Subnetting Workbook B — Design

Companion to Subnetting with CIDR — Part 2 (Module 4). Workbook A drilled the mechanics of a single subnet; here you run the process in reverse — from "how many hosts do we need?" to a complete, non-overlapping addressing plan. This is the arithmetic of real network design, and it feeds the course capstone.

The sizing method

1. Hosts needed → host bits. Find the smallest h where 2^h − 2 ≥ hosts needed. (The −2 pays for the network ID and broadcast.) Exception, worth knowing: the −2 applies to ordinary broadcast subnets like the ones designed here. On a point-to-point link between exactly two routers, RFC 3021 allows a /31 whose two addresses are both usable (no broadcast is needed when there is only one other end); the traditional /30 does the same job with 2 of its 4 addresses spent. A /32 is a single host route, not a subnet.
2. Host bits → prefix. Prefix = 32 − h.
3. Allocate largest subnet first, packing from the start of your block. A subnet must start on a multiple of its own size — largest-first keeps every start aligned automatically and leaves no wasted gaps.
4. Verify: no two subnets overlap, everything fits inside the parent block, and there is room left for growth.

Handy sizes: /30 → 2 hosts · /29 → 6 · /28 → 14 · /27 → 30 · /26 → 62 · /25 → 126 · /24 → 254 · /23 → 510 · /22 → 1,022. Remember the department's router (gateway) consumes one usable address — count it among the hosts.

Worked design — 10.20.0.0/22 for five departments

Requirements: A = 200 hosts, B = 120, C = 60, D = 25, E = 10. The parent block 10.20.0.0/22 spans 10.20.0.0 – 10.20.3.255 (1,024 addresses).

Size each department (smallest h with 2^h − 2 ≥ need):

• A: 200 → h = 8 (254) → /24 · B: 120 → h = 7 (126) → /25
• C: 60 → h = 6 (62) → /26 · D: 25 → h = 5 (30) → /27
• E: 10 → h = 4 (14) → /28

Allocate largest-first from 10.20.0.0:

Each subnet begins exactly where the previous one ended — no gaps, no overlaps — and each start is a clean multiple of its block size. Free for growth: 10.20.1.240 – 10.20.3.255 (one /28 plus all of 10.20.2.0/23 — 528 addresses, over half the parent block).

Design exercises

1. What is the smallest prefix (longest /n) that fits 50 hosts, and how many usable addresses does it give?
2. Pick the right prefix for each requirement: (a) a 2-host router-to-router link, (b) 20 hosts, (c) 500 hosts, (d) 1,000 hosts.
3. Split 192.168.8.0/24 into four equal subnets. List each subnet's network ID, broadcast, and host capacity.
4. From 172.16.40.0/23, allocate largest-first for three groups: 250, 100, and 40 hosts. Show the allocation table and what remains free.
5. A branch office owns 192.168.64.0/26. It needs three VLANs: 25 hosts, 10 hosts, and a 2-host router link. Does everything fit? Produce the allocation.
6. Overlap detection. A junior admin proposes, inside 10.50.0.0/16 — Sales 10.50.16.0/20 · Engineering 10.50.24.0/21 · Guest 10.50.32.0/20 · Servers 10.50.30.0/24. Find every overlap and propose a corrected plan that keeps all four sizes.
7. Growth planning. You hold 172.20.0.0/23 and need four subnets of up to 100 hosts each today; each is expected to double within three years. Show why the obvious plan fails after the doubling and what you should request instead.
8. Full design. Plan 192.168.0.0/23 for: wired staff 180, Wi-Fi 120, VoIP phones 60, servers 20, management 10, plus two 2-host router links. Allocate largest-first; show the table and the leftover space.

Answer key

1. 50 hosts → 2^5 − 2 = 30 is too small, 2^6 − 2 = 62 fits → /26, 62 usable addresses.

2. (a) /30 (2 usable) · (b) /27 (30; a /28's 14 is too small) · (c) /23 (510; a /24's 254 is too small) · (d) /22 (1,022).

3. Four equal parts = borrow 2 bits → /26, block 64:

4. Sizing: 250 → /24 · 100 → /25 · 40 → /26. Parent 172.16.40.0/23 spans 172.16.40.0 – 172.16.41.255:

Free: 172.16.41.192/26 (62 usable addresses).

5. Sizing: 25 → /27 (32 addresses) · 10 → /28 (16) · 2 → /30 (4). Total 32 + 16 + 4 = 52 ≤ 64 → it fits:

Free: 192.168.64.52 – 192.168.64.63 (12 addresses — one /30 at .52 plus one /29 at .56).

6. Write each block's range first — overlaps become obvious:

• Sales 10.50.16.0/20 → 10.50.16.0 – 10.50.31.255
• Engineering 10.50.24.0/21 → 10.50.24.0 – 10.50.31.255 → entirely inside Sales (overlap 1)
• Guest 10.50.32.0/20 → 10.50.32.0 – 10.50.47.255 → clean against Sales
• Servers 10.50.30.0/24 → 10.50.30.0 – 10.50.30.255 → **inside Sales and inside Engineering** (overlaps 2 and 3)

One corrected plan (same sizes, packed cleanly):

7. Today: 100 hosts → /25 (126). Four /25s = 4 × 128 = 512 addresses — the entire /23 with zero spare: already a red flag. After doubling: 200 hosts > 126 → every subnet outgrows its /25. 200 needs /24 (254), and four /24s = 1,024 addresses = a /22. Request 172.20.0.0/22 and allocate 172.20.0.0/24, 172.20.1.0/24, 172.20.2.0/24, 172.20.3.0/24. Design lesson: size for the network you will have, not the one you have today.

8. Sizing: 180 → /24 · 120 → /25 · 60 → /26 · 20 → /27 · 10 → /28 · 2 → /30 (twice). Largest-first from 192.168.0.0:

Allocated: 256 + 128 + 64 + 32 + 16 + 4 + 4 = 504 of 512. Leftover: 192.168.1.248/29 (8 addresses, 6 usable) — thin growth room; flag it.

Network Essentials · Module 4 · Turning Point Academy — backbone: Al-Doori, Network Essentials, Ch. 6.