet style netmasks to split off 8 class C networks beginning at 221.243.242.0 through 221.243.250.0 and lists 128.230.3.1 as the next-hop router.

I looked at it and tried to find summarized subnets

How many supernets in 221.243.242.0 255.255.248.0

taking 3rd octet

xxxxxxx.xxxxxxxx.11110010.xxxxxxxx .242 /20
xxxxxxx.xxxxxxxx.11111000.xxxxxxxx .248 /21

do AND on bits

xxxxxxx.xxxxxxxx.11110000.xxxxxxxx .240 starting subnet


find blocksize

2/(boundary bits used - mask bits)

2/(24-21) = 2/3 = 8 = blocksize (Max Supernets)

221.243.240.0 221.243.240.1 - 221.243.247.254

221.243.248.0
221.243.256.0

Supernet Address Range = 221.243.240.0 - 221.243.247.255

i cant see how they state

"split off 8 class C networks beginning at 221.243.242.0 through 221.243.250.0"

i see the range from 221.243.240.0 through to 221.243.247.255

Supernet Address Range = 221.243.240.0 - 221.243.247.255

I'm not sure where they are coming from ??
I have used CIDR subnet calculator as well .. am i wrong or just missing something ??


regards

Paul

Chris Bloomfield said... 21 August 2010 at 10:27  

Hi Paul,

Apologies for not getting back to you sooner. You are indeed correct. A simple bit of maths from the author wouldn't go amiss. The block size is 8 so you can only start at 240 and not 242 as that is not a multiple of 8.

Good spot and best of luck with your studies!

Munoo said... 24 September 2010 at 18:23  

Hi Chris,

I am quiet an admirer of your skills and i have been wondering for a while if i should bother you with another question but this has been bugging me for so long and i have been having arguments with other students about it guess you can help me? :) you have my sincere gratitude and sorry about disturbing if you were occupied!!

Whats the best way of summerising network having addresses 172.1.50.0 all the way down to 172.1.70.0 could you explain it in a way that even a thick skull like me gets it easily? :)

Regards

Chris Bloomfield said... 25 September 2010 at 13:37  

Hi Munoo.

The answer to your question depends on whether you want to over-summarize or whether you want an exact fit.

172.1.50.0 to 172.1.70.0 does not fit a contiguous block of addresses.

172.1.0.0/17 covers 172.1.0.0 to 172.1.128.0 and would satisy your requirements but over-summarizes a fair bit.

To get exact summarizations you will need a few addresses:

172.1.50.0/23 covers 50 and 51 in 3rd octet
172.1.52.0/22 covers 52 to 55 in 3rd octet
172.1.56.0/21 covers 56 to 63 in 3rd octet
172.1.64.0/22 covers 64 to 67 in 3rd octet
172.1.68.0/23 covers 68 to 69 in 3rd octet
172.1.70.0/24 covers 70 in 3rd octet

Hope this helps to resolve your argument :-)

Good luck with your studies!

Moog said... 7 October 2010 at 17:31  

Hi Chris

I love the way you explain these questions. However when i jsut have a question for exam that says

Given the address 192.168.20.19/28 which of the following are valid hosts on this subnet:

A. 192.168.20.29
B. 192.168.20.16
C. 192.168.20.17
D. 192.168.20.31
E. 192.168.20.0

I dont know if i need to take a break but its confusing me.

Thanks again

Brian

Moog said... 7 October 2010 at 17:32  

Hi Chris

I love the way you explain these questions. However when i jsut have a question for exam that says

Given the address 192.168.20.19/28 which of the following are valid hosts on this subnet:

A. 192.168.20.29
B. 192.168.20.16
C. 192.168.20.17
D. 192.168.20.31
E. 192.168.20.0

I dont know if i need to take a break but its confusing me.

Thanks again

Brian

Chris Bloomfield said... 8 October 2010 at 19:09  

Hi Brian,

This is a nice and easy question and should give you easy points!

You have address 192.168.20.19/28 so let's work out the subnets. Our block size is 16. This is worked out by saying our next boundary is 32 and we have been given a mask of /28. We then plug those into the equation of Block Size = 2 ^ (next_boundary - given_subnet_mask) which in this case is 2 ^ (32 - 28) = 2 ^ 4 = 16.

Our subnet mask is /28 and the 28th bit is in the 4th octet so we will count up from 0 in our block size in the 4th octet.

192.168.20.0
192.168.20.16
192.168.20.32
192.168.20.48
.......etc

Our address is 192.168.20.19/28 so that must sit on 192.168.20.16/28 subnet. The valid host range for this subnet is 192.168.20.17 - 30 (remember 192.168.20.31 is the broadcast address for this subnet).

The answers you were given are:

A. 192.168.20.29
B. 192.168.20.16
C. 192.168.20.17
D. 192.168.20.31
E. 192.168.20.0

We can therefore say that answers A and C are correct.

Hope this helps and good luck with your studies!

Chris

Moog said... 9 October 2010 at 05:24  

Hi Chris

Thanks for that, i think ive got it now.The way i was looking at doing it was looking at:

1) 255.255.255 =24 bits (/24)

2) To get to 28 bits (/28) i would have to add 4 masked bits which would = 16 bits (or subnets)

3) Therefore the increment is 17 - 31 but we use 30 as if you:

-1 from 31 = 30 which = broadcast address

I gather that 32 canot be used as this is already taken?

Can i ask you another question if possible. Its a similar example, but just want to check im on the right track

Thanks

Brian

Moog said... 9 October 2010 at 09:20  

Hi Everyone /Chris

The next question im having trouble with is:

Need to design an addressing scheme with a MINIMUM of 300 sub-networks and a MAXIMUM of 50 hosts per per subnet using 1 class b address.

the answers are

a)255.255.255.0

b)255.255.255.128

c)255.255.255.252

d)255.255.255.224

e)255.255.255.192

f) 255.255.248.0

Confused again here.

I know ticking down (on my chart) the nearest to 300 subnets would be 512 subnet which would be 9 bits.

I guess this would be taken from the 3rd octet(255.255.255)

then 1 bits from the 4th =(128)

Im stuck where to go from here really.

Sorry to hassle you with this Chris.

Thanks

Brian

Chris Bloomfield said... 13 October 2010 at 21:05  

Hi Brian,

I think the use of MAXIMUM and MINIMUM should be swapped round in the question.

They ask for a Class B address and we know that they require 300 subnets. You can only achieve that by borrowing 9 bits or more so that would make the mask 255.255.255.128 (i.e. /16 + 9 = /25). This will give you 126 usable host addresses per subnet.

HTH,

Chris

girl_on_the_billboard said... 20 October 2010 at 20:21  

i need help with a packet tracer lab (3-1 or also known as 3.5.2)

I cant figure the addressing out I seem to be getting the subnet right

the network address is 192.168.9.0/24

need 10 hosts for branch1 lan 1
need 10 hosts for branch1 lan 2
need 10 hosts for branch2 lan 1
need 10 hosts for branch2 lan 2
need 20 host for HQ
hq to branch1 needs ip for each end of the link (2?)
hq to branch2 needs ip for each end of the link (2?)


I have spent hours now and need help badly....THanks

Chris Bloomfield said... 21 October 2010 at 20:59  

Hi,

OK so we have a starter for ten in that our adddress is 192.168.9.0/24

Always start with the subnet that requires the largest amount of hosts - 20 in this question for HQ. We need to reserve 5 bits to achieve enough host addresses for 20 hosts (e.g. (2 ^ 5) - 2 = 30 whereas (2 ^ 4) - 2 = 14 which isn't enough). Subtract those 5 bits from our next boundary which is /32 - 5 = /27 which is our subnet mask. We can now say that our first subnet is for HQ which is 192.168.9.0/27. The next address available is 172.168.9.32.

The first four subnets in the question require 10 hosts. The question is how many host bits should we reserve to get 10 hosts? Well, that is relatively straightforward. 4 bits would give us (2 ^ 4) - 2 host bits which is 14 hosts. If we subtract that from our next boundary (32 in this case) we have a subnet mask of 32 - 4 bits borrowed = /28 mask. We therefore have to start at 192.168.9.32 as we have already taken up the first 32 addresses with the HQ subnet. The next four subnets will therefore be:

192.168.9.32/28
192.168.9.48/28
192.168.9.64/28
192.168.9.80/28

Notice each one has a block size of 16 in the last octet.

The final two subnets are for point-to-point links which are traditionally given /30 masks as that gives us exactly two host addresses (i.e. 2 ^ (32 - 30) - 2 = 2). The question is do we give the next available addresses to the point-to-point links or the last two available addresses? This is a matter of design preference but I tend to give point-to-point links a subnet at the top end of the range. I would therefore give the following two subnets to your point-to-point links:

192.168.9.248/30
192.168.9.252/30

Hope this helps and good luck with your studies!

i need help with a packet tracer lab (3-1 or also known as 3.5.2)

I cant figure the addressing out I seem to be getting the subnet right

the network address is 192.168.9.0/24

need 10 hosts for branch1 lan 1
need 10 hosts for branch1 lan 2
need 10 hosts for branch2 lan 1
need 10 hosts for branch2 lan 2
need 20 host for HQ
hq to branch1 needs ip for each end of the link (2?)
hq to branch2 needs ip for each end of the link (2?)


I have spent hours now and need help badly....THank

Ravan said... 30 October 2010 at 15:41  

hi brother i have didn't get most difficult superneting part for 10 subnet how to do pls be more concise for this why do you mean by break it space in 3 part i didn't get this concept why you chose 10.16.32.0 instead of 10.16.31.0 to get summerization hope you clear my doubts on this topic i am witing for your reply and i must tell you what a easy and best method for summerization you share with word i have no word for you pls post this type of more concept for all topics and pls pls clear my doubt

noteasy said... 3 November 2010 at 22:56  

For those of you who have a problem finding the exponent of an expression for a subnet you can use the following formula : LOG (base 2) of X = Log X / Log 2. in other words supposed you have 128 bits and you need to know what is exponent value of 2 ( to which power 2 is raised to get a value of 128 ) use your regular calculator to fine the log of 128 and divide the result by the log of 2 = (log(128)/log(2)) = 7.. hope this helps

Xaviour said... 24 November 2010 at 04:25  

Hi chris
I dont know how to express my gratitude. Really,,,,,,,, really,,,,,, a fantastic technique

thanq very much

xavioru

torun said... 29 December 2010 at 19:18  

Hi Chris,

Thanx a lot. I really liked the idea for calculating the summary easily. I have a problem and it would be great if you could help me figure out the solution to this problem. The problem is to calculate a summary for a range of subnets from 172.16.20.10/24 to 172.16.20.40/24 . I have to use this range and block them to access a server using ACL. I have tried your method of summarization, but haven't succeeded., maybe because i must have made some mistake. Could you help me to do this calculation??


Regards

Aktar

Chris Bloomfield said... 29 December 2010 at 20:05  

Hi,

I take it that you can use more than a single line in your access list?

Your problem is to calculate a summary for a range of subnets from 172.16.20.10/24 to 172.16.20.40/24.

We know that we have to summarise from 10 in the 4th octet. We have to use a power of 2 that divides into 10. The best one is 2 to the power of 1 = 2 which is 255.255.255.254.

Our first line in our access list is something like this:

access-list 1 deny 172.16.20.10 0.0.0.1

Now we have covered off .10 and .11 we can start at .12. So which power of 2 divides exactly into 12? The answer is 2 as 2 ^ 2 = 4. This is a mask of 255.255.255.252 so our next line of the access-list is:

access-list 1 deny 172.16.20.12 0.0.0.3

We have now covered off .10 to .15 so we now start at .16. So which is the largest power of 2 that divides exactly into 16 but doesn't go above the .40 limit? The best answer is 4 as 2 ^ 4 = 16 which is a mask of 255.255.255.240. Our next line in the access-list is:

access-list 1 deny 172.16.20.16 0.0.0.15

We have now covered .10 through to .31 so we start at .32. So which is the largest power of 2 that divides exactly into 32 but doesn't go above the .40 limit? The best answer is 3 as 2 ^ 3 = 8 which is a mask of 255.255.255.248. Our next line in the access-list is:

access-list 1 deny 172.16.20.32 0.0.0.7

We have now covered .10 through to .39 so that leaves us with the simple entry of:

access-list 1 deny host 172.16.20.40

All in all then your access-list will look like the following:

access-list 1 deny 172.16.20.10 0.0.0.1
access-list 1 deny 172.16.20.12 0.0.0.3
access-list 1 deny 172.16.20.16 0.0.0.15
access-list 1 deny 172.16.20.32 0.0.0.7
access-list 1 deny host 172.16.20.40
access-list 1 permit any

Hope this helps!

Chris

Moog said... 26 May 2011 at 22:22  

Hi Chris

Apologies for the delay getting back to you, but just want to say thanks for your help and great site, i managed to finally pass the ccna after a couple of failed attempts.

I do have a qucik question about increments within VLSM though.

Finally, id say you'd do well in knokcing up a load of questions then selling them for a few pounds i think alot of people would take it up. I know i would.

Thanks again

Moog said... 26 May 2011 at 22:24  

Hi Chris

Apologies for the delay getting back to you, but just want to say thanks for your help and great site, i managed to finally pass the ccna after a couple of failed attempts.

I have a question about VLSM too if thats ok??

Also you should create some questions and then sell them as a download for a few pounds, i think people would snap that up.

Once again Thanks again Chris

Chris Bloomfield said... 13 July 2011 at 14:14  

Great news. Feel free to ask.

cuongtruong said... 12 August 2011 at 04:20  

Please determine the summary address of these networks :
172.148.0.0/13 through 172.156.0.0/13
send result to my email cuongqtruong@yahoo.com . Thanh a lot.

Tim R said... 12 August 2011 at 15:22  

Chris, RE:
Sailesh,

(First of all, thanks to Chris B's excellent technique.)

Your answer of 192.168.8.0/21 is correct if you can live with over-summarization. Because this block includes address blocks 192.168.8.0/24 and 192.168.9.0/24. When you configure your router to advertise the route summary above, the other routers will forward packets destine to those two extra networks to you.

To get the summarization exactly without extras, you'll need two route summaries:

192.168.10.0/23
192.168.12.0/22

Would it not be better to use
192.168.10.0/22
192.168.14.0/23
Please clarify, as I am a Summarization newb and just trying to figure if my technique application is correct. Thanks,
Tim R

I_luv_my_127.0.0.1 said... 7 November 2011 at 22:22  

Thank you Chris, this has really been helpful. I am preparing for my CCNP IP route exam and this will make life easier for me

Chris said... 22 January 2012 at 12:15  

I would like to verify a summarization excercise in the CCNA book I am trying...

Need to find out the summary address and summary mask of these addresses...

172.148.0.0/13 through 172.156.0.0/13

If I count correctly there are 9 nets...
Then when trying 2^4=16 as block size in 2nd octet...

...then the summary address and its mask using your excellent and fast method (oversummarisation assumed OK) is likely to be:

172.144.0.0/9

How come the book says 172.144.0.0/16?
Where does the 7 network bits difference come from, as I know it concerns a Class B network space, but still I don't understand why the summary mask is said to be /16 instead...

Mukom Akong TAMON said... 12 February 2012 at 05:59  

Great post! ...and some great discussion. You might want to check out some techniques and worked examples at http://techxcellence.net/2012/01/29/calculating-exc-v4-summary-addresses/

Jerry said... 27 March 2012 at 04:24  

Chris, I have the same question except I calculated it as 172.144.0.0 /12. I am pretty confused.

Mike said... 8 May 2012 at 05:24  

If it makes you feel any better I'm American and I typically use English spellings as I've never learned "American." ;)

Sri said... 21 August 2012 at 12:59  

Hi Chris,
Thanks a lot for your quicker math. I have a doubt, might be am missing the simple thing.

One of question was summarize below networks
172.1.4.0/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
Above four can be summarized with block size 4, so summarized address would be 172.1.4.0 and now coming to the mask, math is 2^2 = 4, y = 2, subtract 2 from /25 which will be 172.1.4.0/23 , but answer is 172.1.4.0/22. When you have a mix of prefix, from which do you subtract y.

Thanks in Advance

Praveen said... 5 September 2012 at 16:21  

Hi Chris

we learned how to summarize now tell us how do we find subnets which get summarized like suppose we have summmary route 192.168.32.0/20..now i wana determine what subnets this summary hold could u pls tell the math trick for finding address which get summarize in 192.168.32.0/20...........


Tapas Pramanik said... 5 December 2012 at 19:07  

hi
Thank you,this is more helpful for me.before i read your blog i have confuse about summarization .but when i read your blog i got it perfectly .
so,again thanks for that....

Tapas Pramanik said... 5 December 2012 at 19:16  

hi
Thank you, this has been most helpful.The maths and working out of the question is very simple.



thanks bro....

Tapas Pramanik said... 5 December 2012 at 19:19  

hi
Thank you,this is more helpful for me.before i read your blog i have confuse about summarization .but when i read your blog i got it perfectly .
so,again thanks for that....

chelsea said... 11 December 2012 at 17:36  

Hi Chris

can you please help me clarify my doubt. it seems liks simple summarise question but i am confused by the answer provided

need to summarise address

192.168.1.0 through 192.168.120.0

my answer is : 192.168.0.0/17

but in one of the book , answer is mentioned as 192.168.0.0/25

can you please help me clarify this?

Regards
Dilip

David E Smith said... 20 December 2012 at 15:59  

I purchased CCNA eTest Prep. by Todd Lammle. There is a question in the exam that I believe is incorrect but need someone to assist me with figuring out the right answer.

Question (QID: 640-802-Q3094)
Which of the following ie the best summarization of the following networks:
192.168.10.128 /25
192.168.10.0 /25

A) 192.168.10.128 /24
B) 192.168.0 /24
C) 192.168.10.0 /24
D) 192.168.10.10 / 23



My answer is 192.168.10.0 /25

Whatis the right answer

Chris Bloomfield said... 20 December 2012 at 16:17  

192.168.10.0/25 covers 192.168.10.0 through to 192.168.10.127

192.168.10.128/25 covers 192.168.10.128 to 192.168.10.255

Therefore the question is asking you to summarize 192.168.10.0 to 192.168.10.255 which would be 192.168.10.0/24

HTH

David E Smith said... 20 December 2012 at 16:28  

Thanks Chris. Will you please explain what I did wrong? how did you get your answer? I followed you method on your page

Chris Bloomfield said... 20 December 2012 at 16:33  

Hi David,

I've given you the breakdown of the answer and why it is what it is.

You're asked to summarize 192.168.10.0/25 and 192.168.10.128/25.

Clearly the answer you have of 192.168.10.0/25 is wrong as that is one of the constituent subnets you are trying to summarise.

192.168.10.0/24 is the answer.

Remember each /24 is made of up of two /25, or 4 /26, or 8 /27 etc.

HTH

Voltage said... 29 December 2012 at 15:10  

Hi, can you help me with this;
Determine the summary address and the mask to be used that will summarize the subnet 172.148.0.0/13 through 172.156.0.0/13
Explanations will be appreciated.
Thanks

Ilir Cengu said... 11 February 2013 at 10:51  

Hi Voltage ,
It is quite paradoxal this question,cause if you can see there are two B class addresses with a A valid subnet mask address.Even me was surprised by the answer.
but here is my explanation (by myself):
If you can see the difference between networks is exactly 2^3=8,we cannot find a route summarization for both networks,so we need a 2^4=16 networks address space ,that includes both networks:
by counting with 16`s, we come to network 172.144.0.0 - 172.160.0.0
As per the subnet mask,I think since they are B addresses , we need a valid subnet mask address for this,(a /16- /30 mask),the first mask that comes in row is /16 ,that means it is not a subnetted network,it is a classful address.So the address of summarization is: 172.144.0.0 /16.
The summarization topic is something like "try and find".
If I am wrong please correct me.

francis atie said... 26 April 2013 at 04:18  

how do u summarize
192.168.9.96
192.168.9.64
192.168.9.192

and
192.168.9.96
192.168.9.64

Stanio said... 27 June 2013 at 13:49  

Hi,

Can you tell me, please if it is possible to summarize the following range of subnets

010.002.000.128/27
010.002.001.128/27
010.002.002.128/27
010.002.003.128/27
010.002.004.128/27
...... till ......
010.002.048.128/27

The difference is only in the third octet

Thanks

Chris Bloomfield said... 1 July 2013 at 09:28  

010.002.000.128/27
010.002.001.128/27
010.002.002.128/27
010.002.003.128/27
010.002.004.128/27
...... till ......
010.002.048.128/27

Unfortunately not.

Take the first subnet:

10.2.0.128/27

This covers addresses 10.2.0.128 through to 10.2.0.159

As soon as you try to summarise within the 3rd octet you will be encompassing everything in the 4th octet.

Put simply 10.2.0.0/24 covers 10.2.0.0 through to 10.2.0.255 which covers not only the subnet you want but also all of the other subnets.

HTH

sia rey said... 21 August 2013 at 07:02  

I read your post on cidr calculator. Your post is very important to the students for understanding the concept of cidr calculator.

Bakers said... 5 September 2013 at 09:11  

"The best answer is the following:

Break the space down into three ranges. We can summarize 10.16.32.0 to 10.16.39.255 with 3 bits (i.e. there are 8 subnets and 2 to the power of 3 is 8). We simply leave the other two address as they are. We have therefore avoided any wasted address space."
why not 'We can summarize 10.16.31.0 to 10.16.39.255 with 3 bits (i.e. there are 8 subnets and 2 to the power of 3 is 8)'

Unknown said... 19 September 2013 at 17:50  

Hi. I have a slightly different question.
How do you subnet a given address space for a given network topolgy into subnets of given sizes so that you can then perform route summerization on your routers.
Basically, if you have let's say 3 routers connected to several subnets and to each other, how do you assign address spaces in a way that you avoid discontiguous networks which as we all know is the downfall of summerization. Is there a systematic approach for that?
thank you
Ali

Unknown said... 19 September 2013 at 17:51  

Hi. I have a slightly different question.
How do you subnet a given address space for a given network topolgy into subnets of given sizes so that you can then perform route summerization on your routers.
Basically, if you have let's say 3 routers connected to several subnets and to each other, how do you assign address spaces in a way that you avoid discontiguous networks which as we all know is the downfall of summerization. Is there a systematic approach for that?
thank you
Ali

@k@$h said... 29 January 2014 at 13:48  

HI
plz solve this question
Need to determine summarize address and mask that will summarize the address
203.168.6.0/24 and 203.168.60.0/24

as i am not able to understand this question

@k@$h said... 29 January 2014 at 13:53  

Hi
plz provide the solution of below question as i am not able to understand the same

Q: Determine the summary address and the mask to be used that will summarize the subnets.

203.168.6.0/24 and 203.168.60.0/24

Plz suggest me what is the solution of the same as answer in the book is unexpectable

192.168.96.0 255.255.240.0

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