# Calculate the SUMMARY ROUTE without converting the addresses to binary

May 18th, 2010

Hi,
I am  writing to you if you can clarifications about the summary route.

I understand the  operation of the summary route, but I can not calculate it without  having to convert the octets in binary.

Ex:

What single  network address and mask can be used to summarize the networks shown?

172.16.0.0 / 16

172.18.0.0 / 16

172.20.0.0 / 16

172.22.0.0 / 16

I would suffice to convert the figures 16 to 18 - 20 - 22 in  track and find the common elements, obtaining the subnet mask.

10101100 172 = unchanged

16 =  0001 0000

18 = 0001 0010

20 = 0001 0100

22 = 0001 0110

bits in common is 5 + 8 (172) equals 13.

The summary route is 172.16.0.0 / 13.

Instead of converting each time in binary, c is an expeditious  method?

I thought he was doing in this  way:

To calculate the route  summarization:

2x> = N. networks

es.

we have 4 nets  summarizzare:

172.16.10.0 / 16
172.16.11.0 / 16
172.16.12.0 / 16
172.16.13.0 / 16

to calculate  the summary route:

2 high X> =  greater than or equal to 4

X = 2 then 2  to the Second.

Subtract 2-bit subnet.

So the summary route is equivalent:

172.16.10.0 / 14.

but  in reality this is not the exact calculation right?

I would  be happy if someone give me clarification about

Italian translation

Salve,

scrivo per avere se è possibile dei chiarimenti circa la summary route.

Sono a conoscenza del funzionamento della summary route,ma non riesco a calcolarla senza essere costretto a trasformare gli ottetti in binario.

Es:

What single network address and mask can be used to summarize the networks shown?

172.16.0.0  /16

172.18.0.0  /16

172.20.0.0  /16

172.22.0.0  /16

Mi basterebbe convertire le cifre 16 - 18 - 20 - 22 in binario e trovare gli elementi in comune, ricavando la subnet mask.

172 = 10101100 rimane invariato

16 = 0001 0000

18 = 0001 0010

20 = 0001 0100

22 = 0001 0110

i bit in comune sono 5 + 8 (172) è uguale 13.

La summary route è 172.16.0.0  /13.

Invece di convertire ogni volta in binario, c è un metodo sbrigativo ?

Credevo che si facesse in questa maniera:

Per calcolare la summarization route:

2x >= N.reti

es.

abbiamo 4 reti da summarizzare:

172.16.10.0 /16
172.16.11.0 /16
172.16.12.0 /16
172.16.13.0 /16

per calcolare la summary route:

2 elevato a X >= maggiore o uguale a 4

X= 2   quindi    2 alla Seconda.

Sottrarre la subnet di 2 bit.

Quindi la summary route equivale:

172.16.10.0 /14.

ma in realtà non è questo il calcolo esatto giusto?

sarei felice se qualcuno mi desse dei chiarimenti a riguardo

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## Replies

Jon Marshall Tue, 05/18/2010 - 14:09
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Fabrizio

What single  network address and mask can be used to summarize the networks shown?

172.16.0.0 / 16

172.18.0.0 / 16

172.20.0.0 / 16

172.22.0.0 / 16

Firstly if you summarise these networks to 172.16.0.0/13 then you are including networks -

172.16.0.0 -> 172.23.0.0

to work it out without binary -

172.16 is the start

you can summarise 2, 4, 8 , 16, 32, 64 & 128 networks exactly due to the bits in byte. If you want to summarise 3 networks then you need to use 4, if you want to summarise 6 networks then you need to use 8.

So you want to go from 16 -> 22 which is 22 - 16 = 6. So you need an 8. To summarise 8 networks you use a 248 ie.

128 summarised networks = 128  = 1st bit position in byte

64 summarised networks = 192    = 2nd bit position in byte

32 = 224 = 3rd

16 = 240 = 4th

8 = 248   = 5th

4 = 252   = 6th

2 = 254   = 7th

so you need a mask of 255.248.0.0 . From the above 248 is the 5th bit position so =

8 (because the first byte is all 1's ie 255) + 5 (because the next byte is 248) = 13

so you have a summary address of 172.16.0.0/13

it might seem long winded but once you get the hang of it you can do these things in your head very quickly.

Jon

routing01 Tue, 05/18/2010 - 15:25

in Annex I sent  a question about.
In your opinion what are the  steps needed to give a correct answer?
explain  the calculations is to obtain the correct result.

thanks!

Attachment:
Jon Marshall Tue, 05/18/2010 - 15:48
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Fabrizio

Based on my previous thread why don't you try it yourself and post what you think is the answer.

Jon

routing01 Wed, 05/19/2010 - 14:37

hello jon
in accomplishing this task I am in trouble as this exercise  makes me a little confused ideas.

why are  confused is because there are two subnet mask / 30 and / 29 and also IPs  are different in that the third is that the fourth octet change.

correct me if I'm wrong but are grouped Group A 16 Group B  subnet and the subnet 4?

around how many  subnets are there?

Surely at this time are  much smarter than me, because I lack a little midnight

w riting by the United States right? I'm Italian and it is almost midnight.

Jon Marshall Wed, 05/19/2010 - 14:47
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Fabrizio

Actually i am in the UK so it's 22:40 here, guess you are an hour ahead of me. The reason i asked you to try and work it out is that you would learn more trying to do it yourself than simply asking other people to do it for you. I'll go through the answer but you should make the effort to understand this yourself.

So -

192.168.0.0/30
192.168.0.4/30
192.168.0.8/30
192.168.0.16/29
192.168.4.0/30
192.168.5.0/30
192.168.6.0/30
192.168.7.0/30

It doesn't matter about the /30s or /29s. Look at which octet is changing. That is the 3rd octet. So we need to go from 192.168.0 -> 192.168.7 which is 8 networks.

So to summarise 8 networks if you check previous thread that would require a 248 third octet mask ie.  192.168.0.0 255.255.248.0

Again as from previous post 248 = 5th bit in the thord octet byte. So

255 = 8

255 = 8

248 = 5

0 = 0

8 + 8 + 5 = 21 so the answer is 192.168.0.0/21.

Jon

blbennett321 Thu, 03/10/2011 - 10:40

Jon,

Thank you so much for this posting! I do have a question, however. How does this work when you are summarizing the third octet addresses from 172.16.32 -> 172.16.164 (132 networks)? Using this method, I calculate a CIDR block of /18 but the correct answer is a CIDR block of /16. What am I missing?

Bill

Jon Marshall Thu, 03/10/2011 - 13:50
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Bill

Firstly summarising on the 3rd octet is the same principle as summarising on any octet.

Okay you are looking to summarise 132 networks.

/18 = 255.255.192.0

256 - 192 = 64 so you can only summarise 64 networks with that subnet mask which obviously isn't enough ie. from my original post -

64 summarised networks = 192    = 2nd bit position in byte

/17 = 255.255.128.0

256 -128 = 128 so you can summarise 128 networks but again that isn't enough.

So you have to use 255.255.0.0 which can summarise 255 networks in the third octet ie. 172.16.1.0 -> 172.16.255.0. Note that you are including a lot more networks in there than just the 132 networks you actually want to summarise.

I'm not entirely sure how you came to /18 using the method i described. It may well be that i didn't explain it clearly enough or made a typo somewhere although i did reread and i can't find anything amiss. Could you explain how you got a /18 and that might help us clear up any confusion ?

Jon

blbennett321 Thu, 03/10/2011 - 14:27