Subnetting Workbook A — Mechanics

Companion to Subnetting with CIDR — Part 1 (Module 4). Every subnetting problem you will ever meet — in this course, in certifications, on the job — asks for some combination of the same seven things. This workbook drills all seven until they are automatic.

The 7 things every problem asks

1. Network ID — the first address of the subnet (all host bits 0). Names the network; never assigned to a device.
2. Broadcast address — the last address (all host bits 1). Reaches everyone in the subnet; never assigned either.
3. First usable host — network ID + 1.
4. Last usable host — broadcast − 1.
5. Next network — where the following subnet starts (broadcast + 1).
6. Host count — usable addresses = 2^(host bits) − 2 (minus network ID and broadcast).
7. CIDR ↔ mask — translate /n into dotted decimal and back (/26 = 255.255.255.192).

The block-size shortcut answers 1–5 without touching binary. Find the interesting octet — the mask octet that is neither 255 nor 0. Then:

block size = 256 − interesting mask octet

Subnets start at multiples of the block size in that octet. Your network ID is the greatest multiple ≤ your address's value there; the broadcast is one less than the next multiple.

Worked examples

Example 1 — 192.168.10.77/24

/24 = 255.255.255.0 — the whole fourth octet is host bits (8 of them).

• Network ID: zero the host octet → 192.168.10.0
• Broadcast: host octet all 1s → 192.168.10.255
• First / last host: 192.168.10.1 / 192.168.10.254
• Next network: 192.168.11.0
• Hosts: 2^8 − 2 = 254

Example 2 — 172.16.5.200/26

/26 = 255.255.255.192; interesting octet = the fourth; block = 256 − 192 = 64. Multiples: 0, 64, 128, 192. Where does 200 fall? Between 192 and 255.

• Network ID: 172.16.5.192
• Broadcast: next multiple (256) minus 1 → 172.16.5.255
• First / last host: 172.16.5.193 / 172.16.5.254
• Next network: 192 + 64 = 256 → the octet rolls over → 172.16.6.0
• Hosts: 6 host bits → 2^6 − 2 = 62

Example 3 — 10.0.0.100/28

/28 = 255.255.255.240; block = 256 − 240 = 16. Multiples: 0, 16, 32, 48, 64, 80, 96, 112… 100 sits between 96 and 112.

• Network ID: 10.0.0.96
• Broadcast: 112 − 1 → 10.0.0.111
• First / last host: 10.0.0.97 / 10.0.0.110
• Next network: 10.0.0.112
• Hosts: 4 host bits → 2^4 − 2 = 14

Exercises

For each address, give all seven: mask (dotted decimal), network ID, broadcast, first host, last host, next network, host count. Prefixes run /24 → /30, easiest to hardest. Watch for the two classic traps hidden in the set: an address that is the broadcast, and a next-network that rolls into another octet.

1. 192.168.1.50/24
2. 192.168.4.130/25
3. 192.168.7.60/25
4. 10.1.1.90/26
5. 172.16.30.20/26
6. 192.168.100.200/26
7. 10.0.5.37/27
8. 172.16.8.100/27
9. 192.168.50.223/27
10. 10.20.30.40/28
11. 192.168.2.100/28
12. 172.31.4.250/28
13. 10.10.10.10/29
14. 192.168.77.100/29
15. 172.16.99.62/29
16. 10.200.14.66/30
17. 192.168.0.21/30
18. 172.20.5.129/30
19. 192.168.31.158/26
20. 10.99.255.245/28

Answer key

The arithmetic, shown for 1–5

1. 192.168.1.50/24 — mask 255.255.255.0; all 8 fourth-octet bits are host bits. Network 192.168.1.0; broadcast 192.168.1.255; hosts .1–.254; next network 192.168.2.0; 2^8 − 2 = 254 hosts.

2. 192.168.4.130/25 — mask 255.255.255.128; block 256 − 128 = 128; multiples 0, 128. 130 ≥ 128 → network 192.168.4.128; broadcast 128 + 128 − 1 = 255 → 192.168.4.255; hosts .129–.254; next network rolls over to 192.168.5.0; 2^7 − 2 = 126 hosts.

3. 192.168.7.60/25 — same block of 128, but 60 < 128 → network 192.168.7.0; broadcast 192.168.7.127; hosts .1–.126; next network 192.168.7.128; 126 hosts.

4. 10.1.1.90/26 — mask 255.255.255.192; block 256 − 192 = 64; multiples 0, 64, 128, 192. 90 falls in the 64 block → network 10.1.1.64; broadcast 128 − 1 = 127 → 10.1.1.127; hosts .65–.126; next network 10.1.1.128; 2^6 − 2 = 62 hosts.

5. 172.16.30.20/26 — block 64; 20 < 64 → network 172.16.30.0; broadcast 172.16.30.63; hosts .1–.62; next network 172.16.30.64; 62 hosts.

Full results, 1–20

Hosts column = usable hosts. First–last shortens the last octet only; every other octet matches the network ID unless written out.

The traps: in #9 the given address 192.168.50.223 is the broadcast of its subnet — a device configured with it would silently misbehave. In #15 the given address is the last usable host (legal, but no room left). In #20 the next network rolls all the way to 10.100.0.0 — block arithmetic carries across octets exactly like ordinary addition.

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