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route summarization

*route summarization*was created by

*lav_plsb1*

Two routers R1 and R2 connected using serial

the networks connected to R1 router are mentioned below.

172.1.4.0/25

172.1.4.128/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

the R1 advertise its networks to R2 using Route summarization

According to above networks R1 advertise 172.1.4.0 and plz explain how to summarize the subnet mask.

the networks connected to R1 router are mentioned below.

172.1.4.0/25

172.1.4.128/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

the R1 advertise its networks to R2 using Route summarization

According to above networks R1 advertise 172.1.4.0 and plz explain how to summarize the subnet mask.

Replied by

*S0lo*on topic*Re: route summarization*
You can summarize the above networks with 172.1.4.0/22. So how it's done, lets first convert the first two networks to binary (because they have the same netmask of /25):

10101100.00000001.00000100.

10101100.00000001.00000100.

X's here means the host bits which can be 0 or 1. You can see that the first bit of the 4th octet is 0 in the first network and 1 in the second network. This means that those two networks can be summarized by one network looking like this:

10101100.00000001.00000100.

In other words turn the changing bit into an X and subtract 1 from the subnet mask (i.e 25 - 1 = 24). Back to decimal gives us:

172.1.4.0/24

Now we have the following networks:

172.1.4.0/24

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

Again since all have the same subnet mask may be we can summarize further. So again we will find the binary of all:

10101100.00000001.000001

10101100.00000001.000001

10101100.00000001.000001

10101100.00000001.000001

Notice how the two last bits of the 3rd octet are changing. All combinations are there. 00, 01, 10 and 11. There are no more combinations for 2 bits. This means that those four networks can be summerized by one network looking like this:

10101100.00000001.000001

In other words, turn the two changing bits into XX and subtract 2 from the subnet mask (i.e 24 - 2 = 22). Turning it back to decimal we get:

172.1.4.0/22

You can do this for 3,4,5.....etc changing bits. The thing that you have to make sure of is that

This looks too lengthy only because I'm trying to explain it. You can do it much faster once you get used to it.

10101100.00000001.00000100.

**0**XXXXXXX /2510101100.00000001.00000100.

**1**XXXXXXX /25X's here means the host bits which can be 0 or 1. You can see that the first bit of the 4th octet is 0 in the first network and 1 in the second network. This means that those two networks can be summarized by one network looking like this:

10101100.00000001.00000100.

**X**XXXXXXX /24In other words turn the changing bit into an X and subtract 1 from the subnet mask (i.e 25 - 1 = 24). Back to decimal gives us:

172.1.4.0/24

Now we have the following networks:

172.1.4.0/24

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

Again since all have the same subnet mask may be we can summarize further. So again we will find the binary of all:

10101100.00000001.000001

**00**.XXXXXXXX /2410101100.00000001.000001

**01**.XXXXXXXX /2410101100.00000001.000001

**10**.XXXXXXXX /2410101100.00000001.000001

**11**.XXXXXXXX /24Notice how the two last bits of the 3rd octet are changing. All combinations are there. 00, 01, 10 and 11. There are no more combinations for 2 bits. This means that those four networks can be summerized by one network looking like this:

10101100.00000001.000001

**XX**.XXXXXXXX /22In other words, turn the two changing bits into XX and subtract 2 from the subnet mask (i.e 24 - 2 = 22). Turning it back to decimal we get:

172.1.4.0/22

You can do this for 3,4,5.....etc changing bits. The thing that you have to make sure of is that

**ALL COMBINATIONS ARE PRESENT IN THE CHANGING BITS**. Ofcourse, the networks that you want to summarize need to have the SAME subnet mask too.This looks too lengthy only because I'm trying to explain it. You can do it much faster once you get used to it.

Studying CCNP...

Ammar Muqaddas

Forum Moderator

www.firewall.cx

Replied by

*lav_plsb1*on topic*Re: route summarization*
Hi solo,

i am confused with different subnet mask. Thanks for your reply. your explanation is awesome.

Thnks.

i am confused with different subnet mask. Thanks for your reply. your explanation is awesome.

Thnks.

Replied by

*sose*on topic*Re: route summarization*
This question is an extract from Todd Lammle 6th edition

there is what we call block size or magic number, when you get it then you can summarize off hand . the block size give s a subnet mask that captures all listed networks. the only draw back to the block size methodology is that it can advertise route not actually available in the internal network. to solve this you have to choose a smaller block size that will leave fewer networks to be advertised induvidually

block sizes for class b IP

128 -- /17

64 -- /18

32 -- /19

16 -- /20

8 -- /21

4 -- /22

example

172.1.4.0/25

172.1.4.128/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

this is a class b IP, the third octet is call the interesing octet, because it is the octet that numbers start to differ 4,5, 6 etc. so it is the block size of 4 that can capture all the numbers in that octet. therefore your answer is the first IP which is

172.1.4.0/22

you see easier than taking candy

go through your block size again

there is what we call block size or magic number, when you get it then you can summarize off hand . the block size give s a subnet mask that captures all listed networks. the only draw back to the block size methodology is that it can advertise route not actually available in the internal network. to solve this you have to choose a smaller block size that will leave fewer networks to be advertised induvidually

block sizes for class b IP

128 -- /17

64 -- /18

32 -- /19

16 -- /20

8 -- /21

4 -- /22

example

172.1.4.0/25

172.1.4.128/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

this is a class b IP, the third octet is call the interesing octet, because it is the octet that numbers start to differ 4,5, 6 etc. so it is the block size of 4 that can capture all the numbers in that octet. therefore your answer is the first IP which is

172.1.4.0/22

you see easier than taking candy

go through your block size again

sose

Network Engineer

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