🚧 Under construction — VoTechWorks is being built. A free tech program — complimentary, as our commitment to get America working.
voTechWorksGrade 13 · a vocational technology learning center, by urbanicity
newSchoolWorks.com › VoTechWorks › Technology & Design › IT Support / Networking › Networking Fundamentals and Infrastructure › Lesson 2
TEC-ITS43-04 · Lesson 2 · 14 clock hours · theory

IP Addressing and Subnetting

Objective. Students will calculate network address, broadcast address, usable host range, and subnet mask for any given IPv4 CIDR block, and convert between binary, decimal, and hexadecimal without errors on a timed worksheet.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by an industry-certified professional or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

VoTechWorks program approval and accreditation are still in progress. Credentials are issued only by the certifying bodies named in each program, after their own exams.

Why IP Addressing Matters

Every device on a network needs a unique address so data reaches the right destination. IPv4 addresses are 32-bit numbers written in dotted-decimal notation — four groups of digits separated by dots, for example `192.168.10.1`. Each group is called an octet because it represents 8 bits. Understanding how these addresses are structured, divided, and assigned is a core skill tested on CompTIA A+, CompTIA Network+, the Cisco CCST, and the Cisco CCNA.

Binary and the Octet

A computer does not see `192` — it sees `11000000`. Binary (base 2) uses only 0 and 1. Each bit position in an octet has a fixed value, doubling from right to left:

| Bit position | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |

|---|---|---|---|---|---|---|---|---|

| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

To convert decimal 192 to binary, subtract the largest place value that fits and write a 1; write a 0 where nothing fits: 128+64 = 192, so the result is `11000000`. To go from binary back to decimal, add the place values of every position that holds a 1.

Hexadecimal (base 16) uses digits 0–9 and letters A–F, where A=10 through F=15. One hex digit represents exactly 4 bits (a nibble). Split an 8-bit octet into two nibbles and convert each. For example, `11000000` splits into `1100` (=12=C) and `0000` (=0), giving hex `C0`. Hexadecimal appears on the CCNA exam and in IPv6, VLAN IDs, and MAC addresses.

Subnet Masks

A subnet mask is a 32-bit number that identifies which portion of an IP address is the network part and which is the host part. Network bits are all 1s; host bits are all 0s. Written in dotted decimal, a common mask looks like `255.255.255.0`. Written in CIDR (Classless Inter-Domain Routing) notation, that same mask is `/24` — meaning 24 bits are used for the network.

Common masks to memorize:

| CIDR | Dotted-decimal | Host bits | Usable hosts |

|---|---|---|---|

| /24 | 255.255.255.0 | 8 | 254 |

| /25 | 255.255.255.128 | 7 | 126 |

| /26 | 255.255.255.192 | 6 | 62 |

| /27 | 255.255.255.224 | 5 | 30 |

| /28 | 255.255.255.240 | 4 | 14 |

| /30 | 255.255.255.252 | 2 | 2 |

Key Calculations

Number of subnets: When you borrow n bits from the host portion, you create 2ⁿ subnets.

Usable hosts per subnet: If h bits remain for hosts, total addresses = 2ʰ. Subtract 2 (network address and broadcast address) to get usable hosts: 2ʰ − 2.

Network address: Perform a bitwise AND between the IP address and the subnet mask. All host bits become 0.

Broadcast address: Set all host bits to 1 in the network address.

Host range: First usable host = network address + 1; last usable host = broadcast address − 1.

VLSM — Variable Length Subnet Masking

VLSM lets you use different prefix lengths within the same address space so you waste as few addresses as possible. The standard approach is to allocate the largest subnet first, then carve the remaining space for smaller subnets. Subnets must never overlap. VLSM is heavily tested on Network+ and CCNA.

A Practical Design Rule

Always plan for growth. If a department needs 25 hosts today, a /27 (30 usable) leaves only 5 spare addresses. A /26 (62 usable) gives more headroom. Document every decision — a subnet plan with no notes creates serious troubleshooting problems later.

Electrical and Physical Safety Reminders

Subnetting is pencil-and-paper work, but this course also requires hands-on lab time with live equipment. Before touching any server, switch, or patch panel:

These habits protect both you and the equipment, and employers expect them from day one.

Key terms8

IPv4 addressA 32-bit numerical label written as four decimal octets (e.g., 192.168.1.10) that uniquely identifies a device on an IP network.
OctetOne of the four 8-bit groups that make up an IPv4 address; its decimal value ranges from 0 to 255.
Subnet maskA 32-bit number that separates the network portion of an IP address (all 1-bits) from the host portion (all 0-bits), written in dotted-decimal or CIDR notation.
CIDR (Classless Inter-Domain Routing)A notation that expresses the subnet mask as a prefix length (e.g., /24), indicating how many leading bits identify the network.
Network addressThe first address in a subnet, where all host bits are 0; it identifies the subnet itself and cannot be assigned to a host.
Broadcast addressThe last address in a subnet, where all host bits are 1; packets sent to this address are delivered to every host in the subnet.
VLSM (Variable Length Subnet Masking)A technique that allows different subnets within the same address space to use different prefix lengths, minimizing address waste.
Usable host rangeThe contiguous range of IP addresses between the network address and the broadcast address; the formula is 2ʰ − 2, where h is the number of host bits.

Worked examples4

1. Given the address 192.168.5.70/26, find the network address, broadcast address, and usable host range.
  1. Identify the prefix length: /26 means 26 network bits and 6 host bits.
  2. Write the /26 mask in the last octet: 11111111.11111111.11111111.11000000 = 255.255.255.192. The block size in the last octet is 256 − 192 = 64.
  3. Find which block 70 falls in: blocks start at 0, 64, 128, 192. Since 64 ≤ 70 < 128, the network address is 192.168.5.64.
  4. Network address: 192.168.5.64 (all host bits = 0).
  5. Broadcast address: 192.168.5.127 (all host bits = 1; 64 + 64 − 1 = 127).
  6. First usable host: 192.168.5.65 (network address + 1).
  7. Last usable host: 192.168.5.126 (broadcast address − 1).
  8. Number of usable hosts: 2⁶ − 2 = 64 − 2 = 62.
Answer: Network: 192.168.5.64 | Broadcast: 192.168.5.127 | Host range: 192.168.5.65 – 192.168.5.126 (62 usable hosts)
2. A department needs exactly 28 usable host addresses. What is the smallest subnet prefix that meets this requirement, and how many addresses does it waste?
  1. Use the formula 2ʰ − 2 ≥ 28.
  2. Test h = 4: 2⁴ − 2 = 14 — too small.
  3. Test h = 5: 2⁵ − 2 = 30 — meets the requirement.
  4. A /27 mask (32 − 5 = 27) provides 30 usable hosts.
  5. Addresses wasted: 30 available − 28 needed = 2 unused usable addresses, plus the 2 reserved (network + broadcast) = 4 addresses in the block not assigned to active devices.
Answer: Use a /27 (255.255.255.224). It provides 30 usable hosts, meeting the 28-host requirement with 2 usable addresses to spare.
3. Convert the decimal octet 172 to binary and then to hexadecimal.
  1. Write the place values: 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1.
  2. Subtract largest fitting value: 172 − 128 = 44 → bit 7 = 1.
  3. 44 − 32 = 12 → bit 5 = 1; 64 does not fit (44 < 64) → bit 6 = 0.
  4. 12 − 8 = 4 → bit 3 = 1; 16 does not fit → bit 4 = 0.
  5. 4 − 4 = 0 → bit 2 = 1; bits 1 and 0 = 0.
  6. Binary result: 10101100.
  7. Split into nibbles: 1010 and 1100.
  8. 1010 = 10 = A in hex; 1100 = 12 = C in hex.
  9. Hexadecimal result: AC.
Answer: 172 decimal = 10101100 binary = AC hexadecimal
4. A company is given 10.0.0.0/24 and needs four subnets: Dept A = 50 hosts, Dept B = 25 hosts, Dept C = 12 hosts, Dept D = 5 hosts. Assign VLSM subnets in order from largest to smallest.
  1. Start with the largest requirement: Dept A needs 50 hosts. 2⁶ − 2 = 62 ≥ 50, so use /26. Assign 10.0.0.0/26. Network: 10.0.0.0, Broadcast: 10.0.0.63.
  2. Dept B needs 25 hosts. 2⁵ − 2 = 30 ≥ 25, so use /27. Next available address: 10.0.0.64. Assign 10.0.0.64/27. Network: 10.0.0.64, Broadcast: 10.0.0.95.
  3. Dept C needs 12 hosts. 2⁴ − 2 = 14 ≥ 12, so use /28. Next available: 10.0.0.96. Assign 10.0.0.96/28. Network: 10.0.0.96, Broadcast: 10.0.0.111.
  4. Dept D needs 5 hosts. 2⁴ − 2 = 14 ≥ 5; a /28 works. Alternatively /29 gives 2³ − 2 = 6 ≥ 5, which is tighter. Use /29. Next available: 10.0.0.112. Assign 10.0.0.112/29. Network: 10.0.0.112, Broadcast: 10.0.0.119.
  5. Verify no subnets overlap: 0–63, 64–95, 96–111, 112–119 — no overlap. Remaining space: 10.0.0.120–10.0.0.255 is unallocated.
Answer: Dept A: 10.0.0.0/26 (hosts .1–.62) | Dept B: 10.0.0.64/27 (hosts .65–.94) | Dept C: 10.0.0.96/28 (hosts .97–.110) | Dept D: 10.0.0.112/29 (hosts .113–.118)

The math in this lesson2

Learners compute /26 block size 64, network and broadcast addresses, and usable hosts from powers of 2.
8.EE.1F-LE.1
Brush up in the math halls at newSchoolWorks.com:
In class: Instructor-led binary conversion drill: students convert ten decimal octets to binary and back, then self-grade against an answer key before repeating with hexadecimal · CIDR subnetting worksheet set (three levels of difficulty): students manually compute network ID, broadcast, host range, and number of hosts for 20 subnets; instructor circulates to correct errors in real time · Collaborative subnetting scenario: groups receive a fictional company address-space requirement (e.g., four departments with specific host counts) and must design a non-overlapping VLSM scheme on paper, then present their solution to the class
Materials: IPv4 subnetting worksheets, binary/hex conversion tables (for reference until memorized), calculators (as check only), CompTIA Network+ and Cisco CCNA official study guides, whiteboard
Check: Timed 10-question subnetting quiz (25 minutes, no calculator); student must correctly solve at least eight questions including network address, broadcast, and host range; 80% required to continue; remediation worksheet assigned for any misses

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

CompTIA Network+ — IP Addressing
1. A network administrator is assigned the block 172.16.4.0/25. What is the broadcast address for this subnet?
CompTIA A+ — Networking Concepts
2. How many usable host addresses are available in a /28 subnet?
Cisco CCNA — IP Services and Addressing
3. Which subnet mask in dotted-decimal notation corresponds to the CIDR prefix /27?
CompTIA Network+ — IP Addressing
4. An engineer converts the binary value 10110110 to decimal. What is the correct result?
Cisco CCNA — IP Addressing and Subnetting
5. A company is given 10.10.10.0/24 and needs the smallest subnet that supports exactly 60 usable hosts. Which prefix length should be used?
Cisco CCNA — IP Addressing and Subnetting
6. Using VLSM with the address space 192.168.20.0/24, a network engineer assigns 192.168.20.0/26 to Department A. What is the next available network address after this subnet?
Cisco CCST — Networking Basics
7. What is the hexadecimal equivalent of the decimal value 255?
CompTIA Network+ — IP Addressing
8. A technician receives a host address of 10.1.1.200/29. Which of the following is the correct network address for this subnet?
Record this lesson on my transcript. Sign in with your key and enrollment email to record lessons (enroll free if you have no key). This page stays open to read and practise.
← Previous lessonProgram planNext lesson →