- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
03-12-2011 06:14 PM - edited 03-04-2019 11:43 AM
Hi, I imagine I need to build one ACL to match the following addresses:
135.7.183.0
135.7.184.0
I wrote the third octets in binary:
10110111 = 183
10111000 = 184
Then I did AND operation between 183 and 184 in binary:
10110000 = 176
and XOR operation as well to find mask:
10110111 = 183
10111000 = 184
Doing XOR is:
00001111 = 15
So in my calculation (I already used other methods), the access-list would be
access-list 1 permit 135.7.176.0 0.0.15.0
The problem is that this does not seem to match 183 and 184. What am I missing here please? It seems so simple but I do not know why this is failing...
Solved! Go to Solution.
- Labels:
-
Routing Protocols
Accepted Solutions
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
03-12-2011 11:06 PM
Not sure what you are trying to do,
Are you trying to match this-
135.7.183.0/24
135.7.184.0/24
FYI- in 135.7.183.0/24 > 135.7.183.1- 135.7.183.254 is the host range, 135.7.183.0 is subnet id and 135.7.183.255 is the broadcast address.
The easiest way to do the above would be
access-list 1 permit 135.7.183.0 0.0.0.255
access-lsit 1 permit 135.7.184.0 0.0.0.255
Use the foll for better understanding:
http://www.subnet-calculator.com/subnet.php?net_class=B
Please rate the post if it helps.
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
03-12-2011 11:06 PM
Not sure what you are trying to do,
Are you trying to match this-
135.7.183.0/24
135.7.184.0/24
FYI- in 135.7.183.0/24 > 135.7.183.1- 135.7.183.254 is the host range, 135.7.183.0 is subnet id and 135.7.183.255 is the broadcast address.
The easiest way to do the above would be
access-list 1 permit 135.7.183.0 0.0.0.255
access-lsit 1 permit 135.7.184.0 0.0.0.255
Use the foll for better understanding:
http://www.subnet-calculator.com/subnet.php?net_class=B
Please rate the post if it helps.
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
03-13-2011 08:40 AM
In this case mask was not given. It is stricly match the 138.7.183.0 and 138.7.184.0 addresses (not network).
So OK I just wanted to confirm I was not getting crazy. In this case it is not possible to aggregate this into one entry. Thanks.
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
03-13-2011 12:16 AM
Hi
The logic that you used is correct, the only reason that this is not matching the traffic is because you have left the last octect as 0 in the mask. To match all the traffic the in both the subnets you need to have .255 in the mask in the last octet as shown below:
access-list 1 permit 135.7.176.0 0.0.15.255
else it would try to match only the network addresses 135.7.183.0 and 135.7.184.0 and will deny rest of the subnet IPs.
Regards,
Sujit
