Subnetting Practice Problem
Given the network 192.168.1.0/24 (subnet mask 255.255.255.0, supporting 254 hosts), subnet it to create segments where each subnet supports exactly 30 usable hosts. Provide the new subnet mask, number of subnets, and details for all resulting subnets (network address, host range, broadcast)
Determine Required Host Bits
Usable hosts per subnet formula:
$2^h - 2 \ge 30$
Try values:
- $(2^4 - 2 = 14)$ → Not enough
- $(2^5 - 2 = 30)$ → Perfect
So we need 5 host bits.
Original network was /24 → meaning 8 host bits originally.
If 5 bits are for hosts, then:
$32 - 5 = 27$
So the new prefix is:
/27
New Subnet Mask
/27 in dotted decimal:
255.255.255.224
Why?
Last octet in binary:
11100000 = 224
Number of Subnets Created
Originally: /24
Now: /27
Borrowed bits:
$27 - 24 = 3 \text{ bits}$
Number of subnets:
$2^3 = 8 \text{ subnets}$
Block Size
Block size = 256 − 224 = 32
So subnets increase by 32 in the last octet.
Final Subnet Breakdown
Each subnet has:
- 32 total addresses
- 30 usable hosts
- 1 network
- 1 broadcast
Subnet 1
Network: 192.168.1.0/27
Usable: 192.168.1.1 – 192.168.1.30
Broadcast: 192.168.1.31
Subnet 2
Network: 192.168.1.32/27
Usable: 192.168.1.33 – 192.168.1.62
Broadcast: 192.168.1.63
Subnet 3
Network: 192.168.1.64/27
Usable: 192.168.1.65 – 192.168.1.94
Broadcast: 192.168.1.95
Subnet 4
Network: 192.168.1.96/27
Usable: 192.168.1.97 – 192.168.1.126
Broadcast: 192.168.1.127
Subnet 5
Network: 192.168.1.128/27
Usable: 192.168.1.129 – 192.168.1.158
Broadcast: 192.168.1.159
Subnet 6
Network: 192.168.1.160/27
Usable: 192.168.1.161 – 192.168.1.190
Broadcast: 192.168.1.191
Subnet 7
Network: 192.168.1.192/27
Usable: 192.168.1.193 – 192.168.1.222
Broadcast: 192.168.1.223
Subnet 8
Network: 192.168.1.224/27
Usable: 192.168.1.225 – 192.168.1.254
Broadcast: 192.168.1.255
Final Answers
- New Subnet Mask: 255.255.255.224
- CIDR Notation: /27
- Number of Subnets: 8
- Usable Hosts per Subnet: 30
- Total Usable Hosts Across All Subnets: 240