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Course 2: IP Subnetting and VLSMLesson 1.8 (8 of 8 in this course)13 of 98 in the CCNA series

Find the mistake

Each answer below has one error. Finding it is how you check a design, review a colleague's change, or troubleshoot a misconfigured host.

Intermediate · 10 min read · Before this: VLSM for a real office

After this lesson, you can spot the common subnetting errors in someone else's work and explain the correct answer.

Read each task and the answer, find the line that is wrong, and decide what it should be. Then open the explanation.

Exercise 1

Find the network and broadcast of 192.168.5.130/26.

Their answer
Network 192.168.5.128, broadcast 192.168.5.255.
Show the mistake

The broadcast is wrong. They used a block of 128 (a /25). A /26 has a block size of 64, so the block 128–191 ends at .191.

Worked answer for 192.168.5.130/26
StepWorking
1. Find the interesting octet/26 = 255.255.255.192. The 4th octet is neither 255 nor 0: its mask value is 192.
2. Block size256 − 192 = 64. In the 4th octet, subnets start at 0, 64, 128, 192.
3. Network130 falls in the block that starts at 128. Network: 192.168.5.128
4. BroadcastThe next block starts at 192, so this one ends at 191. Broadcast: 192.168.5.191
5. Usable range and hosts192.168.5.129 to 192.168.5.190: 26 − 2 = 62 hosts

Exercise 2

Find the network address of 172.16.45.10/20.

Their answer
172.16.45.0
Show the mistake

They worked as if it were a /24. With a /20 the interesting octet is the 3rd: block size 16, so 45 is in 32–47.

Worked answer for 172.16.45.10/20
StepWorking
1. Find the interesting octet/20 = 255.255.240.0. The 3rd octet is neither 255 nor 0: its mask value is 240.
2. Block size256 − 240 = 16. In the 3rd octet, subnets start at …, 16, 32, 48, 64, ….
3. Network45 falls in the block that starts at 32. The 4th octet becomes 0. Network: 172.16.32.0
4. BroadcastThe next block starts at 48, so this one ends at 47. The 4th octet becomes 255. Broadcast: 172.16.47.255
5. Usable range and hosts172.16.32.1 to 172.16.47.254: 212 − 2 = 4,094 hosts

Exercise 3

How many hosts can a /27 subnet hold?

Their answer
32 − 27 = 5 host bits, 25 = 32 hosts.
Show the mistake

32 is the number of addresses. The network and broadcast addresses can't be given to devices, so a /27 holds 30 hosts.

Exercise 4

A PC is configured as 10.1.1.70/26 with default gateway 10.1.1.64. It can't reach anything off its subnet. Why?

Their answer
The router must be down; the PC's settings are fine.
Show the mistake

10.1.1.64 is the network address of the PC's subnet, so it can't be the router's interface address. The gateway must be a usable address between .65 and .126, and it must match what is actually configured on the router.

Worked answer for 10.1.1.70/26
StepWorking
1. Find the interesting octet/26 = 255.255.255.192. The 4th octet is neither 255 nor 0: its mask value is 192.
2. Block size256 − 192 = 64. In the 4th octet, subnets start at 0, 64, 128, 192.
3. Network70 falls in the block that starts at 64. Network: 10.1.1.64
4. BroadcastThe next block starts at 128, so this one ends at 127. Broadcast: 10.1.1.127
5. Usable range and hosts10.1.1.65 to 10.1.1.126: 26 − 2 = 62 hosts

Exercise 5

A PC is 192.168.10.65/27 with gateway 192.168.10.62.

Their answer
Fine: the gateway is in 192.168.10.0/24, like the PC.
Show the mistake

The mask is /27, not /24. The PC is in 192.168.10.64/27 (.65–.94) and .62 is in the previous block (.32–.63). The PC can't ARP for a gateway outside its own subnet, so everything remote fails.

Exercise 6

A VLSM plan from 192.168.1.0/24: Sales needs 50 hosts, Engineering 25.

Their answer
Sales 192.168.1.0/26, Engineering 192.168.1.32/27.
Show the mistake

The blocks overlap. Sales covers .0–.63, and .32–.63 is inside it. Engineering must start after Sales ends: 192.168.1.64/27.

Exercise 7

A floor will have 300 devices. Choose a mask.

Their answer
255.255.255.0: plenty of room.
Show the mistake

A /24 holds 254 hosts. 300 needs 9 host bits (29 − 2 = 510), so the mask is 255.255.254.0 (/23).

Check yourself

Predict · scenario 1

A colleague says 172.16.20.0 is the network address of 172.16.20.77/22. Are they right?

Predict · scenario 2

A router interface is configured as 192.168.3.255/23. Is that a valid host address?

Keep practising

The subnetting practice drill generates unlimited questions with worked answers. Try Hard (/8 to /30, any octet) once Medium feels easy. To check a real network, enter the address in the subnet calculator.

Learn more: Subnetting OverviewDifferent-Subnet Communication