About this interactive
With a classful mask the network address is easy to read off, but with a custom mask a device works it out by ANDing: it lines up its IP address and its subnet mask in binary and compares them bit by bit. The result bit is 1 only where both bits are 1; 1 AND 0, 0 AND 1 and 0 AND 0 all give 0. Because the mask is ones and then zeros, the network bits of the address come through unchanged and every host bit is cleared, which is exactly the network address.
You only need binary for one octet. Where the mask octet is 255 the address octet copies across, and where it is 0 the result is 0. Take 192.168.0.88 with mask 255.255.255.192: the first three octets copy, and in the last, 88 is 01011000 and 192 is 11000000, so the AND is 01000000, which is 64. The network address is 192.168.0.64. The line can fall in any octet: 10.100.111.5/19 has mask 255.255.224.0, so the third octet does the work (111 AND 224 is 96), the last becomes 0, and the network is 10.100.96.0.
Type the network address in dotted decimal. Anything you miss shows the answer and the AND that gives it. Seven of the twenty questions are drawn each run.
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