** KB ID** 0001138

## Problem

**Note**: Yes I’m spelling Summarisation with an ’S’ I’m English.

Most examples I’ve seen on this give you a bunch of subnets then ask you to come up with a summary route for all of them, (that’s kind of the point of route summarisation, I’ll grant you). However in an exam with a laminated board and the dodgy permanent pen they give you to make notes with, are you seriously expected to convert everything to binary to find the ‘last contigious bit’?

I was studying this today and kept getting it wrong, so I asked a colleague to look and see where I was going wrong. He scribbled on a piece of paper for two minutes and came up with the correct answer. So I’ve stolen his methodology. You can use it for any range of subnets, and if it comes up in an exam, you also have the added bonus that the right answer will be on the screen so even if your maths is off a little, the answer should jump out at you.

## Solution

**Step 1**: You get a range of subnets to summarise;

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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

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Points to note:

**A)** Everything’s happening in the third octet.

**B)** Ignore everything except the lowest and the highest subnet.

**Step 2**: Write down the Highest and Lowest Network (in fact just the third octet).

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10 15

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**Step 3**:Convert those into Binary (use a full 8 bits).

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10 = 00001010 (if you’ve just gone eh! That's an 8, and a 2). 15 = 00001111 (if you’ve just gone eh! That's an 8, a 4, a 2, and a 1).

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**Step 4:** Find the POSITION of last contigious bit, (the same in both).

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00001010 00001111

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Above, the first 5 numbers are the same, thats where the LAST bit of the summarised subnet will be. (If you’re confused, everything to the **left** will be a one, everything to the **right** will be a zero. i.e. 11111111.11111111.11111000.0000000

So the answer will have a /21 mask, (because there’s 21 x 1s).

**Well that’s great, but I still don’t know the subnet address?**

Yes you do! In step 3 you worked out the **LOWEST** subnet, you simply forget all the numbers that are **NOT** contigious, and the subnet is all the bits that are left. (That sounds more complicated than it is), So;

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00001010 00001111

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You ignore the last three, they are not contigious, (010 and 111 are not the same), that leaves you with;

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00001000 = 8

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So the network to summarise is 192.168.8.0/21 (or 255.255.248.0 if you prefer).

You might think, **THAT WAS LONG WINDED!** Well I took pains to explain everything. Once you have the method you can apply it to any list of networks;.

## A More Complicated Example

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192.168.112.0 /24 192.168.113.0 /24 192.168.114.0 /24 192.168.115.0 /24 192.168.116.0 /24 192.168.117.0 /24 192.168.118.0 /24 192.168.119.0 /24Lowest and Higest192.168.112.0 192.168.119.0Convert the changing Octet to Binary01110000 01110111We are the same up to the 5th bit so its a /21Drop the last three bits 01110000 is 112 Answer is 192.168.112.0/21

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## More Exercises to Try (Answers Below)

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1. 10.10.1.0/27 10.10.1.32/28 10.10.1.48/28 10.10.1.64/26 10.10.1.128/29 10.10.1.136/29 10.10.1.144/28 2. 10.22.178.0 /23 10.22.180.0 /23 10.22.182.0 /23 10.22.184.0 /23 10.22.186.0 /23 10.22.188.0 /23 10.22.190.0 /23 3. 172.16.4.0/24 172.16.5.0/24 172.16.6.0/24 172.16.128.0/24 4. 172.16.207.192 /26 172.16.205.64 /26 172.16.206.128 /25 172.16.204.0 /24 5. 172.16.0.0/24 172.16.1.0/24 172.16.2.0/24 172.16.3.0/24

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**ANSWERS**

1. 10.10.1.0 /24

2. 10.22.176.0 /20

3. 172.16.0.0 /16

4. 172.16.204.0 /24

5. 172.16.0.0 /22

## Related Articles, References, Credits, or External Links

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