Standard ACLs that do not have a mask defined default to the class of network.
access-list 10 permit 192.168.200.0
would permit all hosts on the 192.168.200.0/24 subnet.
WOW. My mistake. Sorry. I read the network range of addresses wrong. I didnt realize that the '200' was the 4th octet and that you only wanted 3 subranges of the class C. Time to finally get the surgery on my eye done once and for all.. :-)
You really should read about this because it gets complicated and requires a lot of explaining.
The best way to handle this, especially for you so that you can see exactly what is going on, is to convert the addresses and the ranges to binary.
192.168.80.200 - 207
Lets focus on the last octet range of 200 to 207.
200 base 10 in binary is 11001000
201 base 10 in binary is 11001001
202 base 10 in binary is 11001010
203 base 10 in binary is 11001011
204 base 10 in binary is 11001100
205 base 10 in binary is 11001101
206 base 10 in binary is 11001110
207 base 10 in binary is 11001111
Notice the common bits that never change within that range. They are 11001. The first 5 bits do not change. The ones that do change are the last 3 bits: 000 through 111, and all combinations in between.
Now, 3 bits in binary offers you 8 combinations (as I have just shown), so you can have 8 additional host addresses "added" to the base of 11001 (200).
Do you see that? Stop here and think about it if you dont.
In access lists we use what are called wildcard masks. When converted into binary, a "0" means the match has to be exact. A "1" means that it can vary.
So, if I have an ACL that permits a subnet address and mask of 192.168.80.0 0.0.0.255, it means that, when I convert the mask to binary, I get 00000000.00000000.00000000.11111111. Given that the first 3 octets are represented with "0"s, the first 3 octets must match. However, the 4th octet can be anything. No match. So, all hosts between 192.168.80.0 through 192.168.80.255 will be permitted.
In your case, you want to start matching at 192.168.80.200 and end at 192.168.80.207. This is for the first range of addresses. These are the hosts that will be permitted according to the ACL.
The subnet and mask will be 192.168.80.200 0.0.0.7. The first three octets of the mask (0.0.0) means that the address has to match (192.168.80), and the last octet in the mask of "7" means that the last 3 bits (111), which is 7 (in base 10 - decimal), can vary, from 200 to 207.
I hope this has helped you. As a Russian, you should be laughing at this baby math. :-)
Hi everyone, I would like to thank you in advance for any help you can provide a newcomer like myself!
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