# access-list

Sep 15th, 2008

I saw the following access-list: 'access-list 3 permit 10.1.1.0 0.0.15.255' in a text book. How was the .15 in the wild card mask derived?

Hello!

You can obtain the wildcard mask by replacing the 1 with 0 and 0 with 1 in the netmask binary.

Check the example below.

I know that the 0.0.15.255 in netmask is 255.255.240.0

So, let focus on the 3rd octet: 240

240 in binary is 11110000

we replace the 1 with 0 and 0 with 1 and we get 00001111 which is 15.

After a while with practice you will "see" automatically this netmask / wildcard stuff.

I hope you understand.

Cheers,

Calin

The actual subnet mask was : 255.255.240.0

The invert of this is : 0.0.15.255

To derive wildcard mask subtract the subnet mask's each octet from 255.

1.255-255 = 0

2.255-255 = 0

3.255-240 = 15

4.255-0 = 255

Overall Rating: 5 (3 ratings)

## Replies

satish_zanjurne Mon, 09/15/2008 - 01:28
• Silver, 250 points or more

The actual subnet mask was : 255.255.240.0

The invert of this is : 0.0.15.255

To derive wildcard mask subtract the subnet mask's each octet from 255.

1.255-255 = 0

2.255-255 = 0

3.255-240 = 15

4.255-0 = 255

Calin Chiorean Mon, 09/15/2008 - 01:33
• Silver, 250 points or more

Hello!

You can obtain the wildcard mask by replacing the 1 with 0 and 0 with 1 in the netmask binary.

Check the example below.

I know that the 0.0.15.255 in netmask is 255.255.240.0

So, let focus on the 3rd octet: 240

240 in binary is 11110000

we replace the 1 with 0 and 0 with 1 and we get 00001111 which is 15.

After a while with practice you will "see" automatically this netmask / wildcard stuff.

I hope you understand.

Cheers,

Calin