Page 149 Summarization example 2

[Susmita]

I could not understand the summarization example 2 completely.:confused:
Please anybody more experienced in networking help me understand it.
Thanks in advance.

[crissa]

Hello,
you have the following ip ranges:

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

The two /25 (x.y.4.0 and x.y.4.128) add to a /24, together with the other three /24 this adds to a /22!
Bye, Tore

[tsunamoro]

I'll try to makes it simple.

Let's consider the subnets 172.1.4.0, 172.1.5.0, 172.1.6.0 and 172.1.7.0; they are part of the set of addresses "generated" by the 172.1.4.0/22. Why? Because you have 4, 5 6 and 7 that is a block size of 4 or .252. But the block size of 4 must to be applied to the third octect, that means 255.255.252.0.
What are the subnets of this mask? 172.1.0.0, 172.1.4.0, 172.1.8.0 and so on.
The subnet 172.1.4.0 includes all the addresses from 172.1.4.1 to 172.1.4.255, that means all the addresses of the original subnets.

[Carl]

Page 150 is very weak spot is the book. Most people at this point in the book when they see 172.1.4.0, and a mask of 255.255.252.0, are going to start subnetting, and come up with 64 subnets with 1022 hosts each.

[lildeezul]

i made a excellent post about summarization on another thread here it is.

Yes... but we are talking about route summarization.
you will understand this better once you get to routing protocols, especially EIGRP and OSPF...

todd list the network....

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


now the objective is to summarize these 4 networks into 1, so that the routing table wont be so congested.
again you will learn in routing ospf and eigrp.

but this is what you do...
you find where all 4 networks have in common.... (convert to binary if you want)
you can see that all four have 172.1 in comming. and the third octect is where it starts to variate and differenciate. (spelling) so we are going to look at the third octect and we see that the number range from 4-7....

now think....!!!! what mask will give us a smallest block size that will consist of numbers 4-7... hmm i am thinkin a block size of four... Correct??

yes.. So a block size of four in the third octect is 255.255.252.0 ..

do you get it????

hopefully this helps..

the goal is to see where the networks are in comming, and where they start to differ. in the third octect.....
then you find the range in that third octect. which is 4-7...
then ask youself, what is the smallest block size that will accommadate these number..
a block size of two (255.255.254.0) wont work becuase its to little... and block size of 8,16,32,64,128 would work, but your trying to find the smallest block size that will work..

which is 4..

hope this helps

[Big Evil]

I would do it this way.

As we are going to summarize in the third oct the binary would be
00000100 (.4)
00000101 (.5)
00000110 (.6)
00000111 (.7)

The common are at the 4, count from the left that is 6 . Keep that in mind.

The 172 part is the same all the way through, so give that vaule of 8, the .1 is the same give that value of 8. All parts that are the same give a value of 8.

So add that up.
8+8+ then add the 6 = 22.

Our address is 172.1.4.0 and the mask is a /22 (255.255.252.0)

[lammle]

I would not say this is a weak spot in the book, it is an important part in the book.
We are not subnetting here, we are summarizing.

networks
172.1.4.0, 172.1.5.0, 172.1.6.0 and 172.1.7.0

What is the interesting octet?
the third octet.

We can have a contiguous block of 4, 5, 6 and 7, which is a block size of 4.

Our summary address is always the first one in the range: 172.1.4.0


What mask gives us a block size of 4 in the third octet?

255.255.252.0

that's how you do it.
Todd

[LT72884]

Page 150 is very weak spot is the book. Most people at this point in the book when they see 172.1.4.0, and a mask of 255.255.252.0, are going to start subnetting, and come up with 64 subnets with 1022 hosts each.

i see your point but if you do it in binary you get this

10101100.00000001.00000100.00000000
11111111.11111111.11111100.00000000

now change the parts in bold to all 1's and when you logically "AND" you get

172.1.7.255

so the range is 172.1.4.0 - 7.255 or 4 class C subnets with 256 total IP's per class C subnet which = 1024. Thats if you are sumerizing if not then it is one big subnet with 1022 hosts that just happens to cover from 172.1.4.0 -7.255

i dont see how they would get 64 subnets because the change is in the 3rd octet and if it were 64 subnets then the 1's in the mask would stop at /18. I get 4 subnets cuz the 2^n formula. there are only 2 zeros to worry about in the third octect that change. When i say subnets i am refering to how many class c subnetts i could use. since /16 - /23 will refere to class c subnets.

/16 256 class c subnets
/17 128 class c subnets
/18 64 class c subnets
etc
etc
/23 2 class c subnets (192.168.1.0/23 is 192.168.1.0/24 and 192.168.2.0/24)