How to Calculate Advanced Wildcard Masks
In this blog, we are going to go in depth to understand the real logic behind working of wildcard mask and how you can utilise it more effectively for your configuration. Wildcards masks are used at different places in Cisco IOS CLI and is a very important concept to understand from CCNA and CCNP certification point of view.
For beginners, the easiest method to figure out wildcard mask is still to subtract the subnet mask from 255.255.255.255. For instance, if the subnet mask is 255.255.255.128 the wildcard mask can be calculated as shown below:
–255.255.255.128 <——- Subnet Mask
0.0.0.127 <——-Wildcard Mask
This method is more than sufficient for most of the CCNA level configuration tasks. Basically, wildcard mask is made up of 1’s and 0’s where
|1||Represents the bit that we don’t care about|
|0||Represents the bit that we care about and should not get changed in IP Address that is matched to statement|
Let’s take an example,
We are given a range of 184.108.40.206 – 220.127.116.11 and from this range, I have to match only IP coming in the range of 18.104.22.168 – 22.214.171.124. For that, we have to write down a combination of an IP address and Wildcard mask such that it matches only first 4 IP addresses from the complete range. This can be done as,
|126.96.36.199||00000001 . 00000010 . 00000011 . 00000000||First IP Address in Binary|
|188.8.131.52||00000001 . 00000010 . 00000011 . 00000001||Second IP Address in Binary|
|184.108.40.206||00000001 . 00000010 . 00000011 . 00000010||Third IP Address in Binary|
|220.127.116.11||00000001 . 00000010 . 00000011 . 00000011||Fourth IP Address in Binary|
|Wildcard Mask||00000000 . 00000000 . 00000000 . 00000011||0.0.0.3 – WCM in Decimal|
|Network ID||00000001 . 00000010 . 00000011 . 000000XX||18.104.22.168 – IP Address in Decimal|
As you can see, first 30 bits in all the IP address are exactly same and must be same if we want to match all these IP. So when writing down the IP that we are going to use along with wildcard mask, we don’t have to change the first 30 bits of IP. Write them down as it is. Also, remember that we don’t care about last two bits, so we can write down anything that we can write by changing last two bits only. Matching will be done on the behalf of first 30 bits only. And that’s what we have done there. In place of XX, you can write down whatever you want to use.
First 30 bits must match so while writing down wildcard mask, we have to write 30 zeros and remaining 2 bits don’t care. If you think about this, by not caring about last two bits, we can make only 4 numbers, 00, 01, 10, and 11. Which means 0, 1, 2, 3 in decimal and that’s exactly what we wanted to match.
In order to understand the calculation for wildcard mask for advanced use, we first need to understand two logic gates – AND Gate and XOR Gate.
AND – The output is 1 when both inputs are 1. In all other cases output is 0.
A AND B
XOR- In Exclusive-OR gate the output is 1 when either of inputs A or B is 1, But not if both A and B are 1. Also, the output will be 0 if both are 0. In other words, we can say output will be 1 only when inputs are not same.
A XOR B
Case 1 – “Permit” or “Deny” two different IP Address using one statement in ACL.
In order to find the best match for any specific address and wildcard combination that will match two address we use AND and XOR gates.
For instance, if we have two IP address 22.214.171.124 and 126.96.36.199 and we want to create an ACL that is the most specific match for these two addresses then we will have to use AND gate to figure out the address and XOR operation for finding wildcard mask. For ease of understanding, you can write the address in binary. Below mentioned table illustrates same:-
|188.8.131.52||00000001 . 00000010 . 00000011 . 00000100|
|184.108.40.206||00000100 . 00000011 . 00000010 . 00000001|
|Wildcard Mask in Binary||00000101 . 00000001 . 00000001 . 00000101|
|Network-ID in Binary||00000000 . 00000010 . 00000010 . 00000000|
If you write the same two things in decimal then address becomes 0.2.2.0 and wildcard mask will be 220.127.116.11. So the access-list statement would be something like:
Access-list 10 permit 0.2.2.0 18.104.22.168
This can be utilized to figure out most specific wildcard for ACL in one statement for two different IP Addresses.
This example also illustrates main difference between wildcard mask and subnet mask. When you write a subnet mask you write continuous 1’s followed by continuous 0’s whereas in case of wildcard mask you can write 1’s and 0’s in the discontinuous order.
Case 2 – “Permit / Deny” Odd or Even IP address of a subnet in one statement.
Now, this use example might seem unrealistic and you may never use it but this example gives you further insight into how powerful wildcard mask can be compared to subnet mask. Here it goes, let’s say you have a full subnet in which you just want to permit IP Address whose fourth octet is the odd number. What I mean is that if the subnet is 192.0.0.0/24 then I want only IP Address of 192.0.0.1, 192.0.0.3, 192.0.0.5 and so on to be allowed for the certain thing. Now one easy way is to have lots of statements in ACL having individual entries for these IP Address. This same task can be accomplished in just one statement. For ease of understanding lets write some odd IP Address in binary:
|192.0.0.1||11000000 . 0000000 . 00000000 . 00000001|
|192.0.0.3||11000000 . 0000000 . 00000000 . 00000011|
|192.0.0.5||11000000 . 0000000 . 00000000 . 00000101|
|192.0.0.7||11000000 . 0000000 . 00000000 . 00000111|
|AND Operation for IP Address||11000000 . 0000000 . 00000000 . 00000001|
|XOR Operation for Wildcard Mask||00000000 . 0000000 . 00000000 . 11111110|
If you notice all the IP Addresses with the ODD number in their fourth octet have the last bit as 1. So if this bit is 1 then only IP address has an ODD number in their fourth octet. Relating to the previous statement, here, we only care about the last most bit of the fourth octet. So if we want to accomplish the task of permitting all IP Address with 4th octet as the ODD number then this last bit should remain as it is. We can use a wildcard mask such as it cares about only last most bit and doesn’t care about any other bit in the fourth octet. The statement can be written as:
Access-list 10 permit 192.0.0.1 0.0.0.254
Let’s do the same for even numbers. We have to use an AND operation for IP Address and XOR operation for wildcard mask. Try to understand it in following example,
|192.0.0.0||11000000 . 0000000 . 00000000 . 00000000|
|192.0.0.2||11000000 . 0000000 . 00000000 . 00000010|
|192.0.0.4||11000000 . 0000000 . 00000000 . 00000100|
|192.0.0.6||11000000 . 0000000 . 00000000 . 00000110|
|AND Operation for IP Address||11000000 . 0000000 . 00000000 . 00000000|
|XOR Operation for Wildcard Mask||00000000 . 0000000 . 00000000 . 11111110|
If you notice all the IP Addresses with the EVEN number in their fourth octet have the last bit as 0. So if this bit is 0 then only IP address has an EVEN number in their fourth octet. Here also, we only care about the last most bit of the fourth octet. So to match all IP Address with 4th octet as the EVEN number then this last bit should remain as it is. We can use a wildcard mask such as it cares about only last most bit and doesn’t care about any other bit in the fourth octet. The statement can be written as:
Access-list 10 deny 192.0.0.0 0.0.0.254
Above statement will deny any IP address in given range if it has an EVEN number in 4th octet.
And that is how we can use the wildcard mask for matching multiple IP address in more flexible way than the subnet mask. We couldn’t have done the matching of EVEN/ODD IP addresses with subnet mask as subnet mask is continuous ones, following zeros. While a wildcard mask is not bounded by any such rule.