If in the wildcard mask the 1s are don't cares, and the 0s are matched, and the source address/mask pair is 172.22.75.8 0.0.0.120, then you have:
0xxxx000 ----> you would never have odd numbers that are matched.
That would give you even values of binary:
Which is 8, 16, 24, 32, 40, 48, 56, 64, 72, ... , 120. So I could be wrong, but I think the mask would be 01110111, which translates to 119.
That way, the 4th and 8th bit positions have to be matched as the source address. It would start at 8 and go up to 127.
If your acl statement is
deny 172.22.75.0 0.0.0.120
I think you're blocking hosts where, in the last octet, bits 4-7 can be either 1 or 0, AND bits 1-3 and 8 have to be 0. Besides that, if the only statement in you're ACL is to deny, nothing will pass. I think you need a couple of statements in an ACL to block a range of addresses such as 172.22.75.8 - 172.22.75.127 such as
permit 172.22.75.0 0.0.0.7
permit 172.22.75.128 0.0.0.127
The implicit 'deny any' will block the addresses 172.22.75.8 - 172.22.75.127 so you don't need a deny statement.
Sorry Dean, I'm not sure I understand what you wrote in this last message.
Its all based on matching bits. If you want to start off at 8, thats fine, but if your mask was 120, your source address/mask pair would exclude all odd numbers because of the 0 in the 1st bit of the mask. Every address would then need to have a 0 in the 1st bit in order to be matched by the mask.
Also, since there are zeroes in th 2nd and 3rd bit positions as well, that also makes all sourced addreses a factor of 8. Each source address would need zeroes in the 1st 3 bit positions to be matched.
So what I'm guessing you want, and why I came to 119, is that you want it to start at 8, so that bit is set at sero. Then you want it to be less than 128, so that bit is also set at zero.
Which means the other bit positions are 1s, don't cares. I did this in binary first, then converted to decimal.
Also, you may want to post these types of questions in one of the higher-up forums, that way you get more insight from some of the CCIEs. Most of us never use these types of ACLs, and they are rarely (if ever) tested on in the Associate and Professional certs. It is more theoretical, and not best practice.
I understand now with the 119. but with the zero set in 8 position wouldn't this always have to be a zero? so how would you get 9 or 10...15 then how about 24,25,26,... so on ....
Or if you had it always on how would you get a 17,18,19, and so on
Sorry about this I just trying to understand the best I can. I understand when it jumps from octec like 0.0.15.31 but trying to sum this one has been harder for me to summarize down
Maybe we are not on the same page as far as bit positions go, I'm reading from right to left. On the far left is the 8th bit, and the far right is the 1st bit.
The first bit controls even and odd, as it is either 0 or 1, in binary and decimal.
So if I had 119, then it would be 01110111 in binary.
With the wildcard mask, it would ignore the 1s and focus on the 0s. The 0s would be required to match the source address, which in this case is 8 for the 4th octet, or 00001000.
So the only values that have to be matched are the 8th bit and 4th bit. Like so:
00001000 = 8
01110111 = 119 mask
0xxx1xxx ------> Every source address has to have this bit pattern in the 4th octet. That would give you a range of possible values between 8 and 127.
Message was edited by: DelVonte to avoid confusion