MW Quiz 5 (Spring 2010)
Suppose a network uses the subnet mask 255.255.254.0. To what subnet
would an IP datagram with destination address 128.122.43.193 be
forwarded to?

The first two bytes of the subnet link address will be the same as the
destination address (anding a number will all ones leaves the number
unchanged). The fourth byte will be zero (anding a value with all zeroes
results in all zeroes).

The third byte of the mask in binary:        11111110
The third byte of the destination in binary: 00101011
Anding them together:                        00101010
Complete answer in decimal: 128.122.(2+8+32).0 = 128.122.42.0

Could you have answered this without working out the binary? Sure. 254
is all ones except the bottom (right-most) bit. So, the low bit of the
destination turns from 1 to 0, making the odd number even, and thus
in-effect subtracting 1 from 43.




MW Quiz 5 Other Version
Suppose a network uses the subnet mask 255.255.255.128. To what subnet
would an IP datagram with destination address 128.122.43.193 be
forwarded to?

The first three bytes will be anded with ones, and so unchanged, making
the answer 128.122.43.x. What's x? 128 is 10000000 and 193 is bigger
than 128 (meaning it has its high bit set to one), so the result is
128.122.43.128.




TR Quiz 5
Suppose a network uses the subnet mask 255.255.240.0. To what subnet
would an IP datagram with destination address 128.122.43.193 be
forwarded to?

A quick glance tells us the answer will look like 128.122.y.0. What's y?
In binary, 240 = 255 - 15 = 11111111-1111 = 11110000. This has to be
anded with 43, and 43 = 32 + 8 + 2 + 1, or, in binary, 00101011 (as
above).
Anding the third byte of the destination with the third byte of the
mask:
        11110000
        00101011
        --------
        00100000 = 32 decimal.
So the answer is 128.122.32.0.



TR Quiz 5 Other Version
Suppose a network uses the subnet mask 255.255.240.0. To what subnet
would an IP datagram with destination address 128.122.93.143 be
forwarded to?

The answer will look like 128.122.z.0. What's z? As above, 240 decimal
is 11110000 binary. 93 decimal = 64 + 16 + 8 + 4 + 1 = 01011101
binary. Anding the third bytes of the mask and destination gives us:

        11110000
        01011101
        --------
        01010000 = 80 decimal = z.

So the answer is 128.122.80.0.