Summarization help

Are you serious? Some people are unbelievable.

How can you criticize peoples attempts to answer a question about a subject you admit knowing nothing about?

How about you read up on 'route summarization' and get back to us.

BernieM

ok, look at Doan's posts, he gives an answer (not on his flash cards). And Doan said another (dfiferent) answer (that matched his flash cards) is very good.

Ignore CCNA Nerd's explanation. If you have a look at the responses, you'll see it wasn't rated very highly. And was a bit amateurish, as you noticed .

Now perhaps you should study the topic.

I'm throwing my hat into the ring..... And I realize it's only worth two cents....

This is yet another example of a broken practice test question that are part of all of these products, regardless of the vendor, Test Kind, Boson, I don't care who it is. These people are trying to cash in on the Cert bandwagon and give give a rat's tail on proofing their materials.

It is never a good idea to create a route summarization statement that includes subnet space which is not actually available. As such, there is NO SINGLE summarization statement that can be created which will account for the subnets listed, and ONLY those subnets listed. The only statement which would include all the subnet space listed is a /18. However, doing this would not be wise because dozens of additional subnets would be included in that single statement. This is would be "bad." It's called "over summarizing."

The "best" answer would be to simply leave everything alone and advertise the 6 individual routes as they are and not create any ambiguity downstream.

Use these products with caution!

Better yet, just learn the material and trust in the force.

(LOL, I see I had problems 'proofing,' too. But, what do you want for 'free?' ..... ;-)

indeed, it's a misleading question, but isn't the answer a sensible one. to do the first on individually, and summarize 2-6

looking at ern's summary. networks/routes 2-6 can be summarized so as to cover all networks and only those networks.

ern's summary-

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 -

10.0.191.255 (the same)

I could have sworn the "certification bandwagon" came to a screeching halt about five years ago, right along time the tech boom went down in flames. Did I miss a revival? Does Transcender still command $150 and up per exam simulation?

It would be pretty close to $150...

I have read through this entire thread, and I have seen many of your posts, Mr. Sharp. My observation is this. Your posts are a real distraction. As if you are a grown up class clown. You say a lot, but nothing of any consequence. I would like to suggest that if you have nothing constructive to post, perhaps it would be better to leave things alone.

And I'd like to suggest to you to kiss my rosy red ass.

LOL. I love posts like yours. My observation is this: You, Parr snipped-for-privacy@hotmail.com, have never posted to this newsgroup before. Yet you're already a self-annointed moderator. Moderate THIS, shit for brains!

Glarb W. Shattenstein

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.

Might I suggest you take a look at MCSE instead!

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

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 b> > genki wrote:

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

Yes! You are right, thanks!

The Dude

Sorry - sloppy terminology on my part - what I should really have said is thatth address space we are trying to summarise is contiguous.

Look at the numbers, and the bit about subtract from 256 giving 8. That means for the /21 mask the third octet incerements by eight, but to try to throw you a little they put two four in - looking at the binary helps - two fours are eight, so together they are one /21.

Nope - the first two bits of the third octet are the two that don't change so that's 18 bits altogether.

Practice!

To really get this off pat, you need to *understand* the binary, and that the address field is really a 32 bit integer. subnetting and summarisation are basically logical operations on binary numbers.

If yu are asking that, the best thing I can say is look at the binary, look at the patterns of the bits.