DSE ICT | IP Addressing and Subnetting: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary ICT

Students can recite 255.255.255.0 and then stop dead at a slash 26. This lesson breaks the prefix into three facts: a final octet of 192, blocks of 64 addresses and 62 usable hosts. After that every question is the same arithmetic.

Objectives — Students can: (1) convert a slash 26 prefix into the mask 255.255.255.192 with blocks of 64 addresses; (2) compute 62 usable hosts per block and four subnets in total; (3) identify that 192.168.10.130 slash 26 has network address .128, broadcast .191 and usable range .129 to .190; (4) explain why a slash 30 leaves only two usable hosts and why the minus two covers the network and broadcast addresses.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Binary	How is 200 written in 8-bit binary?	11001000
Bits	How many bits are in an IPv4 address?	32 bits, four octets of eight
Powers of two	What is 2 to the power 6?	64
Division	Splitting a slash 24 block into four gives how many addresses each?	256 divided by 4, so 64

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Opening conflict	Ask how many machines a slash 26 network supports	Answer 64 addresses, 62 usable
10-20 min	Binary	Write 200 and 192 in binary	11001000 and 11000000
20-32 min	The mask	The final octet for slash 26 is 192, with block size 64	Write 255.255.255.192
32-44 min	Network address	Align 192.168.10.130 slash 26 downwards to a multiple of 64	Get .128
44-54 min	Broadcast and range	Add 63 to get .191, usable from .129 to .190	Write all three values
54-60 min	Compare	Contrast usable hosts for slash 25 and slash 30	Write the first exit question

03 Teaching script (read it out)

Scene 1: Count bits first, then talk about masks

IPv4 is 32 bits, so a slash 26 makes the first 26 bits the network part and leaves 6 bits for hosts. The fourth octet holds eight bits, so the mask uses two of them: 11000000, which is 192. Once students remember that the final octet equals 256 minus the block size, 192 and 64 arrive together and no table needs memorising.

Board line: the last octet equals 256 minus the block size.

Scene 2: Finding the network address is aligning downwards

For 192.168.10.130 slash 26 the block size is 64, and the multiples are 0, 64, 128 and 192. Since 130 sits between 128 and 191, the network address is 192.168.10.128, the broadcast is .191 and the usable range is .129 to .190, which is 62 addresses. All three steps use the same block size, so treating them as three separate formulas is where students go wrong.

Board line: all three answers share one block size.

Scene 3: The minus two is two reserved addresses, not superstition

Each block holds 2 to the n addresses, where n is the host bits, but the first is the network address and the last is the broadcast, and neither can be given to a machine. That leaves 62 for a slash 26, 126 for a slash 25 and 2 for a slash 30. Make students say where those two addresses went every time they write the formula.

Board line: the first and last address of every subnet are never assigned.

04 Question types and worked steps

Type	Example	Steps	Answer
Mask from a prefix	What is the subnet mask for slash 26?	Two spare bits in the last octet, write 11000000, keep the rest at 255	255.255.255.192
Usable host count	How many usable hosts per subnet at slash 26?	Host bits are 32 minus 26, so 64 addresses, then subtract two	62
Network and broadcast	For 192.168.10.130 slash 26	Block size 64, align down to .128, add 63 for the broadcast	Network .128, broadcast .191
Same subnet test	Are 192.168.10.65 and 192.168.10.120 on the same slash 26 subnet?	Both align to .64, then confirm the broadcast is .127	Yes, same subnet

Sanity check (teach this): Order check: usable hosts must be below 2 to the power of 32 minus the prefix, so 128 usable hosts at slash 26 means the minus two was skipped. Alignment check: a network address is always a multiple of the block size, so 130 aligns to 128 and not to itself. Range check: the broadcast is one less than the next network address, and .191 plus one is exactly .192. Total check: four blocks of 64 must add to 256, no more and no less.

05 Independent practice (3 questions)

Question	Answer	What to watch
Network and broadcast for 172.16.5.130 slash 25	Network 172.16.5.128, broadcast 172.16.5.255, 126 usable	Block size is 128
Usable hosts at slash 22	1022, since 2 to the 10 minus 2	Host bits are 32 minus 22
Split 10.0.0.0 slash 24 into four slash 26 networks	10.0.0.0, 10.0.0.64, 10.0.0.128 and 10.0.0.192	Each block starts 64 apart

06 Common mistakes and what to say

Mistake	Student writes	Say this
Reading the prefix as a host count	Slash 26 means 26 machines can connect	The 26 is network bits. Host bits are 32 minus 26, and 2 to the 6 minus 2 gives 62 usable addresses.
Skipping the minus two	A slash 25 gives 128 usable hosts	Subtract the network and broadcast addresses, leaving 126. Handing out 128 guarantees a collision with those two.
Leaving the network address unaligned	The network address for 192.168.10.130 slash 26 is .130	Network addresses are multiples of the block size, and the multiple below 130 is .128.
Writing the mask as 255.255.255.26	Slash 26 means the last octet is 26	A mask is a bit pattern, not a copy of the number. Two bits in the last octet give 11000000, which is 192.

07 Homework (paper, section and question numbers)

ICT chapter index — Revise data networks, then do questions 1 and 2
Data networks — Do question 3 on IP versus MAC, then questions 3 to 5
Network security and encryption — Optional: explain how subnetting limits broadcast traffic

08 Exit check (last 5 minutes)

Question	Answer	Decision
Mask for slash 26	255.255.255.192	Error: wrote 255.255.255.26
Usable hosts per slash 26 block	62	Error: said 64
Network address of 192.168.10.130 slash 26	192.168.10.128	Error: said .130
Usable hosts at slash 30	2	Error: said 4

09 Tutor FAQ

Why subtract two?

The first address names the network and the last is the broadcast, so neither can be assigned to a host. That is why 2 to the n becomes 2 to the n minus 2.

Why does the block size equal 256 minus the last octet?

The mask bits decide the block: 192 uses two bits and leaves six, so 2 to the 6 is 64. And 256 minus 192 is also 64, so both routes agree.

How is VLSM different from fixed-length subnetting?

Fixed-length subnets are all the same size, while VLSM assigns lengths by need, for example slash 24 for a backbone and slash 30 for point-to-point links. It saves addresses but each range must be documented.

Find a tutor — free Tutor Registration →

A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.