Summarization question...

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****************Router3-------------Switch---------Network 172.16.1.0/24
/
/
/
*********Router2----Router4-------------Switch-----------Network 172.16.2.0/24
\
\
\
\
*****************Router5-------------Switch-----------Network 172.16.3.0/24

For router3 the sumarized route for networks 172.16.2&3.0/24 is ip route 172.16.2.0 255.255.254.0 (next hope ip address)
For router4 the sumarized route for networks 172.16.1&3.0/24 is ip route 172.16.1.0 255.255.252.0 (next hope ip address)
For router5 the sumarized route for networks 172.16.1&2.0/24 is ip route 172.16.1.0 255.255.252.0 (next hope ip address)

Hello everybody.
I got this setup in PT and now I am going around with static routes instead of using routing protocols. Is my assumption above correct or should I just use one static route per network as I have only two networks to route traffic for, apart from other networks to the left of router2 which now it's not an issue, I am only focussing on these 3 networks.
Any comments well appreciated. I must say that route summarization was one of my favourite subjects when studying for my CCNA but now I am a bit lost as I have the setup done but some networks are just not accessible. Please feel free to see PT attached file.
Cheers.

 

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Are you sure you've done the summarization correctly? I think the summary route for 172.16.1.0 & 172.16.3.0 is 172.16.0.0 255.255.252.0 and similary for 172.16.1&2.0/24.

 

Hello mate. First of all, thanks for your reply. Unfortunately you are wrong and so was I.
After doing more searches on our best friend "Google" I have finally got it.
Let’s have the following IPs:

172.16.1.0 = 1010 1100.0001 0000.0000 0001.0000 0000
172.16.2.0 = 1010 1100.0001 0000.0000 0010.0000 0000

The summarized route for this is 172.16.0.0/22.
As all matching bits are 0 we got the 172.16."0".0 and as we have 6 matching bits we got a /22 or 255.255.252.0.

For the following IPs:
172.16.2.0 = 1010 1100.0001 0000.0000 0010.0000 0000
172.16.3.0 = 1010 1100.0001 0000.0000 0011.0000 0000

The summarized route for these networks is 172.16.2.0/23.
As all matching bits are 0 and a 1 you then have the matching bit 1 coming as 172.16."2".0/23 or 255.255.254.0. These matching bits or the sum of all matching bits equals to the summarized route.

When you get a matching binary 1 on all networks you then add their values and that would give you your summarized route. For example:

172.16.28.0 = 1010 1100.0001 0000.0001 1100.0000 0000
172.16.29.0 = 1010 1100.0001 0000.0001 1101.0000 0000
172.16.30.0 = 1010 1100.0001 0000.0001 1110.0000 0000

As you can see on the third octet there's three matching 0 and three matching 1 bits so, the sum of all those bits equals to "28", as we have;

128 64 32 16 8 4 2 1
0 0 0 1 1 1 0 0 = 16+8+4=28.

I know that I may not have the best way of explaining things but this is how I have work it out.

Hope it helps and once again many thanks for your reply.

 

I think you must have picked up my post during a couple of edits. The last edit did say the summary route for 172.16.1 & 3 is 172.16.0.0 /22 and not 172.16.4.0 /22

 

Yeah I think I did as I am pretty sure that was what I read.
In that case mate you got it.