All practice questions

Question 1Question →

A network administrator is configuring a point-to-point interconnect between two core routers. The interface on Router A is assigned the IPv4 address $10.200.15.14/30$ . What is the directed broadcast address for this subnet?

Show answer & explanation

Answer: 10.200.15.15

Answer

10.200.15.15

For the

/30

network containing the host IP address

10.200.15.14

, the subnet spans from

10.200.15.12

to

10.200.15.15

. The address

10.200.15.12

is the network ID,

10.200.15.13

and

10.200.15.14

are usable host addresses, and

10.200.15.15

is the directed broadcast address.

Step-by-Step Solution

1

Determine the subnet mask and subnet block size from the CIDR notation

A

/30

prefix corresponds to a subnet mask of

255.255.255.252

, which yields a block size of

4

addresses (

256 - 252 = 4

) in the fourth octet.

Understanding the block size identifies subnet boundaries along increments of 4.

2

Calculate the network address for the host IP

For the host IP

10.200.15.14

, dividing

14

by the block size

4

gives

3

with a remainder of

2

. The network ID for this block is

10.200.15.12

.

The network address represents the lowest numerical boundary of the subnet.

3

Determine the directed broadcast address

The broadcast address is the highest address in the subnet block, which is

10.200.15.12 + 3 = 10.200.15.15

.

The last IP address in any IPv4 subnet block is reserved as the directed broadcast address.

Key Concept

IPv4 Subnetting and Broadcast Address Calculation

Question 2Question →

During a security compliance audit of network documentation, a systems engineer needs to standardize an abridged IPv6 address into its complete 128-bit explicit form.

What is the fully expanded, uncompressed representation of the IPv6 address 2001:db8:40::ae0:12?

Show answer & explanation

Answer: 2001:0db8:0040:0000:0000:0000:0ae0:0012; 2001:0DB8:0040:0000:0000:0000:0AE0:0012

Answer

2001:0db8:0040:0000:0000:0000:0ae0:0012

Expanding a compressed IPv6 address requires expanding each block to four hexadecimal digits by restoring omitted leading zeros and replacing the double colon (::) with the required number of 0000 blocks to reach a total of 8 blocks. The address 2001:db8:40::ae0:12 contains 5 explicit blocks (2001, 0db8, 0040, 0ae0, 0012), meaning the double colon represents 3 contiguous zero blocks (0000:0000:0000), producing 2001:0db8:0040:0000:0000:0000:0ae0:0012.

Step-by-Step Solution

1

Count the explicit 16-bit fields present in the compressed address.

There are 5 explicit fields specified: '2001', 'db8', '40', 'ae0', and '12'.

An IPv6 address always consists of 8 total 16-bit fields.

2

Determine the number of missing zero fields represented by the double colon (::).

8 total fields - 5 explicit fields = 3 missing consecutive zero fields (0000:0000:0000).

The double colon symbol expands to fill all omitted 16-bit zero blocks so the address reaches exactly 8 fields.

3

Pad each explicit field with leading zeros to complete 4 hexadecimal digits per field.

'2001' remains '2001', 'db8' becomes '0db8', '40' becomes '0040', 'ae0' becomes '0ae0', and '12' becomes '0012'.

RFC 5952 zero-suppression removes leading zeros within a block, which must be restored during full expansion.

4

Assemble all 8 expanded fields separated by colons.

2001:0db8:0040:0000:0000:0000:0ae0:0012

Combining the padded explicit blocks and the expanded zero blocks yields the full 39-character 128-bit uncompressed IPv6 address.

Key Concept

IPv6 Address Uncompression and Structure Standards

Question 3Question →

A network engineer is configuring a core router interface and needs to enter an IPv6 address adhering strictly to RFC 5952 standard text representation rules. The uncompressed IPv6 address is 2001:0db8:0000:0000:000a:0000:0000:0001. According to RFC 5952 zero-compression and leading-zero suppression standards (including the rule for handling equal-length contiguous zero blocks), what is the correctly compressed IPv6 address string (excluding any prefix length)?

Show answer & explanation

Answer: 2001:db8::a:0:0:1

Answer

2001:db8::a:0:0:1

Applying standard RFC 5952 rules requires three main operations: suppressing all leading zeros in each 16-bit block (0db8 -> db8, 000a -> a, 0001 -> 1), identifying contiguous zero blocks (fields 3–4 and fields 6–7, both of length 2), and applying the tie-breaking rule. Under RFC 5952 Section 4.2.3, when two zero sequences are of equal length, the first sequence must be compressed with '::'. This results in the final canonical address 2001:db8::a:0:0:1.

Step-by-Step Solution

1

Suppress leading zeros in all 16-bit hex fields

2001:db8:0:0:a:0:0:1

RFC 5952 Section 4.2.1 mandates that leading zeros in individual 16-bit fields must be omitted.

2

Identify contiguous sequences of zero blocks

Sequence 1 is '0:0' at fields 3–4 (length 2). Sequence 2 is '0:0' at fields 6–7 (length 2).

Double-colon '::' compression applies to contiguous 16-bit zero fields.

3

Apply the tie-breaking rule for equal-length zero sequences

Compress Sequence 1 ('0:0') to '::' while keeping Sequence 2 expanded as '0:0'.

RFC 5952 Section 4.2.3 requires that when multiple zero sequences of equal maximum length exist, the first sequence must be compressed.

4

Assemble the final compressed RFC 5952 address using lowercase characters

2001:db8::a:0:0:1

Combining zero suppression, first-sequence zero compression, and lowercase formatting yields the canonical RFC 5952 representation.

Key Concept

RFC 5952 IPv6 Address Compression Rules

Question 4Question →

Which autoconfiguration mechanism allows an IPv6 host to dynamically generate its own unicast IP address using network prefix information received in Router Advertisement (RA) messages without requiring a stateful DHCP server?

Show answer & explanation

Answer: SLAAC; Stateless Address Autoconfiguration; stateless address autoconfiguration

Answer

SLAAC (Stateless Address Autoconfiguration)

SLAAC (Stateless Address Autoconfiguration) allows an IPv6 client to automatically generate its own globally unique or link-local address by combining the IPv6 prefix provided by a local router via ICMPv6 Router Advertisement (RA) messages with a locally derived 64-bit interface identifier (such as EUI-64 or random privacy address).

Step-by-Step Solution

1

Identify the IPv6 address assignment requirement

The scenario describes dynamic host address generation using Router Advertisement (RA) messages without a stateful DHCP server.

IPv6 introduces a native mechanism for hosts to autoconfigure network layer parameters directly from local router advertisements.

2

Match the protocol mechanism to its standardized term

Stateless Address Autoconfiguration (SLAAC) is the protocol mechanism that combines the advertised network prefix with an interface identifier.

SLAAC requires no state management on a central server, distinguishing it from stateful DHCPv6.

Key Concept

Stateless Address Autoconfiguration (SLAAC)

Estimated Time:45s

Question 5Question →

A systems engineer is provisioning a dedicated storage area network management VLAN using the IPv4 subnet block $192.168.100.128/26$ . What is the broadcast IP address for this network segment?

Show answer & explanation

Answer: 192.168.100.191

Answer

192.168.100.191

For the network block

192.168.100.128/26

, the subnet mask is

255.255.255.192

, yielding subnets spaced in increments of 64 in the fourth octet (

0

,

64

,

128

,

192

). The subnet starting at

192.168.100.128

spans up to

192.168.100.191

, making

192.168.100.191

the broadcast address.

Step-by-Step Solution

1

Determine the prefix length and host bits

A

/26

prefix leaves

32 - 26 = 6

bits for host addressing.

Calculating host bits identifies the size of each subnet block.

2

Calculate the block size in the fourth octet

The block size is

2^6 = 64

.

Each subnet in this block increments by 64 in the last octet.

3

Find the next network address and determine the broadcast address

The current network address is

192.168.100.128

. The next subnet starts at

128 + 64 = 192.168.100.192

. Subtracting 1 yields the broadcast address

192.168.100.191

.

The broadcast address is always the last IP address in the subnet block immediately preceding the next network boundary.

Key Concept

IPv4 Broadcast Address Calculation

Estimated Time:1m 30s

Question 6Question →

A network operations team is subnetting the enterprise block $10.45.128.0/18$ to create equal-sized subnets capable of supporting at least $500$ usable host IP addresses per subnet. If subnets are allocated sequentially starting from the lowest address in the block, what is the network address (Network ID) of the fifth allocated subnet?

Show answer & explanation

Answer: 10.45.136.0; 10.45.136.0/23

Answer

10.45.136.0

To accommodate at least

500

usable hosts per subnet,

9

host bits are required because

2^9 - 2 = 510

usable addresses. This results in a prefix length of

/23

(

32 - 9 = 23

), which increments the third octet by

2

per subnet (

256 - 254 = 2

). Starting sequentially from

10.45.128.0

, the subnets are: 1st:

10.45.128.0

, 2nd:

10.45.130.0

, 3rd:

10.45.132.0

, 4th:

10.45.134.0

, and 5th:

10.45.136.0

.

Step-by-Step Solution

1

Determine the number of host bits required for 500 usable host IP addresses.

9 host bits are required.

The formula for usable hosts is

2^h - 2 \ge 500

. For

h=8

,

2^8 - 2 = 254

hosts (insufficient). For

h=9

,

2^9 - 2 = 510

hosts (sufficient).

2

Calculate the subnet prefix length and block size.

Prefix length is

/23

and the block size in the third octet is 2.

Subtracting

9

host bits from

32

total bits gives

32 - 9 = 23

(

/23

). The block size per subnet is

2^{24-23} = 2

in the third octet.

3

Calculate the network ID of the fifth subnet allocated sequentially from

10.45.128.0

.

The network address is 10.45.136.0.

The first subnet is

10.45.128.0

. Each subsequent subnet increments the third octet by 2. The 5th subnet corresponds to 4 increments (

4 \times 2 = 8

):

128 + 8 = 136

, yielding

10.45.136.0/23

.

Key Concept

Host bit calculation and sequential IPv4 subnet allocation

Estimated Time:2m 0s

Question 7Question →

A security analyst is investigating a potential incident involving a server assigned the IP address $10.80.35.138/26$ . What is the broadcast IPv4 address of the subnet to which this server belongs?

Show answer & explanation

Answer: 10.80.35.191

Answer

10.80.35.191

The CIDR prefix

/26

indicates a subnet mask of

255.255.255.192

, creating subnet increments of

64

addresses in the fourth octet. The subnets start at

.0

,

.64

,

.128

, and

.192

. The IP address

10.80.35.138

lies within the

.128

subnet block (

10.80.35.128/26

). The broadcast address is the final IP in this block, which is

10.80.35.191

.

Step-by-Step Solution

1

Determine the subnet mask and block size from the CIDR prefix length.

A

/26

prefix corresponds to a subnet mask of

255.255.255.192

. The block size in the fourth octet is

256 - 192 = 64

.

The CIDR notation

/26

uses 26 network bits, leaving 6 host bits (

2^6 = 64

addresses per subnet).

2

Identify the network boundaries for the fourth octet.

Subnet boundaries increment by

64

:

0

,

64

,

128

, and

192

.

Each subnet span covers 64 continuous IP addresses.

3

Locate the given IP address within its subnet range.

The IP address

10.80.35.138

falls within the range

10.80.35.128

through

10.80.35.191

.

Since

128 \le 138 < 192

, the network address for this subnet is

10.80.35.128

.

4

Determine the broadcast address for the subnet.

The broadcast address is

10.80.35.191

.

The broadcast address is the last address in the subnet range, immediately preceding the next subnet network ID (

10.80.35.192 - 1 = 10.80.35.191

).

Key Concept

IPv4 Subnetting and Broadcast Address Calculation

Question 8Question →

A network technician is setting up a new network segment using the IP block $10.0.4.0/24$ . Following standard configuration guidelines, the default gateway will be assigned the last usable host IP address in the subnet. What is the last usable IPv4 host address in the $10.0.4.0/24$ network?

Show answer & explanation

Answer: 10.0.4.254

Answer

The last usable IPv4 host address in the

10.0.4.0/24

network is

10.0.4.254

.

For a

/24

IPv4 subnet, the address space covers 256 total IP addresses (

10.0.4.0

to

10.0.4.255

). The first address (

10.0.4.0

) serves as the network ID and the last address (

10.0.4.255

) serves as the broadcast address. Thus, the assignable host range is

10.0.4.1

through

10.0.4.254

, making

10.0.4.254

the last usable host address.

Step-by-Step Solution

1

Identify the network address and broadcast address for

10.0.4.0/24

.

Network address is

10.0.4.0

and broadcast address is

10.0.4.255

.

A

/24

prefix provides 256 total IP addresses in the final octet (range 0–255).

2

Determine the range of usable host IP addresses.

The usable host address range spans from

10.0.4.1

to

10.0.4.254

.

The network address (

10.0.4.0

) and the broadcast address (

10.0.4.255

) cannot be assigned to hosts.

3

Select the last assignable host address.

10.0.4.254

It is the highest usable IP address immediately preceding the broadcast address.

Key Concept

IPv4 Usable Host Range and Subnet Boundary Calculation

Question 9Question →

During a network security audit, an incident response analyst discovers a compromised host with the IPv4 address $172.16.89.138/22$ . What is the broadcast address for the subnet to which this host belongs?

Show answer & explanation

Answer: 172.16.91.255

Answer

The broadcast address for the subnet is 172.16.91.255.

For an IP address of

172.16.89.138/22

, the subnet mask is 255.255.252.0. The third octet increments in multiples of 4 (

256 - 252 = 4

). The network containing the third octet value of 89 starts at 88 (

172.16.88.0

). The subnet spans up to

172.16.91.255

, making

172.16.91.255

the broadcast address where all 10 host bits are set to 1.

Step-by-Step Solution

1

Determine the subnet mask and block size from the CIDR prefix /22.

A /22 prefix corresponds to a subnet mask of 255.255.252.0. The block size in the third octet is

256 - 252 = 4

.

The prefix /22 leaves 10 host bits across the third and fourth octets, giving

2^{10} = 1024

total IP addresses (

4

blocks of 256).

2

Identify the network address for the host IP address 172.16.89.138.

The third octet boundaries increment by 4 (

0, 4, 8, \dots, 84, 88, 92

). Since 89 lies between 88 and 91, the network address is 172.16.88.0.

The network ID uses the lowest address in the block formed by the multiple of the block size (

88

).

3

Calculate the broadcast address of the subnet.

The broadcast address is 172.16.91.255.

The broadcast address has all 10 host bits set to 1, which corresponds to the last IP address in the

172.16.88.0 - 172.16.91.255

range.

Key Concept

IPv4 Non-Octet Boundary Subnetting and Broadcast Address Calculation

Estimated Time:2m 0s

Question 10Question →

A network infrastructure engineer is dividing the IPv4 address space $172.28.144.0/20$ using Variable Length Subnet Masking (VLSM). The subnets must be allocated contiguously starting from the lowest available network address in descending order of host requirements without wasting address blocks between subnets:

1. Subnet A: Requires $450$ usable host IP addresses.
2. Subnet B: Requires $190$ usable host IP addresses.
3. Subnet C: Requires $60$ usable host IP addresses.

What is the broadcast IP address of Subnet C?

Show answer & explanation

Answer: 172.28.147.63

Answer

172.28.147.63

Subnet C begins at

172.28.147.0

after allocating a

/23

(

512

addresses) for Subnet A and a

/24

(

256

addresses) for Subnet B. A

/26

prefix accommodates

60

hosts (

62

usable, block size of

64

), making the address range

172.28.147.0

through

172.28.147.63

. The last address in this subnet block (

172.28.147.63

) serves as the broadcast IP address.

Step-by-Step Solution

1

Determine prefix length and block size for Subnet A (450 usable hosts)

Subnet A requires

9

host bits (

2^9 - 2 = 510

usable IPs), giving a CIDR mask of

/23

(

32 - 9

) and a block size of

512

addresses (

2^9

). Allocated range:

172.28.144.0

to

172.28.145.255

.

Subnets must be allocated in descending order of required hosts.

2

Determine prefix length, block size, and range for Subnet B (190 usable hosts)

Next available start IP is

172.28.146.0

. Subnet B requires

8

host bits (

2^8 - 2 = 254

usable IPs), giving a CIDR mask of

/24

(

32 - 8

) and a block size of

256

addresses (

2^8

). Allocated range:

172.28.146.0

to

172.28.146.255

.

Subnet B must immediately follow Subnet A contiguously.

3

Determine prefix length, block size, and broadcast address for Subnet C (60 usable hosts)

Next available start IP is

172.28.147.0

. Subnet C requires

6

host bits (

2^6 - 2 = 62

usable IPs), giving a CIDR mask of

/26

(

32 - 6

) and a block size of

64

addresses (

2^6

). Network ID:

172.28.147.0/26

. Broadcast address:

172.28.147.63

.

The broadcast address is the last IP in the

/26

block (

172.28.147.0 + 63

).

Key Concept

Variable Length Subnet Masking (VLSM) contiguous allocation

Question 11Question →

A network technician is configuring an IPv4 interface on a server assigned a network prefix of $/28$ . What is the corresponding subnet mask in dotted-decimal notation?

Show answer & explanation

Answer: 255.255.255.240

Answer

The subnet mask in dotted-decimal notation for a

/28

prefix is

255.255.255.240

.

A

/28

CIDR prefix specifies that

28

bits are allocated for the network portion of the address. The first three octets account for

24

bits (

255.255.255

). The remaining

4

network bits in the fourth octet (

11110000_2

) sum to

128 + 64 + 32 + 16 = 240

, giving the dotted-decimal subnet mask

255.255.255.240

.

Step-by-Step Solution

1

Determine the distribution of network bits across the four 8-bit octets.

A

/28

prefix allocates

28

network bits out of

32

total bits, leaving

4

bits for host addressing (

32 - 28 = 4

).

An IPv4 address consists of four 8-bit octets (

32

bits total).

2

Write the subnet mask in binary form.

The binary representation is

11111111.11111111.11111111.11110000

.

The first three octets are completely filled (

8 \times 3 = 24

bits), and the fourth octet contains

4

network bits.

3

Convert each binary octet to decimal notation.

The fourth octet binary value

11110000_2

equals

128 + 64 + 32 + 16 = 240

, resulting in

255.255.255.240

.

Converting binary bit positions (

128, 64, 32, 16

) to decimal yields the final subnet mask.

Key Concept

CIDR prefix notation to dotted-decimal subnet mask conversion

Alternative Method

Calculate the host block size by using

2^{\text{host bits}} = 2^4 = 16

. Subtract the block size from

256

in the interesting octet:

256 - 16 = 240

. Thus, the subnet mask is

255.255.255.240

.

Estimated Time:45s

Question 12Question →

A network administrator is configuring a host on an enterprise network segment using Stateless Address Autoconfiguration (SLAAC). The local router advertises the IPv6 prefix $2001:\text{db8}:0000:00\text{ab}::/64$ . The host network interface controller has the MAC address `00:50:56:A1:B2:C3`. Assuming the host generates its 64-bit Interface Identifier using the standard EUI-64 mechanism, what is the resulting host IPv6 address expressed in strict compliance with RFC 5952 zero-compression and leading-zero suppression rules?

Show answer & explanation

Answer: 2001:db8:0:ab:250:56ff:fea1:b2c3; 2001:DB8:0:AB:250:56FF:FEA1:B2C3

Answer

2001:db8:0:ab:250:56ff:fea1:b2c3

Generating the SLAAC IPv6 address involves forming the EUI-64 interface ID from MAC address `00:50:56:A1:B2:C3` by inserting `FF:FE` between the third and fourth bytes (`00:50:56:FF:FE:A1:B2:C3`) and inverting the 7th bit of the first byte (`00` becomes `02`), yielding `0250:56ff:fea1:b2c3`. Combining this with prefix `2001:0db8:0000:00ab::/64` gives `2001:0db8:0000:00ab:0250:56ff:fea1:b2c3`. Under RFC 5952, leading zeros in each field are omitted (`0db8` -> `db8`, `0000` -> `0`, `00ab` -> `ab`, `0250` -> `250`). Because there is only one contiguous 16-bit zero field, `::` cannot be applied, resulting in `2001:db8:0:ab:250:56ff:fea1:b2c3`.

Step-by-Step Solution

1

Split the 48-bit MAC address into two 24-bit halves and insert FF:FE in the middle.

Inserting `FF:FE` into `00:50:56` and `A1:B2:C3` yields `00:50:56:FF:FE:A1:B2:C3` (or formatted as hextets: `0050:56ff:fea1:b2c3`).

EUI-64 expands a 48-bit MAC address to a 64-bit interface identifier by inserting the reserved 16-bit hex sequence `FF-FE`.

2

Invert the 7th bit (Universal/Local bit) of the first byte of the MAC address.

The first byte `00` in binary is `0000 0000`. Inverting the 7th bit (the U/L bit) gives `0000 0010`, which is `02` in hexadecimal. The modified interface ID becomes `0250:56ff:fea1:b2c3`.

The IEEE EUI-64 standard specifies that the 7th bit of the first octet must be inverted to indicate whether the address is universally unique (0 changed to 1).

3

Combine the /64 network prefix with the 64-bit EUI-64 interface identifier.

Combining `2001:0db8:0000:00ab` with `0250:56ff:fea1:b2c3` produces the uncompressed 128-bit address `2001:0db8:0000:00ab:0250:56ff:fea1:b2c3`.

The full IPv6 unicast address is formed by concatenating the subnet prefix and host interface ID.

4

Apply RFC 5952 compression rules strictly.

Suppress leading zeros in all hextets (`0db8` -> `db8`, `0000` -> `0`, `00ab` -> `ab`, `0250` -> `250`). Note that double-colon `::` cannot be used to replace a single zero hextet (`0`). The final compressed address is `2001:db8:0:ab:250:56ff:fea1:b2c3`.

RFC 5952 mandates lower-case hexadecimal digits, leading zero suppression in each 16-bit field, and explicitly forbids using double-colon `::` for a single isolated zero field.

Key Concept

EUI-64 Interface ID Generation & RFC 5952 Address Compression Rules

Question 13Question →

A network administrator is allocating IP addresses for a newly created VoIP segment assigned the IPv4 subnet block $172.31.80.0/21$ . What is the broadcast address of this subnet?

Show answer & explanation

Answer: 172.31.87.255

Answer

172.31.87.255

For the subnet

172.31.80.0/21

, the subnet mask is

255.255.248.0

. The block size in the third octet is 8. The network range extends from

172.31.80.0

to

172.31.87.255

, making

172.31.87.255

the broadcast address.

Step-by-Step Solution

1

Determine the subnet mask and block size for a /21 prefix.

A /21 prefix corresponds to a subnet mask of

255.255.248.0

. The block size in the third octet is

256 - 248 = 8

.

Prefix length /21 leaves 3 subnet bits in the third octet (

2^3 = 8

block size).

2

Calculate the next subnet network address.

Adding the block size of 8 to the third octet of

172.31.80.0

yields the next network address:

172.31.88.0

.

The next subnet boundary defines the end of the current subnet range.

3

Subtract 1 from the next network address to find the broadcast address.

172.31.88.0 - 1 = 172.31.87.255

.

The broadcast address is the highest address within the subnet range where all host bits are set to 1.

Key Concept

Subnet Boundary and Broadcast Address Calculation

Estimated Time:1m 0s

Question 14Question →

A network engineer is configuring a core router interface assigned the IPv4 address $10.16.42.138/22$ . What is the broadcast address for this subnet?

Show answer & explanation

Answer: 10.16.43.255

Answer

The broadcast address for the

10.16.42.138/22

subnet is

10.16.43.255

.

For the host IP

10.16.42.138/22

, the subnet mask is

255.255.252.0

. In the third octet, subnets increment by 4 (

0, 4, 8, \dots, 40, 44

). The network address containing

10.16.42.138

is

10.16.40.0/22

, and the subnet range spans from

10.16.40.0

to

10.16.43.255

. Therefore, the broadcast address for this subnet is

10.16.43.255

.

Step-by-Step Solution

1

Determine the subnet mask and block size from the CIDR prefix.

A

/22

CIDR prefix corresponds to a subnet mask of

255.255.252.0

. The block size in the third octet is

256 - 252 = 4

.

The CIDR prefix

/22

allocates 22 bits to the network portion, leaving 2 network bits in the third octet.

2

Calculate the network boundary (Network ID) for the host IP address

10.16.42.138

.

The third octet network boundaries in increments of 4 are

0, 4, 8, \dots, 40, 44

. Since 42 lies between 40 and 44, the Network ID is

10.16.40.0

.

The network ID is determined by rounding down to the nearest multiple of the block size in the third octet.

3

Calculate the broadcast address of the subnet.

The next network boundary is

10.16.44.0

. Subtracting 1 from this address gives the broadcast address

10.16.43.255

.

The broadcast address is the highest possible IP address within the subnet range, setting all 10 host bits to 1.

Key Concept

IPv4 Subnetting and Broadcast Address Calculation

Question 15Question →

A network engineer is documenting the IPv6 address allocation for an enterprise router interface. The fully expanded, uncompressed address assigned to the interface is 2001:0db8:0000:0042:0000:0000:0000:0007. According to RFC 5952 rules, what is the correctly compressed and canonicalized representation of this IPv6 address?

Show answer & explanation

Answer: 2001:db8:0:42::7

Answer

2001:db8:0:42::7

According to RFC 5952 standards for IPv6 address canonicalization: 1) Leading zeros within each 16-bit field must be removed (0db8 -> db8, 0042 -> 42, 0007 -> 7). 2) A single all-zero field is written as 0. 3) The double-colon (::) must be used to compress the longest contiguous sequence of all-zero fields. In this address, the sequence of three zeros (0000:0000:0000) is longer than the single zero field, making 2001:db8:0:42::7 the only correct canonical compressed address.

Step-by-Step Solution

1

Suppress leading zeros in each 16-bit hexadecimal field (hextet).

2001:0db8 becomes 2001:db8; 0000 becomes 0; 0042 becomes 42; 0000:0000:0000 becomes 0:0:0; 0007 becomes 7. The intermediate string is 2001:db8:0:42:0:0:0:7.

RFC 5952 requires removing leading zeros within each hextet while preserving a single zero for any all-zero hextet.

2

Identify all contiguous sequences of all-zero hextets and select the longest sequence to compress using double-colon (::).

Sequence 1 is at index 3 (length 1: '0'). Sequence 2 is at indices 5-7 (length 3: '0:0:0'). The longer sequence (length 3) is replaced with '::'.

RFC 5952 specifies that double-colon (::) must replace the longest contiguous run of all-zero hextets. Shorter single-zero hextets must remain as a single '0'.

3

Combine the compressed sections into the final canonical IPv6 address.

2001:db8:0:42::7

Ensures full compliance with RFC 5952 formatting and lowercase hexadecimal standards.

Key Concept

RFC 5952 IPv6 Address Compression and Canonical Representation

Question 16Question →

A network administrator is documenting the configuration of a enterprise web server. The server interface is assigned the uncompressed IPv6 address $2001:0\text{db}8:0000:0001:0000:0000:0000:0020$ . What is the correctly formatted, compressed IPv6 address according to RFC 5952 standard rules?

Show answer & explanation

Answer: 2001:db8:0:1::20; 2001:DB8:0:1::20

Answer

2001:db8:0:1::20

According to RFC 5952 standard rules for IPv6 address representation: 1) Leading zeros within each 16-bit field must be removed (`0db8` -> `db8`, `0001` -> `1`, `0020` -> `20`). 2) The longest contiguous sequence of all-zero 16-bit fields must be replaced with `::`. The sequence of three zero fields at hextets 5, 6, and 7 is longer than the single zero at hextet 3. Therefore, `0000:0000:0000` is compressed to `::`, while hextet 3 is represented as a single `0`. The resulting canonical address is `2001:db8:0:1::20`.

Step-by-Step Solution

1

Suppress leading zeros in each 16-bit hexadecimal field.

The fields `0db8`, `0000`, `0001`, and `0020` become `db8`, `0`, `1`, and `20` respectively.

RFC 5952 requires that leading zeros in any hextet must be omitted (e.g., `0db8` -> `db8`, `0001` -> `1`).

2

Identify all zero fields and locate the longest consecutive run of all-zero fields.

There is a single zero field at hextet 3 (`0`) and a run of three consecutive zero fields at hextets 5, 6, and 7 (`0000:0000:0000`).

RFC 5952 states that the double colon `::` must replace the single longest contiguous sequence of all-zero fields.

3

Apply double-colon `::` compression to the longest zero run.

Hextets 5, 6, and 7 are replaced by `::`, while hextet 3 remains as a single `0`.

The single zero at hextet 3 cannot be replaced with `::` because a double colon can only appear once in an address, and the run of three zeros takes precedence.

4

Combine the fields into the finalized compressed address string.

`2001:db8:0:1::20`

This represents the strict RFC 5952 canonical representation of the IPv6 address.

Key Concept

RFC 5952 IPv6 Address Text Representation Rules

Question 17Question →

A network technician is performing an audit on core router configuration files and identifies an uncompressed Unique Local IPv6 Address (ULA) recorded in full hexadecimal format as `fd12:0000:0000:a1b2:0000:0000:0000:0099`. To ensure compliance with enterprise documentation guidelines, the technician must format this address according to standard RFC 5952 rules. What is the correctly compressed RFC 5952 representation of this IPv6 address?

Show answer & explanation

Answer: fd12:0:0:a1b2::99

Answer

fd12:0:0:a1b2::99

Under RFC 5952 rules, leading zeros in every 16-bit hexadecimal field are suppressed, reducing `0000` to `0` and `0099` to `99`. When multiple zero runs exist, the double colon `::` must replace the single longest contiguous run of zero fields. In `fd12:0000:0000:a1b2:0000:0000:0000:0099`, the first zero run spans 2 fields (`0:0`) and the second zero run spans 3 fields (`0:0:0`). Compressing the longer 3-field sequence yields `fd12:0:0:a1b2::99`.

Step-by-Step Solution

1

Suppress leading zeros in each 16-bit hexadecimal field.

fd12:0:0:a1b2:0:0:0:99

RFC 5952 Section 4.1 mandates that leading zeros in a 16-bit field must be removed (e.g., '0000' becomes '0' and '0099' becomes '99').

2

Identify all contiguous sequences of all-zero 16-bit fields.

Sequence 1 has 2 zero fields ('0:0'); Sequence 2 has 3 zero fields ('0:0:0').

To apply double-colon ('::') compression properly, the lengths of all consecutive zero runs must be evaluated.

3

Apply double-colon ('::') compression strictly to the longest sequence of zero fields.

fd12:0:0:a1b2::99

RFC 5952 Section 4.2.2 requires that '::' must replace the single longest run of consecutive zero fields. Sequence 2 (3 fields) is longer than Sequence 1 (2 fields).

Key Concept

RFC 5952 IPv6 Address Text Representation and Zero Compression Rules

Estimated Time:1m 0s

Question 18Question →

A network administrator is configuring a interface on a syslog server with the IPv4 address $192.168.45.138/26$ . What is the network ID (subnet address) of the subnetwork to which this server belongs?

Show answer & explanation

Answer: 192.168.45.128; 192.168.45.128/26

Answer

192.168.45.128

With a

/26

prefix, the subnet mask is

255.255.255.192

. The block size in the fourth octet is

256 - 192 = 64

. Subnets in this range start at

.0

,

.64

,

.128

, and

.192

. Since the host address fourth octet is

138

, it resides within the

192.168.45.128/26

subnet (covering addresses

192.168.45.128

through

192.168.45.191

). The network ID is

192.168.45.128

.

Step-by-Step Solution

1

Determine the subnet mask block size from the CIDR prefix length.

A

/26

prefix leaves

32 - 26 = 6

host bits in the fourth octet. The block size (increment value) is

2^6 = 64

.

Knowing the block size allows identification of the subnet boundaries in the relevant octet.

2

Calculate the subnet boundaries in the fourth octet.

The fourth octet subnet ranges start at increments of

64

:

.0

,

.64

,

.128

, and

.192

.

Subnet IDs always begin on multiples of the block size starting from 0.

3

Locate the range containing the host IP fourth octet (

138

).

The fourth octet value

138

falls between

128

and

191

. Therefore, the network ID is

192.168.45.128

.

The network ID is the lowest address (boundary start) of the subnet containing the host.

Key Concept

Determining Network Address from IPv4 Host Address and CIDR Prefix

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