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JDmailNY

join:2007-12-02
Pearl River, NY

[CCNA] Calculating VLSM summary for ICEND2

Can some tell me why this answer is correct for the questions below and why ???
These are from Cisco ICND2 part 2 book.

How can the most effectively summarize the IP range of addresses from 10.1.32.0 to 10.1.35.255.
ans : 10.1.32.0 / 22

How can you most effectively summarize the IP range of addresses from 172.168.12.0 / 24 to 172.168.13.0 /24.

ans: 172.168.12.0 /23


frostycpu

join:2007-05-08
Quebec, QC

1 edit

I'll do the first one and let you think out the second one from what I'll tell you.

You want to summarize different smaller ranges into a larger one (supernetting). Take the ranges you have, turn them to binary for easier viewing:

0000 1010 . 0000 0001 . 0010 0000 . 0000 0000 (10.1.32.0)
0000 1010 . 0000 0001 . 0010 0011 . 1111 1111 (10.1.35.255)

The easiest way to do this is to start from the left and go right until the numbers don't match up. If you look at the 23rd bit, I have a 0 for the starting range and a 1 for the end range. It doesn't match up, so I know I need to use the 22nd bit for network part of the address, the rest of the bits will be used for hosts. To put it visually;

0000 1010 . 0000 0001 . 0010 00|00 . 0000 0000 (10.1.32.0)
0000 1010 . 0000 0001 . 0010 00|11 . 1111 1111 (10.1.35.255)

Zero out the host part of either range and you end up with 0000 1010 . 0000 0001 . 0010 00|00 . 0000 0000. The seperation tells you the CIDR mask to use (/22).

Turn it back into decimal and you get 10.1.32.0/22. Now the example is made to be an optimal situation where you have a big chunk of contiguous addresses. Be careful not to summarize addresses that you don't have (if say, you didn't own 10.1.34.0/24 for example).


JDmailNY

join:2007-12-02
Pearl River, NY

frostycpu ; Thanks so much, I think I need a smack in the back of the head. I had this concept a while ago and you just reminded me of it; The differance is you explained better than my instructor did. have a great day


aryoba
Premium,MVM
join:2002-08-22
kudos:4
reply to frostycpu

frostycpu See Profile described the answer using long way. It is probably preferred to use shorter and quicker way, especially when you are doing exam with limited time.

Using concept of doubling and half-sizing, following is the quick way.

10.1.32.0 - 10.1.35.255
= 10.1.32.0 - 10.1.32.255
+ 10.1.33.0 - 10.1.33.255
+ 10.1.34.0 - 10.1.34.255
+ 10.1.35.0 - 10.1.35.255
= 10.1.32.0/24 + 10.1.33.0/24 + 10.1.34.0/24 + 10.1.35.0/24
= 10.1.32.0/23 + 10.1.34.0/23
= 10.1.32.0/22

From 172.168.12.0/24 to 172.168.13.0/24
= 172.168.12.0/24 + 172.168.13.0/24
= 172.168.12.0/23



FLengineer
CCNA, CEH, MCSA
Premium
join:2007-06-26
Leesburg, FL

2 edits
reply to JDmailNY

The easiest way I found is .....

10.1.32.0 to 10.1.35.255

Take the lowest number which is 10.1.32.0

/24 is only 10.1.32.0
/23 is 2 subnets so 10.1.32.0 to 10.1.33.255
/22 is 4 subnets so 10.1.32.0 to 10.1.35.255
/21 is 8 subnets so 10.1.32.0 to 10.1.39.255

just keep subtracting 1 from the mask and doubling the number of subnets.