Computer Networks · TEE · read

Equal-size subnet design · example 2

Subnetting: first address, last address and host range

TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023

Practice question. Divide 172.16.0.0/16 into at least 8 equal-size ordinary subnets. Give the new prefix, number of available subnets, usable hosts per subnet, and every resulting address range. [3 practice marks]

Answer

Given

Parent=172.16.0.0/16; required equal-size subnets=8.

To find

Borrowed bits, new mask/prefix and complete subnet ranges.

Formula and why

2^s\geq N_{\mathrm{subnets}}LaTeX: 2^s\geq N_{\mathrm{subnets}}
p_{\mathrm{new}}=p_{\mathrm{old}}+sLaTeX: p_{\mathrm{new}}=p_{\mathrm{old}}+s

Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.

Step-by-step solution

Step 1. Choose borrowed bits

s=3, because 23=8 is the smallest power of two at least 8.

Step 2. Find the new prefix and size

/16+3=/19; mask=255.255.224.0. Each subnet contains 8192 total and 8190 usable addresses.

Step 3. List each subnet

SubnetFirst/networkLast/broadcastUsable range
172.16.0.0/19172.16.0.0172.16.31.255172.16.0.1–172.16.31.254
172.16.32.0/19172.16.32.0172.16.63.255172.16.32.1–172.16.63.254
172.16.64.0/19172.16.64.0172.16.95.255172.16.64.1–172.16.95.254
172.16.96.0/19172.16.96.0172.16.127.255172.16.96.1–172.16.127.254
172.16.128.0/19172.16.128.0172.16.159.255172.16.128.1–172.16.159.254
172.16.160.0/19172.16.160.0172.16.191.255172.16.160.1–172.16.191.254
172.16.192.0/19172.16.192.0172.16.223.255172.16.192.1–172.16.223.254
172.16.224.0/19172.16.224.0172.16.255.255172.16.224.1–172.16.255.254

Step 4. State spare subnet capacity

8 available−8 required=0 extra equal-size subnet(s).

Final answer: /19; 8 subnets; 8190 usable hosts each.