From: Michael Snyder (msnyder@revolutioncomputer.com)
Date: Wed Aug 22 2007 - 09:47:22 ART
Hey Guys,
Been putting off setting this message to the groupstudy board. Around 2001
I started studying for my CCIE. Took the test five times, with my highest
score being 80%.
So, when I was thinking of taking the test the sixth time, I looked at all
my finances and said gee, I could have gotten a college degree with the all
money I spend on the racks, remote classes, and my study rack!
So instead of the taking the test the sixth time, I went to college!
Long story short, it is now 2007, I'm 20 credit hours away from getting an
undergrad degree in physics. I have found it's easier to pass a quantum
mechanics class than passing the ccie test!
My classmates turn about a bit green during our three hour physics finals,
and I'm just sitting there smiling, thinking gee, a least it isn't an 8 hour
ccie test!
Looking back at my old emails, I did come up with a method on how to use
windows calc to do bit based route summarization. Found out later that it
is the same method computers use, its called bit based erosion in my present
math classes. My version of the method is reposted below.
Also, while in college I've been filming the magnetic field, here's a video
and an undergrad paper of my present work.
http://www.youtube.com/watch?v=2MHIncd1rSY
http://www.esnips.com/doc/2e05744f-aa5d-40b3-937b-73c22d6eaa6f/Magnetic-Cont
ours
I've been putting off unjoining the list, but I need to move on, and those
40,000 messages from groupstudy is starting to freak out my outlook email.
So sadly, please remove me from the list.
I thank you guys and the groupstudy host for everything, it turned out to be
a really good thing learning how to study and work together. My five failed
tests were not wasted, they taught me how to study and I now use those
skills daily!
I encourage you guys, keep going on your studies; but please remember you
can change the game if the rules ain't working for you.
Michael Snyder
--------------------------
Using the windows calc to do summarization. The reason that it works is
that logical operations work in any number base. Of course you should know
how to do it in binary, but speed is what we need in the lab; not to mention
less mistakes.
The method is as follows:
Assuming you have networks you want to summarize in octet columns
a1.b1.c1.d1
a2.b2.c2.d2
You do the logical operations in columns, of example all the a`s then the
b`s then the c`s, etc. Of course it only applies to the octets you wish to
summarize. If all the a`s and b`s and d`s are the same, you only have to do
is the c'`s. Wildcard zeros will do an exact match, Wildcard 255 is a any
match.
`and` all the c`s and call it result-1
`or` all the c`s and call it result-2
The finished access list will use the results in this way
Access-list 1 permit a.b.(result-1).d 0.0.(result-1 xor result-2).255
So the wildcard octet is really result-1 xor result-2 = wildcard
It's not as complex as it looks; I'll do some examples with real network
numbers.
-------------------------
Let's deny the odd subnets of 192.168.x.x/32
R2#show ip route 192.168.0.0 255.255.0.0 longer-prefixes
192.168.9.0/32 is subnetted, 1 subnets
O E1 192.168.9.9 [110/1128] via 172.16.56.6, 1d16h, Serial0.56
192.168.4.0/32 is subnetted, 1 subnets
D 192.168.4.4 [90/2297856] via 172.16.24.2, 1d16h, Serial0.24
192.168.5.0/32 is subnetted, 1 subnets
O IA 192.168.5.5 [110/65] via 172.16.56.5, 1d16h, Serial0.56
192.168.6.0/32 is subnetted, 1 subnets
O IA 192.168.6.6 [110/65] via 172.16.56.6, 1d16h, Serial0.56
192.168.7.0/32 is subnetted, 1 subnets
O 192.168.7.7 [110/75] via 172.16.56.6, 1d16h, Serial0.56
[110/75] via 172.16.56.5, 1d16h, Serial0.56
192.168.1.0/32 is subnetted, 1 subnets
R 192.168.1.1 [120/1] via 172.16.12.1, 00:00:22, Serial1
192.168.2.0/32 is subnetted, 1 subnets
C 192.168.2.2 is directly connected, Loopback0
R2#
192.168.1.1
192.168.5.5
192.168.7.7
192.168.9.9
So, the third and fourth octets are same, let`s just do one of them.
First `and` the values
(1&5&7&9) = 1
'or' the values
(1|5|7|9) = 15
`xor` the results
1 xor 15 = 14
Answer in permit format (the show `ip route list` list command didn't like
my deny format)
access-list 12 permit 192.168.1.1 0.0.14.14
Let`s try it on the router.
R2#show ip route list 12
O E1 192.168.9.9 [110/1128] via 172.16.56.6, 1d17h, Serial0.56
O IA 192.168.5.5 [110/65] via 172.16.56.5, 1d17h, Serial0.56
O 192.168.7.7 [110/75] via 172.16.56.6, 1d17h, Serial0.56
[110/75] via 172.16.56.5, 1d17h, Serial0.56
R 192.168.1.1 [120/1] via 172.16.12.1, 00:00:17, Serial1
The final form would be
acess-list 12 deny 192.168.1.1 0.0.14.14
acess-list 12 permit any
Ok, some notes. The math works, but think about the answer it's giving you.
Permit 10.0.0.0 128.255.255.255 isn't much better than a default route. If
the single line answer has too much scope, try a multiline answer. I've
included a more complex example from a previous email below.
--------------------------
Summarize the following list:
133.6.11.0
135.16.171.0
172.60.51.0
121.15.120.0
112.59.9.0
Now using windows calc in decimal mode, lets do some octet equations.
First will check the first octet for a common network. If there isn't a
common network, then granddaddy of all summaries is the single line answer.
0.0.0.0/0
133&135&172&121&112=0, which means there's no common network for a one line
answer, other than a default network.
There's only 5 networks, so lets check pairs for common networks.
133&135 = 133, there's common network.
Just checking against the others, 133&172=132, another common network.
Note that we're using the result of the preceding common network check to
check against the next network.
Using 132&121=0; no good.
Checking 132&112=0; also no good.
Maybe 121&112 are common to each other. 121&112=112, which means we can
have a two line solution. The first three networks, then the next two.
A summary is defined as the networks `and` together for the common network,
then the values `or` together. Then take the two results and `xor` for the
wildcard mask.
You do one octet column at a time.
133.6.11.0
135.16.171.0
172.60.51.0
(133&135&172) xor (133|135|172)
answer 132, 132 xor 175
answer network 132 wildcard 43
Next octet,
(6&16&60) xor (6|16|60)
Network 0, 0 xor 62
Answer network 0 wildcard 62
Third octet
(11&171&51) xor (11|171|51)
Network 3, wildcard 184
Putting the answers together,
132.0.3.0 43.62.184.255
Applying the same treatment to
121.15.120.0
112.59.9.0
results as
112.11.8.0 9.52.113.255
My final answer
access-list 10 permit 132.0.3.0 43.62.184.255
access-list 10 permit 112.11.8.0 9.52.113.255
-----------------------------------
Good Luck and thanks for the Fish!
This archive was generated by hypermail 2.1.4 : Sat Sep 01 2007 - 11:32:12 ART