Past ExamsRwanda ICT preparation

Network Engineer calculations with worked answers

Updated 15 September 2026 · 2 min read

An engineer's arithmetic is mostly addressing and capacity. Addressing plans have to fit every site without overlaps or wasted space, and capacity estimates have to be good enough to justify spending money on a link. The examples below show each method step by step, with every figure recalculated by independent code before publication.

How this guide was made

Written for this site with AI assistance, then checked automatically before publishing. The topic table, sample questions, worked examples and study plan are generated from this site’s question bank: each worked example is recalculated by separate code, and each sample answer says where it came from.

The guide deliberately states no salaries, grades, pass marks or exam dates. Those change, and they could not be checked here, so take them from the advert itself.

This site is independent and is not affiliated with MIFOTRA or any employer.

Worked examples

Try each one on paper first. Every figure is generated and then recalculated by separate code before it is published.

  1. Subnet details for a /27 address

    A host is configured as 192.168.175.29/27. Find the subnet mask, network address, broadcast address, usable host range and number of usable hosts.

    Show the working
    1. /27 means 27 network bits, so the mask is 255.255.255.224.
    2. The interesting octet is the fourth, where the mask is 224. Block size = 256 − 224 = 32.
    3. The network address is the multiple of 32 at or below the address in that octet: 192.168.175.0.
    4. The broadcast is one below the next network: 192.168.175.31.
    5. Usable hosts run from 192.168.175.1 to 192.168.175.30: 2^5 − 2 = 30 addresses.

    Answer: Mask 255.255.255.224, network 192.168.175.0, broadcast 192.168.175.31, hosts 192.168.175.1–192.168.175.30 (30 usable)

  2. Choosing the smallest subnet that fits

    A department needs 60 usable IPv4 host addresses on one subnet. What is the longest prefix (smallest subnet) that fits, and how many usable addresses does it give?

    Show the working
    1. Usable hosts = 2^(host bits) − 2, because the network and broadcast addresses cannot be assigned.
    2. Find the fewest host bits h where 2^h − 2 ≥ 60: h = 6 gives 62, while h = 5 gives only 30.
    3. Prefix = 32 − 6 = /26.

    Answer: /26, giving 62 usable addresses

  3. How long a file takes to transfer

    How long does a 700 MB file take to copy over a 1000 Mbps link? Use decimal units (1 GB = 1,000 MB) and ignore protocol overhead.

    Show the working
    1. Storage is sized in bytes, link speed in bits. Multiply megabytes by 8 to get megabits.
    2. 700 MB × 8 = 5600 Mb
    3. Time = 5600 Mb ÷ 1000 Mbps = 5.6 s ≈ 5.6 seconds
    4. Real transfers take longer: TCP/IP headers, retransmissions and disk speed all eat into the link rate.

    Answer: 5.6 seconds

  4. Converting an octet from binary to decimal

    Convert the binary octet 00100000 to decimal.

    Show the working
    1. Write the place values under each bit: 128 64 32 16 8 4 2 1.
    2. Add the place values where the bit is 1: 32
    3. = 32

    Answer: 32

  5. Subnet details for a /29 address

    A host is configured as 172.28.39.6/29. Find the subnet mask, network address, broadcast address, usable host range and number of usable hosts.

    Show the working
    1. /29 means 29 network bits, so the mask is 255.255.255.248.
    2. The interesting octet is the fourth, where the mask is 248. Block size = 256 − 248 = 8.
    3. The network address is the multiple of 8 at or below the address in that octet: 172.28.39.0.
    4. The broadcast is one below the next network: 172.28.39.7.
    5. Usable hosts run from 172.28.39.1 to 172.28.39.6: 2^3 − 2 = 6 addresses.

    Answer: Mask 255.255.255.248, network 172.28.39.0, broadcast 172.28.39.7, hosts 172.28.39.1–172.28.39.6 (6 usable)

  6. Choosing the smallest subnet that fits

    A department needs 12 usable IPv4 host addresses on one subnet. What is the longest prefix (smallest subnet) that fits, and how many usable addresses does it give?

    Show the working
    1. Usable hosts = 2^(host bits) − 2, because the network and broadcast addresses cannot be assigned.
    2. Find the fewest host bits h where 2^h − 2 ≥ 12: h = 4 gives 14, while h = 3 gives only 6.
    3. Prefix = 32 − 4 = /28.

    Answer: /28, giving 14 usable addresses

  7. How long a file takes to transfer

    How long does a 700 MB file take to copy over a 20 Mbps link? Use decimal units (1 GB = 1,000 MB) and ignore protocol overhead.

    Show the working
    1. Storage is sized in bytes, link speed in bits. Multiply megabytes by 8 to get megabits.
    2. 700 MB × 8 = 5600 Mb
    3. Time = 5600 Mb ÷ 20 Mbps = 280 s ≈ 4.67 minutes
    4. Real transfers take longer: TCP/IP headers, retransmissions and disk speed all eat into the link rate.

    Answer: 4.67 minutes

  8. Converting an octet from binary to decimal

    Convert the binary octet 00000011 to decimal.

    Show the working
    1. Write the place values under each bit: 128 64 32 16 8 4 2 1.
    2. Add the place values where the bit is 1: 2 + 1
    3. = 3

    Answer: 3

Where these calculations appear in design work

Choosing the smallest prefix that fits a site's hosts is the core step of a VLSM plan, and getting it wrong either wastes addresses or leaves a site no room to grow. Subnet boundaries decide which addresses can be summarised into a single route. Transfer-time estimates turn a vague complaint that backups are slow into a number you can compare with the link speed.

Building in room to grow

The smallest prefix that fits today is not always the right one to deploy. A site that needs 50 addresses now may need 80 later, and renumbering a live network is disruptive. Engineers usually choose the next size up for sites expected to grow, and leave gaps between allocations so neighbouring subnets can expand without breaking summarisation.

Common questions

How do I check a VLSM plan quickly?

Write each subnet's network and broadcast address in order. Every network address must be higher than the broadcast address before it; if any pair overlaps, the plan is wrong.

Why are link speeds and file sizes in different units?

Links are measured in bits per second and storage in bytes, with eight bits in a byte. Multiply megabytes by eight before dividing by the link speed.

More Network Engineer guides

The same guide for related positions