09-15-2008 12:53 AM - edited 03-03-2019 11:32 PM
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?
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09-15-2008 01:28 AM
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
HTH..rate if helpful..
09-15-2008 01:33 AM
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
09-15-2008 02:14 AM
09-15-2008 01:28 AM
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
HTH..rate if helpful..
09-15-2008 07:07 AM
Thanks you.
09-15-2008 01:33 AM
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
09-15-2008 02:14 AM
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