Route Summarization Question

[jmcoreymv]

Ok lets suppose Im running a classless protocol like OSPF or EIGRP. Ive been giving the following subnets to summarize together 172.16.168.0/24 through 172.16.175.0/24. How do you find the answer (172.16.168.0/21) without using a piece of paper and binary math? Is there any other possible method so I can do it in my head.

 

[pcmark]

This is how I do it:If you count the /24 blocks you get 8 all together(172.16.168.0=1, 172.16.169.0=2, 172.16.170.0=3,....172.16.175.0=8). Now,from the far right hand place(binary)you start at 1 and double every time:256-128-64-32-16-8-4-2-1. You see that the number 8 is four places in? Thats what you subtract from your single block of /24,counting /24 of course. You get /21.(Count backwards 4 places;/24-/23-/22-/21) My description probably confused it somewhat. It's real easy once you figure it out.

 

[jmcoreymv]

I think I understand, Thanks!

On a side note. When you list the numbers like 8 is the fourth number in, it might be easier to say 2^x = 8 and since x=3 /24 - 3 = /21

Thanks ALOT!

 

[pcmark]

Yeah,what you said😀 I'm not that mathematically inclined.

 

[spidey07]

yeah, count backwards in binary. summarizing is EXACTLY like figuring out hosts/per subnet except you deal in networks (/24, /21, /18) instead of hosts.

172.16.168.0/21 summerarizes

172.16.168.0 - 172.16.175.0 and any masks inbetween

172.16.176.0 would be on the next /21 boundary (group of eight class C) and be described as 172.16.176.0/21
172.16.160.0 woulbe be the previous /21 boundary = 172.16.160/21

How to do it in your head? Count just like described above.

practice, practice, practice.

 

[jmcoreymv]

Are you allowed scientific calcs on the CCNP?

 

[Damaged]

Practice is the key. After you see enough of them you just learn to recognize certain things. Like that 168 is divisible evenly by 8, and that the difference b/w 175 and 168 is 8, and finally that 2^3=8, and 24-3=21. Voila, it's a slash 21. I guess more appropriately in a mathematical sense it would be (2^24)-(2^3)=2^21 (or /21).

Okay, maybe my reasoning is oddly shortcutted, but it works for me.

Oh yeah, never forget that everything it is a multiple of 2 and you're okay. People always seem to forget that one. 🙂

 

[TheCandyman]

<<Are you allowed scientific calcs on the CCNP?>>

No. it's all based on memorization.

 

[pcmark]

Which CCNP test is this for? I thought all the subnetting stuff was covered by the CCNA test,and the CCNP tests were on switching and routing topics,like BGP,HSRP,etc.

 

[jmcoreymv]

Its for the BSCN (routing exam) the CCNA covered subnetting and stuff, but this is sort of like the reverse..i still dont know how to do this route summarization in my head completely. Does anyone know of any sites that can help?

 

[spidey07]

try cramsession.com. You won't get but a few summarization questions on the exam. BUT, if you want to build networks it is a MUST HAVE skill. You will not be able to engineer large networks without summarization.

PM me if you have any other questions. The CCNP exams are not that hard, the routing one being the toughest. Make sure you know OSPF and BGP like the back of your hand (about 60% of exam). Also, you have to know the commands as there are a lot of type in the command to do this...no multiple choice here.

there are forums on cramsession to help you. Counting in binary always helped me. 128-192-224-240-248-252-256-512-1024-2048-4096...count the number of bits as you count upwards. I've been reamed by so called trainers for teaching this method, but it is alot quicker and easier than 2^x=65536. what is X? method.