Here's your answer:
permit 1.1.1.1 0.0.0.0
permit 2.1.1.1 158.0.0.0
permit 100.100.0.0 0.0.14.0
You can't do anything with 1.1.1.1 so it takes up a line by itself.
Now, take 150.1.1.1 and 10.1.1.1. Their first octets in binary are:
150 = 10010100
10 = 00001010
Now, consider each bit in turn, starting from left (bit 0) to right (bit 7).
Bit 0 can be a don't care bit since it needs to match both 0 and 1
Bit 1 has to be 0 for both
Bit 2 has to be 0 for both
Bit 3 can be a don't care bit since it needs to match both 0 and 1
Bit 4 can be a don't care bit since it needs to match both 0 and 1
Bit 5 can be a don't care bit since it needs to match both 0 and 1
Bit 6 has to be 1 for both
Bit 7 has to be 0 for both
Therefore, the wildcard mask is: 10011110 (158). The corresponding network octet is: 00000010 (2)
That gives you your second line.
Now for the last one. We need to allow 0,2,4,6,8,10,12,14 for the third octet. In binary, they are:
2 = 00000010
4 = 00000100
6 = 00000110
8 = 00001000
10 = 00001010
12 = 00001100
14 = 00001110
Bits 0-3 are common and should be zero
Bits 4-6 can be either 0 or 1 so they should be don't care bits
Bit 7 has to be 0 for both.
Therefore, the wildcard mask is: 00001110 (14). The corresponding network octet is: 00000000 (0)
Hope that helps - pls rate the post if it does.
Paresh
