Summarization help

Have a question or want to start a discussion? Post it! No Registration Necessary.  Now with pictures!

Threaded View
Can someone explain to me exactly how you get to these answers from this
question? Its a question from my Cisco Press CCNP flash cards.

What is the best summarization for:
10.0.152.0/21
10.0.160.0/21
10.0.168.0/21
10.0.176.0/21
10.0.184.0/22
10.0.188.0/22

I know the answers are 10.0.152.0/21 and 10.0.160.0/19, but I am a bit
confused as to how you work out the problem to find these solutions,
especially for the 10.0.160.0/19. Any help is appreciated.

Re: Summarization help
Just do it in binary.

10.0.152.0/21 = 10.0.128+0+0 +0 +16 (+8+0+0.0)
10.0.160.0/21 = 10.0.128+0+32+0 +0
10.0.168.0/21 = 10.0.128+0+32+0 +8
10.0.176.0/21 = 10.0.128+0+32+16+0
10.0.184.0/22 = 10.0.128+0+32+16+8+0
10.0.188.0/22 = 10.0.128+0+32+16+8+4

common bits are 10.0.128+0. Therefor the answer is 10.0.128.0/18

Doan

On Thu, 8 Jun 2006, Michael wrote:

Quoted text here. Click to load it


Re: Summarization help
Quoted text here. Click to load it

Which naturally does not correspond to any of the answers he said were in
the flash book.  How do you get the 18?



Re: Summarization help
On Fri, 9 Jun 2006, A Sharp wrote:

Quoted text here. Click to load it
I believe the book is wrong.  It is better to understand the concept than
blindly believe in the book.  The /18 comes from 8+8+2 (8 in the first
octet, 8 in the second octet and 2 in the third octet).

Doan



Re: Summarization help
Write down the third octet in binairy for each IP-address and but them in a
list:

10011000
10100000
10101000
10110000
10111000
10111100

They all differ, accept the last two numbers. The last two numbers are
always 00. Put a line in this:

100110|00
101000|00
101010|00
101100|00
101110|00
101111|00

Now you can see that you're summerisation is 10.0.152.0/22.. I don't see how
you get 10.0.152/21 and 10.0.160.0/19...

Quoted text here. Click to load it



Re: Summarization help
Quoted text here. Click to load it

And this thread just gets more confusing, doesn't it?  Now we have four
right answers.



Re: Summarization help
This confermes my opinion:

http://www.ciscopress.com/articles/article.asp?p=330807&seqNum=5&rl=1

Quoted text here. Click to load it



Re: Summarization help

It said:

"For summarization to work properly, addresses must be carefully assigned
in a hierarchical fashion so that summarized addresses share the same
high-order bits."

High-order bits are on the left, not right.

Doan

On Fri, 9 Jun 2006, CCNA Nerd wrote:

Quoted text here. Click to load it


Re: Summarization help

Aren't you suppose to look at the common bits from left to right?

Doan


On Fri, 9 Jun 2006, CCNA Nerd wrote:

Quoted text here. Click to load it


Re: Summarization help
Michael napisa=B3(a):
Quoted text here. Click to load it

It's simple ;)

First - let's write down network ranges:
1  10.0.152.0/21    10.0.152.0 - 10.0.159.255
2  10.0.160.0/21    10.0.160.0 - 10.0.167.255
3  10.0.168.0/21    10.0.168.0 - 10.0.175.255
4  10.0.176.0/21    10.0.176.0 - 10.0.183.255
5  10.0.184.0/22    10.0.184.0 - 10.0.187.255
6  10.0.188.0/22    10.0.188.0 - 10.0.191.255

2,3,4,5 and 6 can be summarized to 10.0.160.0/19    range: 10.0.160.0 -=20
10.0.191.255 (the same)

all these networks can be summarized to 10.0.128.0/18   range:=20
10.0.128.0 - 10.0.191.255 but the question is tricky.
We have only those network and nothing else - especially networks=20
10.0.128.0 to 10.0.151.0 can be somwhere else.

So the aswer in the flash cards is OK

------------------------
Ern

Re: Summarization help
How do you guys do this quickly ?



Ern wrote:
Quoted text here. Click to load it


Re: Summarization help
genki wrote:

Quoted text here. Click to load it

Practice.

The real nuts and bolts. To start, you are probably best going to the binary.


Quoted text here. Click to load it

Another little hint that you prbably know is the subtract from 256 - /21
255.255.248.0 256-248 is 8 so we know there are no missing bits. The last two
are a subnetting of .184.0 /21

Drop the third octet to binary and we get

152 10011000
160 10100000
168 10101000
176 10110000
184 10111000

When we look at the bits that don't change we get 10xxxxxx so the summary that
includes them all is 10.0.128.0/18


So we have a little problem. If what we wanted to summarise started at 128, we
would have a simple summary, but we are including quite a bit more in the
summary than we need.

Back to the binary then for a closer look


The mask is
248 11111000

152 10011000

Split here and we get

160 101 00 000
168 101 01 000
176 101 10 000
184 101 11 000

Where we have a portion that does not change - 101 then we get two bits where
we get all values from all 0s to all ones, then we get the host portion. We can
thus summarise this octet as 10100000 mask 11100000 - 10.0.160.0 /19

While learning it is quite important to go to the binary. I have 20 years in
networks, and still drop to the binary once in a while.

Once you *understand* what you are doing with the actual bits, you will get
used to the number patterns.
--
Paul Matthews                                         CCIE #4063
Please post questions to the NG, NOT by e-mail.

Re: Summarization help
Thanks for your response Paul !

Paul Matthews wrote:
Quoted text here. Click to load it


Re: Summarization help

Quoted text here. Click to load it

Then, the last two miss 4 bits each. What do you mean by missing bits,
please?


The last two
Quoted text here. Click to load it

And why is that?



Quoted text here. Click to load it

Did you decide on /18  from the substraction: 256 - 128?

Quoted text here. Click to load it


Is there any way to summarize by not writing down bits?

I can easily tell that

the first 4 subnetworks are multiple of 8:
10.0.152.0/21
10.0.160.0/21
10.0.168.0/21
10.0.176.0/21

and the last 2 networks are multiple of 4:
10.0.184.0/22
10.0.188.0/22

How can we go from here, please?

Thanks!

The Dude



Re: Summarization help
I have just finished this capter (I am on CCNA sem3)INtro to CLassless
routing
One of the assessment questions was a s folows

What is the summariztion address for the networks  172.21.136.0/24 and
172.21.143.0/24 ?
(ok forgot the "a")

The answer given is 172.21.136.0/21

10001000 -is the binary for 136
Is this how you work the mask out?
10001111 -143 ,so do you just take whichever 1's match and include the
zeros in your mask, or am I getting this tits up?
Thanks- if Mr MAtthews could answer I would be very grateful.
The Dude wrote:
Quoted text here. Click to load it


Re: Summarization help

Quoted text here. Click to load it

According to CISCO the summarization consists of the common leftmost bits.
In your case the first two octets are the same, so no need of writing them
down in binary.
The third octets in binary are as follows:

10001000 (136)
10001111 (143)

the common leftmost bits are:

10001xxx  (136)

It's just a coincidence that 136 is the same number.

So, the summarization of the two IPs will be 172.21.136.0.
What I do not get is the format: how did it change from /24 to /21? What's
the logic behind it?

The Dude



Re: Summarization help
My understanding is that as its the 5 bits in the 3rd Octet (and the 16
in the 1st 2 octets) this makes for a mask of 21? 5+16....
Is this correct?
TIA
The Dude wrote:
Quoted text here. Click to load it


Re: Summarization help

Quoted text here. Click to load it

Yes! You are right, thanks!

The Dude


Quoted text here. Click to load it



Re: Summarization help
Thanks -Lebowski! Hows the rug?
The Dude wrote:
Quoted text here. Click to load it


Re: Summarization help
gregg johnstone wrote:

Quoted text here. Click to load it

Yup. You know more than you think <G>

The thing about subnetting and summarisation is that it is NOT difficult. Some
people struggle, unfortunately some people that teach struggle. The people they
teach then struggle and so on, and we have lots of people struggling.

P.
--
Paul Matthews                                         CCIE #4063
Please post questions to the NG, NOT by e-mail.

Site Timeline