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How to calculate Summary addr.?

Unanswered Question
Jul 28th, 2012
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Hi,

I'm trying to understand Summarization aka Route Aggregation. I don't understand what is it that I'm mistaking in this exercise:

"Find the summary address and mask of the following set of networks

a) 172. 148. 0. 0/13 through 172. 156. 0. 0/13"


MY ATTEMPT:

Clearly it's the 2nd octet that is changing, so the mask will be 255. ___.0.0

The number of networks are: (156 -148) +1= 9 nets

I need to cover at least 9 nets. Which block size will I use? Not 8 but 16.

16 means 4 bits on (starting from the left, from the most significant bit) i.e. 240.

So the mask should ultimately be:

255. 240. 0.0

and that means (8 +4)=12 bits on therefore it's going to be a /12

As to the summary addr., it's the 1st of the block, right?

Maybe 172. 148. 0. 0 /12 ?


However this answer is wrong.


CORRECT ANSWER

172. 144. 0. 0 /16


Where did I go wrong?

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John Blakley Sat, 07/28/2012 - 05:33
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Personally, I think the 'correct' answer is wrong: partially. If you work this out manually, you'd get 172.144.0.0, but you'd get a mask of 255.240.0.0. Here's why:


Top is 148 and bottom is 156:


10010100

10010101

10010110

10010111

10011000

10011001

10011010

10011011

10011100


You can see that the first 4 bits match, so you have to stop at that bit for your starting network:

1001


You make the rest of the bits '0' since they're not part of the new network:

0000


You take the first 4 common bits and the last 4 0's that you just created and come up with:

10010000


This is your 2nd octet of 144. So your new address is 172.144.0.0.


To get the mask, you put in 1's for the common bits (remember the first 4 bits) and then 0s for the other 4 bits:

11110000


This gives you 240. So your mask is 255.240.0.0. The correct answer should be 172.144.0.0/12.


The 'correct' answer that they give would only cover the network 172.144.0.0/16 and nothing else.


HTH,

John

nkarthikeyan Sat, 07/28/2012 - 07:56
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Hi Giovanna,


The actual answer is 172. 144. 0. 0/12 the major subnet would be which should cover 172.144.0.0 - 172.159.0.0.


When you say /13 that subnet id should be in multiples of 8. The same as /29 subnets should be in multiples of 8.


0,8.16,24,32,40,48,56,64,72,80....144,152,160,168....248.



1286432168421








1286432168421








1286432168421








1286432168421


Hope this clears you the doubt.


Please do rate if the given information helps.


By


Karthik

nkarthikeyan Sun, 07/29/2012 - 05:37
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Hi Giovanna,


Yes your calculation is correct. One small suggestion If your total number of subnets in your above example is 31+1. So you are considering the 3rd octet in the subnetting. So you follow the regular numbering for the decimal values for the binary. I.e 128 64 32 16 8 4 2 1. So your value 31 resides when we calculate the last 5. So we are considering the 1st 3 octets so 24 (total value for 3 octets) - 5 (Calculated for 31+1 subnets) because the change in value will be only for 3rd octet.


Hope this will clear your calculation option. However its up to us how do we calculate. This is a very mandate in our technology. But i do see always most of the guys look for subnet calculator. Its good to know that you are trying to be with your manual calculation.


But your question seems to be confused which you have asked in your above post.

c) 203.168.6.0/24 AND 203.168.60.0/24 which has 2 no.s of /24 which cannot be combined together as /23.


Please do rate if the given information helps.


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Karthik

WobblyWindows Mon, 07/30/2012 - 01:52
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My problem is that I can't understand how the book can say the summary address (s.a.) is 192.168.96.0 with a mask of 255.255.240.0

That would mean a prefix of /20, right? Why is the s.a. starting with 192.168. ...? Shouldn't it be at least starting with 203.168. ...? Notice it's just 2 networks that need to be summarized. Does this little detail impact on the s.a.?


Please, anyone, help if you can. I'm told it's very important to understand and get these exercises right, to pass the CCNA exam.

Thank you.

Alessio Andreoli Mon, 07/30/2012 - 03:01
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hi giovanna,

don't get so worried about the subnet masks because if it is true that they are very important (actually very very very very important) it is also true that:


a) books can make mistakes

b) it is much easier to compute a mask in binary



In order t do this just remember the AND true table:


0 and 0 = 0

0 and 1 = 0

1 and 0 = 0

1 and 1 = 1



Now, exploding the 255.255.240.0 (as example) in binary you get




11111111.11111111.11110000.00000000



following the rules that the other guys gave you above, you will see that if your address start with 203, let's say 203.168.0.0


11001011.10101000.00000000.00000000


there is no way that the result of your summurisation will start with 192!!! because that would be



(192) 11000000.whatever.whatever.whatever wrong !!!!


and it is wrong because for the first 20 positions(bits of 255.255.240.0) the value of the network portion cannot change due to the AND operation. Essentially the bit set to 1 means "ignore me"


At the beginning the subnet masks are a little bit confusing but the more you practise in binary the easier it gets.

Do not use subnet calculator !!!!!


HTH

Alessio

      


PS: remember that summarization is a valid technique only if  used with contiguos networks:


203.168.0.0/24

203.168.1.0/24

203.168.2.0/23


that would be

203.168.0.0/22 if i am not wrong



sometimes you can waste some network (especially with private ip adresses):


10.0.0.0/24

10.0.1.0/24

10.0.4.0/22


that could be


10.0.0.0/21 = 10.0.0.0 up to 10.0.7.255


where 10.0.2.0/24 and 10.0.3.0/24 wheren't used at all but they are advertised in the summarisation..  generally this is not a good idea for many reasons, security in first place.


Take Care and ejoy your studies


Alessio

WobblyWindows Tue, 07/31/2012 - 09:59
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Ok,

Basically you are saying the book is wrong. I thought about that, it's just that there are too many mistakes in one exercise. For anyone who's following the same book I mentioned above, it's Lab 5, at the end of Chapter 5. Anyways, thanks to everyone for your insights.

WobblyWindows Sun, 07/29/2012 - 03:04
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Hi,


Thank you both for your inputs. I was hoping there was a faster & easier way to work out the problem without having to "translate" into binary.


So here's a 10-step procedure I came up with, when trying to find out a Summary Address and its Mask. Correct me if I'm wrong.


1. Identify the changing octet.

2. Calculate n° of networks to summarize (usually do a simple SUBTRACTION, remember to add 1 because you also have to count the starting network).

3. Choose appropriate network Block Size (2, 4, 8, 16 ...)

4. Associate correct mask to chosen BS (8 → 248, 16 → 240, 32 → 224 ...)

5. Write the mask, including the Prefix (i.e. /22 or /17 ...)

6. Consider the 1st network in the block.

7. Write its changing octet in binary.

8. Starting from left, keep the bits that correspond to 1s in the mask, change to zero the remaining bits.

9. Calculate the decimal value of the entire octet.

10. Write your full summary address.


Continuing with the exercise...


Find the summary address and mask of the following set of networks

b) 192.168.32.0 through 192.168.63.0


WORKOUT

The changing octet is the 3rd one (.32 and .63).

N. of networks: (63 - 32) + 1= 32

I choose a BS= 32 (it's exactly what I need, this time!)

This means Mask= 255.255.224.0  (also 8 + 8+ 3 = /19), 3 bits on in the 3rd octet.

Now I consider the 3rd octet of the 1st network i.e. 32 = 0010 0000

I leave the first 3 bits unaltered and zero everything else. Result: 001 00000= 32 (again!)

The summary address will be 192.168.32.0 /19 with a mask of 255.255.224.0

I checked the answer and it's correct.


However I am puzzled again with exercise c) :


c) 203.168. 6.0/24 AND 203.168.60.0/24


WORKOUT

So the changing octet is once again the 3rd octet.

N. of networks: 2 (notice the AND in bold between the network addresses)

I choose a BS= 2 → 254, threfore 7 out of 8 bits are on in 3rd octet

Mask= 255. 255. 254. 0 or /23 (= 8+8+7)

Now, I consider the 1st network's 3rd octet and write it in binary: 6= 4+2 = 0000 0110

I keep the first 7 bits intact, turn to 0 the remaining bit and get: 0000011 0 = 6

Therefore my answer would be: s.a.= 203.168.6.0 /23, mask of 255. 255. 254. 0

Right? Wrong! The book says it should be 192.168.96.0 with mask 255.255.240.0


I don't get it. Which is the correct answer? Is the book right? How reliable is Todd Lammle's CCNA Study Guide by Wiley (that's the book I'm studying on...)?

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