## Question

1) 172.77.32.0 – 172.77.63.255

2) 172.77.112.1 – 172.77.127.254

3) 172.77.144.1 – 172.77.159.254

4) 172.77.176.1 – 172.77.191.254

5) 172.77.178.1 - 172.77.191.254

6) 172.77.208.1 – 172.77.223.254

7) 172.77.224.1 – 172.77.241.254

## Solution Preview

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The provided network ID corresponds to a class B IP address since the first octet, namely 172, is in the [128,191] range. This means we have 16 bits on network part of the address and 16 bits on the host part of the address.Since the subnet mask is 255.255.240.0, this means we can borrow 4 bits from the host part to the network part when building subnets. This comes from the binary representation of this subnet mask for each of the four octets. It also means we have 20 bits for mask and the remaining 32-20=12 bits for host.

255=11111111(because 255=128+64+32+16+8+4+2+1) and 240=11110000 (since 240=128+64+32+16). These observations lead to:

255.255.240.0 = 11111111.11111111.11110000.00000000

This gives us 16 possible values in the 3rd octet, namely 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, and 1111.

We are required to assign 6 subnets, but the provided subnet mask makes possible to build 16 subnets (2 raised to power 4). It is needed to see which the range is (for each of them) in order to decide the correct choices. This is the tricky part of the question, since for n bits borrowed from host to network it is possible to build 2 raised to power n subnets. Since 4<6<8 (we locate 6 between two consecutive powers of two) => we are tempted to say that 3 borrowed bits are enough; but in that case the subnet mask doesn’t remain 255.255.240.0. It would be:

11111111.11111111.11100000.00000000 which is 255.255.224.0 instead!...

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