Subnetting a School Network with Five Laboratories
A school in Irbid is rebuilding its computer network on the private address 192.168.20.0/24. The network must serve five separate laboratories with these current device counts: 50, 25, 10, 5, and 2 hosts. A student volunteer proposes giving every laboratory a /26 subnet, arguing that /26 is simple and provides plenty of addresses. The IT teacher points out that the plan must be checked against both the number of laboratories and the number of hosts, and that unused addresses should be left as a contiguous free block for future growth. The school also wants to keep the design easy to document, with each laboratory on its own subnet and no overlap. After the plan is agreed, the science laboratory announces that it will grow to 70 devices next year. The teacher asks whether the original address plan can still accommodate that growth without changing the other four laboratories, and if so, where the new subnet should be placed.
1. Explain why a /26 for every laboratory fails, giving the number of /26 subnets and usable hosts.
Read the worked answer
A /24 has 8 host bits. A /26 borrows 2 bits for subnetting, so it creates [2^2=4] subnets. Each subnet has (2^6-2=62) usable host addresses.
The school needs five separate laboratory subnets. Four /26 subnets cannot cover five laboratories, so at least one laboratory would have no subnet. In addition, giving a /26 to the 2-host and 5-host laboratories wastes most of their 62 addresses. The /26 plan fails on the number of subnets, even though it has enough hosts per subnet for the 50-host laboratory.
2. Design a VLSM plan for the five laboratories, largest first. List the network address, prefix, usable range, and broadcast address for each.
Read the worked answer
Allocate the largest requirement first, using the smallest prefix that fits.
- 50 hosts: needs 62 usable addresses, so a /26. Network (192.168.20.0/26). Usable (192.168.20.1)–(192.168.20.62). Broadcast (192.168.20.63).
- 25 hosts: needs 30 usable addresses, so a /27. Network (192.168.20.64/27). Usable (192.168.20.65)–(192.168.20.94). Broadcast (192.168.20.95).
- 10 hosts: needs 14 usable addresses, so a /28. Network (192.168.20.96/28). Usable (192.168.20.97)–(192.168.20.110). Broadcast (192.168.20.111).
- 5 hosts: needs 6 usable addresses, so a /29. Network (192.168.20.112/29). Usable (192.168.20.113)–(192.168.20.118). Broadcast (192.168.20.119).
- 2 hosts: needs 2 usable addresses, so a /30. Network (192.168.20.120/30). Usable (192.168.20.121)–(192.168.20.122). Broadcast (192.168.20.123).
The block (192.168.20.124)–(192.168.20.255) remains free as one contiguous range. Every subnet is aligned on its prefix boundary, and there is no overlap.
3. The 50-host laboratory grows to 70 devices. Can the existing plan absorb it without moving the other four laboratories? Give the new subnet and check for overlap.
Read the worked answer
A /26 gives only 62 usable addresses, so it cannot hold 70 devices. The grown laboratory needs at least 126 usable addresses, which means a /25. A /25 has 7 host bits and provides [2^7-2=126] usable addresses.
The free block is (192.168.20.124)–(192.168.20.255). A /25 must start on a multiple of 128, so the valid choice is [192.168.20.128/25.] Its usable range is (192.168.20.129)–(192.168.20.254), and its broadcast address is (192.168.20.255).
This does not overlap the other four laboratories, which occupy (192.168.20.64)–(192.168.20.123). The old (192.168.20.0/26) is freed, and the addresses (192.168.20.124)–(192.168.20.127) remain unused. So the other four laboratories can stay exactly where they are; the grown laboratory is moved to the new /25.