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PostPosted: Fri Jun 22, 2007 12:52 pm 
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Give it another try, Pat.

Remember, binary 1s in wildcard masks signify "don't care" bits, meaning anything can be there, hence the name wildcard. 0s in wildcard masks signify that there must be a match.

Given that wildcard masks don't have to be contiguous, think about an 8 bit binary number that only allows even numbers using wildcard type bits (1s).

And for the sake of simplicity, just focus on the subnet 192.168.0.0/24. So create an acl that only permits addresses like 192.168.0.2, 192.168.0.4, etc.

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PostPosted: Sat Jun 23, 2007 9:35 am 
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dpocoroba wrote:
Your on the right track. The subnet mask does not need to be contigious. Thats one of the neat things with subneting.



Hmm, notwithstanding the great example and application, I'm finding it a bit scary that you seem to be referring to Cisco wildcard masks as subnet masks.

Of course what may look like a subnet mask can be used as a wildcard mask in an ACL (or OSPF or EIGRP config) - but it is functioning as a Cisco wildcard mask not a subnet mask.

However, using what looks like a wildcard mask (any mix of 0s then 1s) as a subnet mask is not going to give you any joy whatsoever.

Aubrey

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PostPosted: Sat Jun 23, 2007 2:18 pm 
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Eh sometimes my brain types out the wrong words.

As long as you understand the point I was trying to make then thats all that matters.

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PostPosted: Sat Jun 23, 2007 11:41 pm 
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dpocoroba wrote:
Eh sometimes my brain types out the wrong words.
.


That's ok - I know the feeling - you gotta watch those brains - they have a mind of their own sometimes 8)

Actually applying wildcard patterns like 0.0.0.128 or 0.0.0.10 in ACLs to given networks is a good way to reinforce the operation of binary.

Aubrey

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PostPosted: Tue Jun 26, 2007 2:51 pm 
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access-list 102 permit 192.168.0.0 0.0.0.254 any

I really dont see how that is any different than my last try though

Burgalveist wrote:
As for the even routes only, lemme take a stab at that:

access-list 102 permit 0.0.0.0 255.255.255.254 any


except that was for a 0.0.0.0/0 network (every possible even ip address) and this is for a 192.168.0.0/24 network.


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PostPosted: Tue Jun 26, 2007 2:58 pm 
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It's not really... I just wanted you to specify the subnet...

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PostPosted: Sat Sep 22, 2007 6:31 pm 
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Here are some affordable and free subnetting resources
http://www.easysubnet.com
http://www.subnettingquestions.com/custom/bren/


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 Post subject: Subnetting problem
PostPosted: Thu Nov 01, 2007 5:52 pm 
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Hey guys, Im having problem say finding the 20th and 50th subnet of subnets having a problem with it.

Have my test today in 6 hours, and still havent learned lol

*sigh*

Are there any online tutorials that easily show how to find subnets that are in the tens or hundreds without having to write them all out?


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PostPosted: Thu Nov 01, 2007 5:55 pm 
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subnet-calculator.com...? Memorize them all? Hehehe

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PostPosted: Thu Nov 01, 2007 6:38 pm 
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haha, we all learn lessons in life. Think I will learn one today :P


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PostPosted: Mon Feb 16, 2009 1:48 pm 
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Steve wrote:
As mentioed above, a class C address uses the final 8 bits to represent the host portion of an address. With our example subnet of 255.255.255.192 we borrowed 2 bits, giving us two usable subnets. The formula to figure out the useable hosts or subnets it to raise 2 to the number of bits available and minus 2 ((2^n)-2)).

...
...
...

As well, when subnetting you may not borrow 1 bit, or leave 1 bit for the host portion. At least two bits must occupy either network or host portion at all times).
Can you explain this part a little more clearly. I couldn't really fathom what you mean by that. And can you elaborate on how to calculate the bits that got borrowed?


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PostPosted: Fri Feb 20, 2009 12:52 am 
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TravisT wrote:
I told you I'd feel stupid. Looking at what I wrote down (5 = 00001001), I found that my error was from rushing.

If I have this right now, this should work

access-list 10 permit 192.168.5.0 0.0.2.255
access-list 10 permit 192.168.13.0 0.0.2.255

Is this right? On paper it seems to work...


hmm, I get a different answer..

0000 01|01 - 5
0000 01|11 - 7

that means that these addresses are identical till the 22nd bit, so that leaves 255-252 which is 3(1111 1100)... ones are dont care bits

access-list 10 permit 192.168.5.0 0.0.3.255 =)


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PostPosted: Sun Mar 08, 2009 11:25 pm 
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it took me ages to figure this out, then i realised that "AND" isnt the same as "ADD"

Heres a nice explaination of the AND function.
http://www.xcprod.com/titan/XCSB-DOC/binary_and.html

Thanks
Lee


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PostPosted: Thu Jul 23, 2009 3:14 am 
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i have to say this makes it easier to understand for people like me that are coming in fresh.God bless.but please if any one has a material or site that can elaborate more on subnetting and VLSm cos its really giving a great challenge.tanx


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PostPosted: Tue Jun 15, 2010 3:33 am 
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**EDIT

removing all the quote and skimmed down i guess im stuck onhow to:

find host number for IP,
total number of subnets,
and total number hosts


if anyone has anything that can simplify this for me id be grateful!!

ta

dave


Last edited by daver1982 on Tue Jun 15, 2010 6:02 am, edited 2 times in total.

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PostPosted: Tue Jun 15, 2010 4:29 am 
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I'd help if you'd just quote the part you're having trouble with.

I've subscribed to the topic so I'll get notified when you reply. :wink:


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PostPosted: Tue Jun 15, 2010 5:40 am 
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sure il rattle it down!

ok so i understand the "AND" calculation to get the network address.
what i dont understand is how you know which numbered host it is. also the borrowing of bits and "raise 2 to the number of bits avaliable then -2"
if this is too vague which reading back it maybe let me know , and il try to help you help me more!

so basically i cant:
find host number for IP,
total number of subnets,
and total number hosts

thanks

dave


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PostPosted: Tue Jun 15, 2010 6:14 am 
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daver1982 wrote:
so basically i cant:
find host number for IP,
dave


I work out the number of IP addresses by building up to the point were the subnet mask starts.

If you're subnet mask is 255.255.255.0 then how many IP addresses are in the last portion? 256
If you're subnet mask is 255.255.254.0 then how many IP addresses are in the last portion? 512

If you can work out how I got those numbers you're on the right track.


daver1982 wrote:
so basically i cant:
total number of subnets,


http://packetlife.net/media/library/15/ ... etting.pdf

Check out how many IP addresses there are for each subnetmask. Subnets work in the reverse order.

daver1982 wrote:
so basically i cant:
and total number hosts
dave


For the hosts, work out the range of usable IP addresses eg. 256 and take away one IP address for the Network address and one for the broadcast address.


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PostPosted: Tue Jun 15, 2010 6:48 am 
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Project2501 wrote:
daver1982 wrote:
so basically i cant:
find host number for IP,
dave

I work out the number of IP addresses by building up to the point were the subnet mask starts.

If you're subnet mask is 255.255.255.0 then how many IP addresses are in the last portion? 256
If you're subnet mask is 255.255.254.0 then how many IP addresses are in the last portion? 512

If you can work out how I got those numbers you're on the right track.




ok, from looking at the pdf link i have come up with the following patterns and conclusions.
from 255 to 0 there are 8 parts which i guess are the bits ? 255 counts down binary, so -1 -2 -4 etc until - 128 to = .0 .
so if there is 8 parts to each part of the ip address then there are 32 levels of binary = total ammount of ip's possible?

so trying to explain where i am with this in my working out, if the subnet mask is 255.252.0.0 , there would be 18 levels of binary to = the amount of ip addresses. so in basic maths, 1*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2=262144.
i know i have the answers on the pdf, i just want to get it straight in a way i can understand!


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PostPosted: Tue Jun 15, 2010 6:52 am 
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daver1982 wrote:
so basically i cant:
and total number hosts
dave

For the hosts, work out the range of usable IP addresses eg. 256 and take away one IP address for the Network address and one for the broadcast address.


so am i right in saying there would be a possible total of 254 hosts- ranging from 255.255.255.0 to 255.255.255.255 ?


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