Subnetting, VLSM and Summarization

[khalillaghari]

hi all
this is my 1st post here
please can anyone tell me easy way to understand subnetting vlsm summrization

[Xtopher62705]

khalillaghari;5875]hi all
this is my 1st post here
please can anyone tell me easy way to understand subnetting vlsm summrization
Yes,I can tell u the easy and simple way to memorise subnetting vlsm
summerization.
1.Implenment what u read asap.
2.Answer similar questions anywhere u see it.Eg there are thousands
of questions about subnetting vlsm summerization.
I pray it helps because I am new when it comes to networking.
Thanks.
Magnus.

[khalillaghari]

Xtopher62705 /hi all
this is my second post here
please lammle help me about summrization!! i have problem in summrization

i have these networks how can i summrize these n/w into one summry
192.168.10.0/25
host range is 192.168.10.1 to 126 and 127 is brodcats
mask is 255.255.255.128
192.168.0.126/26
host range 192.168.10.129 last host is 190 and broadcats is 191
mask is 255.255.255.192
192.168.10.208/28
host range 192.168.10.209 to 192.168.10.222 and broadcast is 223
mask is 255.255.255.240

can anyone tell me how can i summrize these into one summary
thanks
Khalil

[gabrielshorn]

These three networks don't really summarize.

Do you mean for the second network to be 192.168.20.128/26?

If so, they summarize as 192.168.10.0/24.

To summarize these, you have to find the right-most bit in all three network addresses that are identical to one another. Looking at the fourth octet for these three networks, you'd have:

192.168.10.0/25 = 0
192.168.10.128/26 = 10
192.168.10.208/29 = 11010

Because the top address has a 25 bit mask, only the first bit in the last octet can be compared between the three networks. Since they don't match, you have to move to the left one bit, to the 24th bit. This one matches on all three. Since it does, you build your mask to mask away all the bits to the right of it. So the summary address is 192.168.10.0/24.

[Fuzz]

Be careful when using that method to summarise, as you may also include other subnets in that range that you do not want part of the summarisation. It's best to summarise networks that are contiguous to prevent problems with routing protocols later on. This means designing your network with expansion in mind!

[gabrielshorn]

Thanks for mentioning that Fuzz. If you were to:

1. summarize the networks mentioned using this method, and
2. there was another subnet that in any way used the 192.168.10.224/28 network, and
3. the path to that subnet was different than the path to the summarized networks, then

you'd have a routing mess on your hands. If you must summarize these networks, DON'T use any or all of 192.168.10.224/28.

[Commander_Keen]

Originally Posted by khalillaghari

hi all
this is my 1st post here
please can anyone tell me easy way to understand subnetting vlsm summrization

What part of it has you confused?

[khalillaghari]

i have little bit confusion in summrization ..
A routter 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 ?

- 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 because I do not have the whole range of subnets.

Please clarify me in detial
Thank You

[gabrielshorn]

Of the answers provided, 192.168.32.0 - 192.168.47.0 would be the only correct answer.

All address bits that that fall under the subnet mask (leftmost 20 bits) must remain the same as the summary address for a network address to be a subnet of the summary address.

Summary Address and Subnet Mask
11000000 10101000 00100000 00000000 - 192.168.32.0
11111111 11111111 11110000 00000000 - 255.255.240.0

Of all the networks listed in the answers, the following all have different bits in the masked portion:

192.168.0.0 - 19th bit different- This makes answers A & B wrong.
192.168.48.0 - 20th bit different - This makes answer D wrong.
192.168.63.0 - 20th bit different - This makes answer E wrong.

C is the only possible answer.

192.168.32.0 - all masked bits identical
192.168.47.0 - all masked bits identical

[khalillaghari]

thanks dear but i have other question if u can asnwer
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

[khalillaghari]

192.168.16.0 for this one
summary wud be 192.168.16.0/20
am i correct