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**posted on**

June 8, 2006, 7:01 pm

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.

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

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:

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...

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...

Re: Summarization help

Michael napisa=B3(a):

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

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

Practice.

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

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

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

please?

The last two

And why is that?

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

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:

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:

Re: Summarization help

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

gregg johnstone wrote:

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.

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.

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