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.
Bu deste hakkında
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.
Bu destedeki kartlar
Kart 1
Soru
In IPv4 CIDR notation, what does the
/27specify?Cevap
The first 27 bits are the network prefix. The remaining
32 − 27 = 5bits identify positions within that address block.Kart 2
Soru
Which dotted-decimal IPv4 subnet mask matches
/27?Cevap
255.255.255.224. The final octet is11100000in binary: three prefix bits after the first 24.Kart 3
Soru
How many total IPv4 addresses are in a
/29block, before excluding any addresses?Cevap
8 addresses. There are
32 − 29 = 3host bits, so the block has2^3 = 8positions.Kart 4
Soru
To find an IPv4 network address from an address and prefix, what values must the host bits have?
Cevap
All host bits must be zero. Keep the prefix bits unchanged; clearing the remaining bits gives the block’s first address.
Kart 5
Soru
For an ordinary IPv4 LAN with a
/24–/30prefix, which two addresses are excluded from the traditional usable host count?Cevap
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.Kart 6
Soru
Which dotted-decimal IPv4 subnet mask matches
/23?Cevap
255.255.254.0. Seven prefix bits in the third octet give11111110; the fourth octet contains only host bits.Kart 7
Soru
What is the network address of
198.51.100.77/27?Cevap
198.51.100.64. The block has 32 addresses;77lies between the boundaries64and96.Kart 8
Soru
What is the subnet broadcast address for an ordinary IPv4 LAN containing
203.0.113.149/28?Cevap
203.0.113.159. This block starts at.144; the next starts at.160, so the broadcast is one address before.160.Kart 9
Soru
What is the first traditionally usable host address in the ordinary IPv4 LAN containing
192.0.2.174/26?Cevap
192.0.2.129. The network address is.128; the first host is the next address.Kart 10
Soru
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?
Cevap
/29. Its 8 total addresses leave 6 traditional host addresses. A/30leaves only 2.Kart 11
Soru
On an ordinary IPv4 LAN, is
198.51.100.127/27a network address, subnet broadcast address, or traditionally usable host address?Cevap
Subnet broadcast address. The block runs from
.96through.127;.127has every host bit set to one.Kart 12
Soru
A learner says an ordinary IPv4
/25LAN has 128 traditionally usable host addresses. What is the corrected usable count?Cevap
126 host addresses. The block has 128 total addresses; exclude its network and broadcast addresses:
128 − 2 = 126.Kart 13
Soru
What is the network address of
10.46.19.203/23?Cevap
10.46.18.0. Third-octet blocks start at even values; this block covers third octets18and19.Kart 14
Soru
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?
Cevap
/28. Its 16 total addresses leave 14 traditional host addresses; a/29leaves 6.Kart 15
Soru
What is the last traditionally usable host address in the ordinary IPv4 LAN containing
192.0.2.219/29?Cevap
192.0.2.222. The block is.216–.223;.223is broadcast, so.222is the last host.Kart 16
Soru
Do
203.0.113.126/26and203.0.113.129/26belong to the same numerical IPv4 subnet?Cevap
No. Their network addresses are
203.0.113.64and203.0.113.128. A shared first three octets does not settle a/26comparison.Kart 17
Soru
A learner gives
198.51.100.192as the broadcast for an ordinary IPv4 LAN containing198.51.100.173/27. What is the corrected broadcast?Cevap
198.51.100.191. This block starts at.160;.192starts the next block. Broadcast is the address immediately before that next boundary.Kart 18
Soru
What is the subnet broadcast address for an ordinary IPv4 LAN containing
172.22.35.9/23?Cevap
172.22.35.255. The block starts at172.22.34.0and ends just before172.22.36.0.Kart 19
Soru
Which IPv4 CIDR prefix length matches the subnet mask
255.255.255.240?Cevap
/28. The first three octets contribute 24 one bits;240is11110000, contributing four more.Kart 20
Soru
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?
Cevap
/27. It leaves 30 traditional host addresses. A/28leaves 14, which is one short.Kart 21
Soru
On an ordinary IPv4 LAN, is
10.61.13.0/23a network address, subnet broadcast address, or traditionally usable host address?Cevap
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.Kart 22
Soru
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?Cevap
172.24.71.254. The block’s broadcast is172.24.71.255; the last traditional host address is one before it.Kart 23
Soru
What is the network address of
192.168.47.201/22?Cevap
192.168.44.0. Third-octet boundaries are multiples of 4;47falls in the block covering44–47.Kart 24
Soru
On an ordinary IPv4 LAN, is
203.0.113.184/29a network address, subnet broadcast address, or traditionally usable host address?Cevap
Network address.
/29boundaries advance by 8 in the fourth octet, and184is a multiple of 8. This block ends at.191.Kart 25
Soru
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?
Cevap
/26. The demand is30 + 1 = 31;/27leaves only 30 traditional host addresses, while/26leaves 62.Kart 26
Soru
What is the first traditionally usable host address in the ordinary IPv4 LAN containing
10.84.51.230/23?Cevap
10.84.50.1. The network is10.84.50.0; this block spans third octets50and51.Kart 27
Soru
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?Cevap
10.73.28.0. The mask is255.255.252.0, so boundaries advance by 4 in the third octet; this block covers28–31.Kart 28
Soru
An IPv4 CIDR block contains exactly 128 total addresses, before exclusions. What is its prefix length?
Cevap
/25. Since128 = 2^7, there are 7 host bits;32 − 7 = 25prefix bits.Kart 29
Soru
What is the subnet broadcast address for an ordinary IPv4 LAN containing
192.0.2.46/30?Cevap
192.0.2.47. The 4-address block starts at.44; the next starts at.48.Kart 30
Soru
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?
Cevap
/26. Its 64 total addresses leave 62 traditional host addresses;/27leaves 30.Kart 31
Soru
Do
10.95.24.250/23and10.95.25.7/23belong to the same numerical IPv4 subnet?Cevap
Yes. Both have network address
10.95.24.0; a/23here covers third octets24and25.Kart 32
Soru
What is the last traditionally usable host address in the ordinary IPv4 LAN containing
172.26.86.9/22?Cevap
172.26.87.254. The block runs from172.26.84.0through broadcast172.26.87.255.Kart 33
Soru
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?Cevap
2 interface addresses. The block has 4 total addresses. Exclude network and broadcast once; a gateway, if present, occupies one of the remaining two.
Kart 34
Soru
On an ordinary IPv4 LAN, is
192.168.100.255/23a network address, subnet broadcast address, or traditionally usable host address?Cevap
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.Kart 35
Soru
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?
Cevap
/25. It leaves 126 traditional host addresses. A/26leaves 62, so it is one short.Kart 36
Soru
What is the network address of
203.0.113.233/25?Cevap
203.0.113.128. The two fourth-octet blocks start at0and128;233is in the second.Kart 37
Soru
A learner rounds
198.51.100.107/28up to network address198.51.100.112. What is the corrected network address?Cevap
198.51.100.96. Network boundaries advance by 16;107is inside.96–.111. Choose the containing block, not the next boundary.Kart 38
Soru
On an IPv4 point-to-point link using RFC 3021, how many endpoint addresses does
10.106.8.40/31provide?Cevap
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.Kart 39
Soru
An IPv4 address lies in a subnet’s traditional usable host range. Does that alone prove it is free to assign?
Cevap
No. It may already be assigned or reserved by local policy. Arithmetic gives a candidate range; address management determines availability.
Kart 40
Soru
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?
Cevap
/24. It has 256 total addresses and 254 traditional host addresses;/25leaves 126.Kart 41
Soru
What is the subnet broadcast address for an ordinary IPv4 LAN containing
10.117.58.177/22?Cevap
10.117.59.255. The block starts at10.117.56.0, covers third octets56–59, and ends just before10.117.60.0.Kart 42
Soru
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?Cevap
30 interface addresses. The gateway was already included. Count it once; the stated demand fits the
/27block’s 30 traditional host addresses.Kart 43
Soru
What destination address does the IPv4 host route
10.128.9.17/32match?Cevap
10.128.9.17only. All 32 bits are specified, so the route matches one address. It is not an ordinary multi-host LAN block.Kart 44
Soru
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?
Cevap
/23. The demand is 255;/24leaves only 254 traditional host addresses./23leaves 510.Kart 45
Soru
On an ordinary IPv4 LAN, is
172.28.92.0/22a network address, subnet broadcast address, or traditionally usable host address?Cevap
Network address. The third octet
92is a multiple of 4 and all host bits are zero. The block covers third octets92–95.Kart 46
Soru
What is the first traditionally usable host address in the ordinary IPv4 LAN containing
198.51.100.154/30?Cevap
198.51.100.153. The block starts at.152; its traditional hosts are.153and.154, with broadcast.155.Kart 47
Soru
Two IPv4 interfaces have addresses in the same numerical subnet. Does that alone prove that they can communicate?
Cevap
No. Matching subnet arithmetic does not establish a working link, compatible configuration, or permitted traffic. Those require checks on the actual network.
Kart 48
Soru
On an ordinary IPv4 LAN, is
192.0.2.193/26a network address, subnet broadcast address, or traditionally usable host address?Cevap
Traditionally usable host address. The block starts at
.192and broadcasts at.255;.193is its first traditional host address.
48 kart
IPv4 Subnetting Practice Flashcards: Boundaries & Host Ranges
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