Networks and the Internet: Question 8

Worked solution

Part (a): Host bits in the original /24 network

An IPv4 address is 32 bits long, split into a network portion and a host portion by the subnet mask. The mask 255.255.255.0 sets the first three octets (24 bits) to all 1s, marking them as network bits, and leaves the last octet (8 bits) as all 0s, marking it as the host portion. So there are 8 host bits.

With this default mask, the whole 192.168.20.0 block is treated as a single network: it has one continuous range of host addresses (.0 to .255 in the last octet) and no bits are reserved to divide that range into separate, independent blocks. To create 4 distinct subnets, some of these 8 host bits must be deliberately “borrowed” and reassigned as extra network bits. The default mask alone does not reserve any bits for this.

[3 marks]: [1] for stating 8 host bits, [1] for identifying that the default mask defines only one continuous network/range, [1] for explaining that bits must be borrowed from the host portion to divide it into subnets.

Part (b): Bits borrowed and subnets created

The new mask’s last octet is 192. In binary:

192 = 128 + 64 = 11000000

Two of the 8 bits in this octet are set to 1 beyond the original all-network first three octets, so 2 bits have been borrowed from the host portion and turned into extra network bits (the mask is now a /26: 24 + 2 = 26 network bits).

Borrowing n bits creates 2^n subnets. With 2 bits borrowed:

2^2 = 4 subnets

This produces exactly 4 subnets, one for each of Sales, Accounts, IT and Warehouse.

[3 marks]: [1] for correctly identifying 192 = 11000000 in binary, [1] for stating 2 bits borrowed, [1] for correctly calculating 4 subnets from 2^2.

Part (c): Usable host addresses per subnet

After borrowing 2 bits for subnetting, the number of host bits left in each subnet is:

8 (original host bits) - 2 (borrowed) = 6 host bits

The number of addresses in each subnet is:

2^6 = 64 addresses

Of these 64 addresses, 2 cannot be given to hosts: the first is the subnet’s own network address (identifying the subnet itself), and the last is the subnet’s broadcast address (used to send to every host on that subnet at once). So the number of usable host addresses is:

64 - 2 = 62 usable host addresses

Since each department needs at most 50 host addresses, and each /26 subnet provides 62 usable addresses, this is enough, with 12 addresses to spare in each subnet for future growth.

[3 marks]: [1] for correctly calculating 2^6 = 64 total addresses, [1] for subtracting 2 reserved addresses to reach 62 usable addresses, [1] for correctly stating that 62 is enough for a department needing at most 50.

Final answers

• (a) 8 host bits; the default /24 mask keeps the whole block as one single network, so bits must be borrowed to create subnets.
• (b) 2 bits borrowed (from 192 = 11000000), producing 2^2 = 4 subnets.
• (c) 64 total addresses per /26 subnet, minus 2 reserved addresses = 62 usable host addresses, which is enough for a department needing at most 50.