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Novice networker here.

I can't seem to find any straightforward answers online.

Example:

  IP             SUBNET           NEXT HOP
> 129.10.112.0   255.255.255.0    R1
> 129.10.80.0    255.255.255.0    R1
> 129.10.0.0     255.255.0.0      R2
> 129.10.63.0    255.255.255.0    R4
> 129.10.63.0    255.255.255.0    R4
> 129.10.64.0    255.255.192.0    R3
> 129.10.65.0    255.255.255.0    R4
> 129.10.66.0    255.255.255.0    R4

What is the proper method to aggregate these entries into the minimum number of entries?

user5032
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  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can post and accept your own answer. – Ron Maupin Jan 05 '21 at 02:31

1 Answers1

11

First, separate the entries by next hop. You have to summarize them separately:

IP             SUBNET           NEXT HOP
129.10.112.0   255.255.255.0    R1
129.10.80.0    255.255.255.0    R1

129.10.0.0     255.255.0.0      R2

129.10.64.0    255.255.192.0    R3

129.10.63.0    255.255.255.0    R4
129.10.63.0    255.255.255.0    R4
129.10.65.0    255.255.255.0    R4
129.10.66.0    255.255.255.0    R4

Then for each next next hop, convert all the network addresses to binary. Here is just the first one:

10000001.00001010.01110000.00000000 = 129.10.112.0 
10000001.00001010.01010000.00000000 = 129.10.80.0

Now find all the identical digits, starting from the left. In this case, the digits are all the same up to the 18th position. So your new mask is /18. Now, using either address and the /18 mask, find the network address by ANDing the address and the mask:

10000001.00001010.01110000.00000000 = 129.10.112.0
11111111.11111111.11000000.00000000 = /18 (255.255.192.0)
-------------------------------------
10000001.00001010.01000000.00000000 = 129.10.64.0 /18

So the best summarization of the first two routes is 129.10.64.0/18.

(the rest is left as a exercise for the reader)

Ron Trunk
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  • Using that method, you need to take care that you don't inadvertently aggregate unwanted destinations between other prefixes (e.g. 129.10.81/24 in the example above). You should make sure that all combinations for the bits subtracted from the initial prefixes may be of any value for the same next hop. – Zac67 Feb 01 '24 at 14:19
  • If there are unwanted subnets, then you can't aggregate at all. – Ron Trunk Feb 01 '24 at 15:53