IPv4 Subnetting Practice Flashcards: Boundaries & Host Ranges
48 original IPv4 subnetting flashcards on network and broadcast boundaries, usable host endpoints, address roles, capacity, and common mistakes, with brief worked reasoning.
Об этой колоде
Practice IPv4 subnetting with 48 original English flashcards. Each card asks for one result: a CIDR subnet mask or count, a network or broadcast address, a usable host-range endpoint, an address role, a numerical same-subnet decision, a capacity prefix, or one corrected value in a subnetting mistake. Two inverse cards retrieve a unique prefix from a mask or a total address count.
The exercises mainly use /23–/30, with a few /22 cross-octet cases. Ordinary LAN questions use the traditional network-and-broadcast exclusions. Every capacity question states whether its interface demand already includes the gateway and excludes growth and extra reservations. A /31 card applies only to an RFC 3021 point-to-point link; a /32 card concerns a single-address host route. Other cards separate numerical membership from connectivity and usable range from assignment availability.
The first six cards establish prefix, mask, and counting anchors. The remaining sequence mixes boundary calculations, roles, capacity choices, and repairs, moving into third-octet boundaries and later exceptions. Related variants are separated by at least three cards that ask for other facts; the review scheduler handles long-term spacing.
This is a compact recall and error-repair companion. It excludes non-unique reverse questions such as guessing an input address from a network boundary, every prefix and reverse permutation, classful shortcuts, VLSM design projects, IPv6, and device configuration. Recalling fixed examples does not establish skill on unfamiliar calculations or a working network. Learn the method with IPv4 subnetting practice, then change the addresses and solve fresh exercises. The networking flashcard study guide gives broader study context.
All exercises use artificial private or documentation addresses. Facts were checked against RFC 4632, RFC 3021, RFC 5737, RFC 1918, and Cisco’s addressing reference. The prompts, explanations, sequence, metadata, and abstract cover were independently created with AI assistance and reviewed; no source wording, exercises, diagrams, or exam material were copied. Original elements are released under CC0 1.0 to the extent applicable rights exist. This general topic deck claims no certification alignment or endorsement.
Карточки в этой колоде
Карточка 1
Вопрос
In IPv4 CIDR notation, what does the
/27specify?Ответ
The first 27 bits are the network prefix. The remaining
32 − 27 = 5bits identify positions within that address block.Карточка 2
Вопрос
Which dotted-decimal IPv4 subnet mask matches
/27?Ответ
255.255.255.224. The final octet is11100000in binary: three prefix bits after the first 24.Карточка 3
Вопрос
How many total IPv4 addresses are in a
/29block, before excluding any addresses?Ответ
8 addresses. There are
32 − 29 = 3host bits, so the block has2^3 = 8positions.Карточка 4
Вопрос
To find an IPv4 network address from an address and prefix, what values must the host bits have?
Ответ
All host bits must be zero. Keep the prefix bits unchanged; clearing the remaining bits gives the block’s first address.
Карточка 5
Вопрос
For an ordinary IPv4 LAN with a
/24–/30prefix, which two addresses are excluded from the traditional usable host count?Ответ
The network address and the subnet broadcast address. Traditional usable capacity is
2^(32 − prefix length) − 2; this rule is not the RFC 3021/31point-to-point rule.Карточка 6
Вопрос
Which dotted-decimal IPv4 subnet mask matches
/23?Ответ
255.255.254.0. Seven prefix bits in the third octet give11111110; the fourth octet contains only host bits.Карточка 7
Вопрос
What is the network address of
198.51.100.77/27?Ответ
198.51.100.64. The block has 32 addresses;77lies between the boundaries64and96.Карточка 8
Вопрос
What is the subnet broadcast address for an ordinary IPv4 LAN containing
203.0.113.149/28?Ответ
203.0.113.159. This block starts at.144; the next starts at.160, so the broadcast is one address before.160.Карточка 9
Вопрос
What is the first traditionally usable host address in the ordinary IPv4 LAN containing
192.0.2.174/26?Ответ
192.0.2.129. The network address is.128; the first host is the next address.Карточка 10
Вопрос
An ordinary IPv4 LAN needs 6 interface addresses in total, including its gateway. With no growth allowance or extra reservations, which prefix gives the smallest address block that fits?
Ответ
/29. Its 8 total addresses leave 6 traditional host addresses. A/30leaves only 2.Карточка 11
Вопрос
On an ordinary IPv4 LAN, is
198.51.100.127/27a network address, subnet broadcast address, or traditionally usable host address?Ответ
Subnet broadcast address. The block runs from
.96through.127;.127has every host bit set to one.Карточка 12
Вопрос
A learner says an ordinary IPv4
/25LAN has 128 traditionally usable host addresses. What is the corrected usable count?Ответ
126 host addresses. The block has 128 total addresses; exclude its network and broadcast addresses:
128 − 2 = 126.Карточка 13
Вопрос
What is the network address of
10.46.19.203/23?Ответ
10.46.18.0. Third-octet blocks start at even values; this block covers third octets18and19.Карточка 14
Вопрос
An ordinary IPv4 LAN needs 14 interface addresses total, gateway included. Ignore growth and extra reservations. Which prefix gives the smallest address block that fits?
Ответ
/28. Its 16 total addresses leave 14 traditional host addresses; a/29leaves 6.Карточка 15
Вопрос
What is the last traditionally usable host address in the ordinary IPv4 LAN containing
192.0.2.219/29?Ответ
192.0.2.222. The block is.216–.223;.223is broadcast, so.222is the last host.Карточка 16
Вопрос
Do
203.0.113.126/26and203.0.113.129/26belong to the same numerical IPv4 subnet?Ответ
No. Their network addresses are
203.0.113.64and203.0.113.128. A shared first three octets does not settle a/26comparison.Карточка 17
Вопрос
A learner gives
198.51.100.192as the broadcast for an ordinary IPv4 LAN containing198.51.100.173/27. What is the corrected broadcast?Ответ
198.51.100.191. This block starts at.160;.192starts the next block. Broadcast is the address immediately before that next boundary.Карточка 18
Вопрос
What is the subnet broadcast address for an ordinary IPv4 LAN containing
172.22.35.9/23?Ответ
172.22.35.255. The block starts at172.22.34.0and ends just before172.22.36.0.Карточка 19
Вопрос
Which IPv4 CIDR prefix length matches the subnet mask
255.255.255.240?Ответ
/28. The first three octets contribute 24 one bits;240is11110000, contributing four more.Карточка 20
Вопрос
An ordinary IPv4 LAN needs 15 interface addresses in total, gateway included. Exclude growth and extra reservations. Which prefix gives the smallest address block that fits?
Ответ
/27. It leaves 30 traditional host addresses. A/28leaves 14, which is one short.Карточка 21
Вопрос
On an ordinary IPv4 LAN, is
10.61.13.0/23a network address, subnet broadcast address, or traditionally usable host address?Ответ
Traditionally usable host address. This block is
10.61.12.0–10.61.13.255;.13.0is inside it. A final.0does not by itself identify the network address.Карточка 22
Вопрос
A learner gives
172.24.71.255as the last usable host in an ordinary LAN containing172.24.70.18/23. What is the corrected last host address?Ответ
172.24.71.254. The block’s broadcast is172.24.71.255; the last traditional host address is one before it.Карточка 23
Вопрос
What is the network address of
192.168.47.201/22?Ответ
192.168.44.0. Third-octet boundaries are multiples of 4;47falls in the block covering44–47.Карточка 24
Вопрос
On an ordinary IPv4 LAN, is
203.0.113.184/29a network address, subnet broadcast address, or traditionally usable host address?Ответ
Network address.
/29boundaries advance by 8 in the fourth octet, and184is a multiple of 8. This block ends at.191.Карточка 25
Вопрос
An ordinary IPv4 LAN needs addresses for 30 endpoint interfaces plus 1 gateway interface. With no growth allowance or extra reservations, which prefix gives the smallest address block that fits?
Ответ
/26. The demand is30 + 1 = 31;/27leaves only 30 traditional host addresses, while/26leaves 62.Карточка 26
Вопрос
What is the first traditionally usable host address in the ordinary IPv4 LAN containing
10.84.51.230/23?Ответ
10.84.50.1. The network is10.84.50.0; this block spans third octets50and51.Карточка 27
Вопрос
A learner treats
/22as fourth-octet blocks and reports10.73.31.0as the network for10.73.31.88/22. What is the corrected network address?Ответ
10.73.28.0. The mask is255.255.252.0, so boundaries advance by 4 in the third octet; this block covers28–31.Карточка 28
Вопрос
An IPv4 CIDR block contains exactly 128 total addresses, before exclusions. What is its prefix length?
Ответ
/25. Since128 = 2^7, there are 7 host bits;32 − 7 = 25prefix bits.Карточка 29
Вопрос
What is the subnet broadcast address for an ordinary IPv4 LAN containing
192.0.2.46/30?Ответ
192.0.2.47. The 4-address block starts at.44; the next starts at.48.Карточка 30
Вопрос
An ordinary IPv4 LAN needs 62 interface addresses in total, including the gateway. Ignore growth and extra reservations. Which prefix gives the smallest address block that fits?
Ответ
/26. Its 64 total addresses leave 62 traditional host addresses;/27leaves 30.Карточка 31
Вопрос
Do
10.95.24.250/23and10.95.25.7/23belong to the same numerical IPv4 subnet?Ответ
Yes. Both have network address
10.95.24.0; a/23here covers third octets24and25.Карточка 32
Вопрос
What is the last traditionally usable host address in the ordinary IPv4 LAN containing
172.26.86.9/22?Ответ
172.26.87.254. The block runs from172.26.84.0through broadcast172.26.87.255.Карточка 33
Вопрос
A learner subtracts network, broadcast, and a separate gateway reservation from a
/30LAN and reports 1 usable interface address. What is the traditional usable capacity, counting a gateway as an interface?Ответ
2 interface addresses. The block has 4 total addresses. Exclude network and broadcast once; a gateway, if present, occupies one of the remaining two.
Карточка 34
Вопрос
On an ordinary IPv4 LAN, is
192.168.100.255/23a network address, subnet broadcast address, or traditionally usable host address?Ответ
Traditionally usable host address. The block is
192.168.100.0–192.168.101.255; its broadcast is.101.255. A final.255alone does not identify broadcast.Карточка 35
Вопрос
An ordinary IPv4 LAN needs 63 interface addresses in total, gateway included. Exclude growth and extra reservations. Which prefix gives the smallest address block that fits?
Ответ
/25. It leaves 126 traditional host addresses. A/26leaves 62, so it is one short.Карточка 36
Вопрос
What is the network address of
203.0.113.233/25?Ответ
203.0.113.128. The two fourth-octet blocks start at0and128;233is in the second.Карточка 37
Вопрос
A learner rounds
198.51.100.107/28up to network address198.51.100.112. What is the corrected network address?Ответ
198.51.100.96. Network boundaries advance by 16;107is inside.96–.111. Choose the containing block, not the next boundary.Карточка 38
Вопрос
On an IPv4 point-to-point link using RFC 3021, how many endpoint addresses does
10.106.8.40/31provide?Ответ
2 endpoint addresses. They are
10.106.8.40and10.106.8.41. In this point-to-point context, both are endpoint addresses; do not subtract network and directed broadcast.Карточка 39
Вопрос
An IPv4 address lies in a subnet’s traditional usable host range. Does that alone prove it is free to assign?
Ответ
No. It may already be assigned or reserved by local policy. Arithmetic gives a candidate range; address management determines availability.
Карточка 40
Вопрос
An ordinary IPv4 LAN needs 254 interface addresses in total, including the gateway. With no growth allowance or extra reservations, which prefix gives the smallest address block that fits?
Ответ
/24. It has 256 total addresses and 254 traditional host addresses;/25leaves 126.Карточка 41
Вопрос
What is the subnet broadcast address for an ordinary IPv4 LAN containing
10.117.58.177/22?Ответ
10.117.59.255. The block starts at10.117.56.0, covers third octets56–59, and ends just before10.117.60.0.Карточка 42
Вопрос
An ordinary IPv4
/27LAN has a requirement of 30 interface addresses total, already including its gateway. A learner increases the requirement to 31. What is the corrected required interface count?Ответ
30 interface addresses. The gateway was already included. Count it once; the stated demand fits the
/27block’s 30 traditional host addresses.Карточка 43
Вопрос
What destination address does the IPv4 host route
10.128.9.17/32match?Ответ
10.128.9.17only. All 32 bits are specified, so the route matches one address. It is not an ordinary multi-host LAN block.Карточка 44
Вопрос
An ordinary IPv4 LAN needs addresses for 254 endpoint interfaces plus 1 gateway interface. Ignore growth and extra reservations. Which prefix gives the smallest address block that fits?
Ответ
/23. The demand is 255;/24leaves only 254 traditional host addresses./23leaves 510.Карточка 45
Вопрос
On an ordinary IPv4 LAN, is
172.28.92.0/22a network address, subnet broadcast address, or traditionally usable host address?Ответ
Network address. The third octet
92is a multiple of 4 and all host bits are zero. The block covers third octets92–95.Карточка 46
Вопрос
What is the first traditionally usable host address in the ordinary IPv4 LAN containing
198.51.100.154/30?Ответ
198.51.100.153. The block starts at.152; its traditional hosts are.153and.154, with broadcast.155.Карточка 47
Вопрос
Two IPv4 interfaces have addresses in the same numerical subnet. Does that alone prove that they can communicate?
Ответ
No. Matching subnet arithmetic does not establish a working link, compatible configuration, or permitted traffic. Those require checks on the actual network.
Карточка 48
Вопрос
On an ordinary IPv4 LAN, is
192.0.2.193/26a network address, subnet broadcast address, or traditionally usable host address?Ответ
Traditionally usable host address. The block starts at
.192and broadcasts at.255;.193is its first traditional host address.
48 карточек
IPv4 Subnetting Practice Flashcards: Boundaries & Host Ranges
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