Top 28 Subnetting Interview Questions & Answers (2026)

Subnetting is a core networking skill tested heavily at Cisco, Juniper, network-focused roles at TCS, Infosys, and Wipro, and in GATE examinations. Candidates report that subnetting questions range from straightforward CIDR calculations to complex VLSM design scenarios. Based on public preparation resources and candidate-reported interview accounts, questions on subnet mask arithmetic, VLSM allocation, and supernetting appear most frequently at networking interviews.

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Binary and IP Addressing Foundations

Q1. Convert the subnet mask 255.255.240.0 to CIDR prefix length notation. (Easy)

Convert each octet to binary:

• 255 = 11111111
• 255 = 11111111
• 240 = 11110000
• 0 = 00000000

Count the 1-bits: 8 + 8 + 4 + 0 = 20

CIDR notation: /20

Quick reference for common masks:

Subnet Mask	CIDR	1-bits Pattern
255.0.0.0	/8	8 ones
255.255.0.0	/16	16 ones
255.255.255.0	/24	24 ones
255.255.255.128	/25	25 ones
255.255.255.192	/26	26 ones
255.255.255.224	/27	27 ones
255.255.255.240	/28	28 ones
255.255.255.248	/29	29 ones
255.255.255.252	/30	30 ones

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Q2. What are the classes of IPv4 addresses and their default subnet masks? (Easy)

Class	First Octet Range	Default Mask	Private Ranges (RFC 1918)
A	1-126	/8 (255.0.0.0)	10.0.0.0/8
B	128-191	/16 (255.255.0.0)	172.16.0.0/12
C	192-223	/24 (255.255.255.0)	192.168.0.0/16
D	224-239	Multicast	N/A
E	240-255	Experimental	N/A

Special addresses:

• 127.0.0.0/8: Loopback (127.0.0.1 is localhost)
• 169.254.0.0/16: Link-local (APIPA when DHCP fails)
• 0.0.0.0/0: Default route (all destinations)
• 255.255.255.255: Limited broadcast

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Q3. Given IP address 192.168.10.75 with subnet mask 255.255.255.192, find the network address, broadcast address, and valid host range. (Easy)

Step 1: Convert to binary and apply AND operation.

192.168.10.75:

• 192 = 11000000
• 168 = 10101000
• 10 = 00001010
• 75 = 01001011

255.255.255.192:

• 255 = 11111111
• 255 = 11111111
• 255 = 11111111
• 192 = 11000000

AND result (network address):

• 11000000 AND 11000000 = 11000000 = 192

Network address: 192.168.10.64

Step 2: Broadcast address - set all host bits to 1.

• Host bits = last 6 (192 = /26, so 32-26=6 host bits)
• 64 in binary = 01000000, set last 6 bits to 1: 01111111 = 127 (wrong, let me redo)
• 64 = 01000000, OR with 00111111 = 01111111 = 127...
• Actually: 64 | (64-1) = 64 | 63 = 127. Broadcast: 192.168.10.127

Step 3: Host range.

• First host: 192.168.10.65
• Last host: 192.168.10.126
• Usable hosts: 2^6 - 2 = 62

Summary:

• Network: 192.168.10.64/26
• Broadcast: 192.168.10.127
• Host range: 192.168.10.65 - 192.168.10.126
• Usable hosts: 62

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Q4. How many subnets and hosts are created when you subnet 10.0.0.0/8 with a /12 mask? (Easy)

Original prefix: /8. New prefix: /12. Bits borrowed: 12 - 8 = 4.

Number of subnets: 2^4 = 16

Host bits remaining: 32 - 12 = 20 Hosts per subnet: 2^20 - 2 = 1,048,576 - 2 = 1,048,574

The 16 subnets are:

• 10.0.0.0/12
• 10.16.0.0/12
• 10.32.0.0/12
• 10.48.0.0/12
• ...continuing in increments of 16 in the second octet
• 10.240.0.0/12

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CIDR and Subnet Calculations

Q5. What is CIDR and how does it differ from classful addressing? (Easy)

CIDR (Classless Inter-Domain Routing, RFC 4632) removes the rigid Class A/B/C boundaries. With classful addressing, a network had a fixed prefix length based on the first octet. With CIDR, any prefix length from /0 to /32 is valid.

Key differences:

Feature	Classful	CIDR
Prefix length	Fixed by class	Arbitrary
Routing table size	Large (every Class C separate)	Smaller (route aggregation)
Address waste	High	Reduced
Example	200.10.5.0 is always /24	200.10.4.0/22 aggregates 4 Class C blocks

CIDR enables route aggregation (supernetting): multiple contiguous classful networks advertised as a single CIDR block, reducing routing table entries.

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Q6. A company needs 500 hosts in one subnet and 50 hosts in another. Allocate subnets from 172.16.0.0/16. (Medium)

This is a VLSM allocation problem.

For 500 hosts: Need at least 502 addresses (500 + network + broadcast). Smallest power of 2 >= 502 is 512 = 2^9, so 9 host bits. Prefix: /23 (32 - 9 = 23). Subnet: 172.16.0.0/23

• Network: 172.16.0.0
• Broadcast: 172.16.1.255
• Hosts: 172.16.0.1 - 172.16.1.254 (510 usable)

For 50 hosts: Need at least 52 addresses (50 + network + broadcast). Smallest power of 2 >= 52 is 64 = 2^6, so 6 host bits. Prefix: /26 (32 - 6 = 26). Subnet: 172.16.2.0/26

• Network: 172.16.2.0
• Broadcast: 172.16.2.63
• Hosts: 172.16.2.1 - 172.16.2.62 (62 usable)

VLSM allocation table:

Purpose	Subnet	Usable Hosts	Block Size
Large segment	172.16.0.0/23	510	512
Small segment	172.16.2.0/26	62	64
Next available	172.16.2.64	-	-

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Q7. What is VLSM? How does it improve IP address utilization? (Medium)

VLSM (Variable Length Subnet Masking) allows a single classful network to be divided into subnets of different sizes by using different prefix lengths within the same major network.

Without VLSM (fixed-length subnetting): If a network needs subnets of 100, 50, 10, and 2 hosts, fixed subnetting wastes addresses. Using /25 for all subnets (126 hosts each) wastes 76 addresses in the 50-host subnet, 116 in the 10-host subnet, and 124 in the 2-host subnet.

With VLSM:

Network: 192.168.1.0/24

Requirement        Subnet            Block  Usable
100 hosts  ->  192.168.1.0/25       128     126
50 hosts   ->  192.168.1.128/26      64      62
10 hosts   ->  192.168.1.192/28      16      14
2 hosts    ->  192.168.1.208/30       4       2
(WAN link)

Remaining space: 192.168.1.212 - 192.168.1.255 (44 addresses for future use).

Rule: Allocate largest subnets first, then carve smaller ones from remaining space. Always subnet on binary boundaries.

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Q8. Explain the concept of a "magic number" in subnetting calculations. (Medium)

The magic number (also called block size or subnet increment) is the size of each subnet in terms of addresses. It tells you the starting address of each subsequent subnet.

Calculation: Magic number = 256 - (interesting octet value of the subnet mask)

The "interesting octet" is the last non-255, non-0 octet of the subnet mask.

Examples:

Subnet Mask	Interesting Octet	Magic Number	Subnets start at...
255.255.255.192	192	64	0, 64, 128, 192
255.255.255.224	224	32	0, 32, 64, 96, 128...
255.255.255.240	240	16	0, 16, 32, 48...
255.255.240.0	240 (3rd octet)	16	0.0, 16.0, 32.0...

Using the magic number to find subnet: For IP 10.20.35.100 with mask 255.255.240.0 (magic = 16, interesting in 3rd octet):

• 35 / 16 = 2 remainder 3. So subnet is at 10.20.32.0 (2 * 16 = 32).
• Broadcast: 10.20.47.255 (32 + 16 - 1 = 47).

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Q9. Is 172.16.5.0/20 a valid subnet? What is its network address? (Medium)

Check if /20 is valid for this address. A /20 mask has block size 16 in the third octet (256 - 240 = 16). Valid network addresses in the third octet are: 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240.

Third octet = 5. Is 5 a multiple of 16? No (5 / 16 = 0 remainder 5).

172.16.5.0/20 is NOT a valid subnet. It is a host address within the 172.16.0.0/20 network.

The network that contains 172.16.5.0 with /20:

• Third octet: 5 / 16 = 0 (integer division). Network third octet = 0 * 16 = 0.
• Network address: 172.16.0.0/20
• Broadcast: 172.16.15.255
• Host range: 172.16.0.1 - 172.16.15.254

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Q10. Divide 192.168.0.0/24 into exactly 8 equal subnets. List all of them. (Medium)

8 subnets = 2^3, so borrow 3 bits. New prefix = 24 + 3 = /27. Block size = 32.

Subnet	Network Address	Broadcast	Host Range	Usable Hosts
1	192.168.0.0	192.168.0.31	.1 - .30	30
2	192.168.0.32	192.168.0.63	.33 - .62	30
3	192.168.0.64	192.168.0.95	.65 - .94	30
4	192.168.0.96	192.168.0.127	.97 - .126	30
5	192.168.0.128	192.168.0.159	.129 - .158	30
6	192.168.0.160	192.168.0.191	.161 - .190	30
7	192.168.0.192	192.168.0.223	.193 - .222	30
8	192.168.0.224	192.168.0.255	.225 - .254	30

Total hosts: 8 * 30 = 240 usable (original had 254 usable; 14 lost to subnet/broadcast overhead).

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Supernetting and Route Aggregation

Q11. What is supernetting? Aggregate these networks: 192.168.8.0/24, 192.168.9.0/24, 192.168.10.0/24, 192.168.11.0/24. (Medium)

Supernetting (route aggregation, CIDR summarization) combines multiple smaller networks into a single larger network advertisement. This reduces routing table size.

Conditions for aggregation:

1. Networks must be contiguous (adjacent in address space)
2. Block size must be a power of 2
3. The first network address must be on a boundary aligned to the aggregate block size

Example aggregation: Networks: 192.168.8.0, 192.168.9.0, 192.168.10.0, 192.168.11.0

Third octets in binary:

• 8 = 00001000
• 9 = 00001001
• 10 = 00001010
• 11 = 00001011

Common bits: 00001000 through 00001011 share the first 6 bits: 000010xx

Total prefix: first 2 octets (/16 base) + 6 bits from third octet = 22 bits.

Aggregate: 192.168.8.0/22

Verification:

• Network: 192.168.8.0
• Broadcast: 192.168.11.255
• Covers exactly the 4 original /24 networks.

Route summarization impact: 4 routing table entries compressed to 1.

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Q12. Can you always aggregate contiguous networks? What can go wrong? (Hard)

No, aggregation requires strict alignment. Problems arise when:

1. Not power-of-2 count: Three networks (192.168.1.0, 192.168.2.0, 192.168.3.0) cannot be aggregated into one prefix. You need either aggregate to 192.168.0.0/22 (4 networks, including 192.168.0.0/24 which you may not own) or use two separate prefixes.

2. Misaligned start: 192.168.3.0/24 and 192.168.4.0/24 cannot aggregate to /23 because 192.168.3.0/23 = third octet 3, which is odd. /23 boundaries are on even numbers. They must aggregate differently.

Actually: 192.168.3.0/23 would contain 192.168.3.x and 192.168.4.x only if 3 is on a /23 boundary. Check: 3 is odd, so not on /23 boundary. 192.168.4.0 is on /23 boundary (4 is even). These two cannot form a /23.

3. Ownership boundaries: An ISP cannot aggregate customer address blocks that span addresses owned by different customers without creating incorrect routing.

Discontiguous supernet problem: If a router advertises a supernet but does not own all addresses within it, traffic for the "gaps" gets black-holed. Solution: null-route the unowned gaps pointing to Null0.

ip route 0.0.0.0 0.0.0.0 [next-hop]          # default
ip route 192.168.8.0 255.255.252.0 Null0       # black-hole the supernet gaps
ip route 192.168.8.0 255.255.255.0 [interface] # specific routes win

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WAN Links and Special Cases

Q13. What subnet mask do you use for a point-to-point WAN link? Why? (Easy)

Use /30 (255.255.255.252) for point-to-point links.

A /30 provides 4 addresses:

• 1 network address
• 2 usable host addresses (one per endpoint)
• 1 broadcast address

This is the minimum subnet that supports two routers with IP addresses.

Example: WAN link: 10.0.0.0/30

• Router A: 10.0.0.1
• Router B: 10.0.0.2
• Network: 10.0.0.0
• Broadcast: 10.0.0.3

Even more efficient: /31 (RFC 3021) provides only 2 addresses with no network or broadcast address, usable specifically for point-to-point links. Some routers support this.

Loopback interfaces typically use /32 (single host) for router IDs in OSPF and BGP.

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Q14. What is the difference between a network address and a broadcast address? (Easy)

Network address: First address in a subnet, all host bits set to 0. Identifies the subnet itself. Cannot be assigned to a host. Used in routing table entries.

Broadcast address: Last address in a subnet, all host bits set to 1. Packets sent to this address are delivered to all hosts on the subnet. Cannot be assigned to a host.

Example for 192.168.5.0/25:

• Subnet mask: 255.255.255.128
• Network: 192.168.5.0 (last 7 bits = 0000000)
• Broadcast: 192.168.5.127 (last 7 bits = 1111111)
• Host range: 192.168.5.1 - 192.168.5.126

Types of broadcast:

• Limited broadcast: 255.255.255.255 (not forwarded by routers)
• Directed broadcast: network's broadcast address (e.g., 192.168.1.255 for 192.168.1.0/24); can be forwarded; disabled by default on Cisco routers (no ip directed-broadcast)

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Q15. A /30 subnet starting at 10.1.1.4 -- what are the usable host addresses? (Easy)

/30 = 255.255.255.252. Block size = 4. Valid /30 subnets starting at multiples of 4:

• 10.1.1.0/30: network .0, broadcast .3, hosts .1-.2
• 10.1.1.4/30: network .4, broadcast .7, hosts .5-.6
• 10.1.1.8/30: network .8, broadcast .11, hosts .9-.10

For 10.1.1.4/30:

• Network: 10.1.1.4
• Usable hosts: 10.1.1.5 and 10.1.1.6
• Broadcast: 10.1.1.7

Verify: 4 is a multiple of 4? 4/4 = 1. Yes, valid subnet boundary.

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VLSM Design Problems

Q16. Design a subnetting scheme for this network: 10.10.0.0/16. Requirements: Segment A needs 1000 hosts, Segment B needs 500 hosts, Segment C needs 250 hosts, Segment D needs 100 hosts, and 3 WAN links. (Hard)

Process: allocate largest first.

Segment A (1000 hosts): Need 1002 addresses. 2^10 = 1024 > 1002. Prefix: /22 (32-10=22). Subnet: 10.10.0.0/22

• Range: 10.10.0.1 - 10.10.3.254 (1022 hosts)
• Next available: 10.10.4.0

Segment B (500 hosts): Need 502. 2^9 = 512 > 502. Prefix: /23 (32-9=23). Subnet: 10.10.4.0/23

• Range: 10.10.4.1 - 10.10.5.254 (510 hosts)
• Next available: 10.10.6.0

Segment C (250 hosts): Need 252. 2^8 = 256 > 252. Prefix: /24 (32-8=24). Subnet: 10.10.6.0/24

• Range: 10.10.6.1 - 10.10.6.254 (254 hosts)
• Next available: 10.10.7.0

Segment D (100 hosts): Need 102. 2^7 = 128 > 102. Prefix: /25 (32-7=25). Subnet: 10.10.7.0/25

• Range: 10.10.7.1 - 10.10.7.126 (126 hosts)
• Next available: 10.10.7.128

Three WAN links (/30 each):

• WAN1: 10.10.7.128/30 (hosts .129, .130)
• WAN2: 10.10.7.132/30 (hosts .133, .134)
• WAN3: 10.10.7.136/30 (hosts .137, .138)
• Next available: 10.10.7.140

Allocation summary:

Segment	Subnet	Required	Allocated	Waste
A	10.10.0.0/22	1000	1022	22
B	10.10.4.0/23	500	510	10
C	10.10.6.0/24	250	254	4
D	10.10.7.0/25	100	126	26
WAN1	10.10.7.128/30	2	2	0
WAN2	10.10.7.132/30	2	2	0
WAN3	10.10.7.136/30	2	2	0

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Q17. Given 172.20.0.0/14, how many /24 networks can be carved out? (Medium)

/14 to /24: difference = 10 bits. Number of /24 networks = 2^10 = 1024.

Verification:

• /14 block size: 2^(32-14) = 2^18 = 262,144 addresses
• /24 block size: 2^(32-24) = 2^8 = 256 addresses
• Count: 262,144 / 256 = 1024 subnets. Confirmed.

Address range of 172.20.0.0/14:

• Mask: 255.252.0.0
• Network: 172.20.0.0
• Broadcast: 172.23.255.255
• /24 subnets from 172.20.0.0/24 through 172.23.255.0/24

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IPv6 Subnetting

Q18. How does IPv6 subnetting differ from IPv4? (Medium)

IPv6 addresses are 128 bits vs IPv4's 32 bits. CIDR notation still applies but the default for a LAN is /64 (standard practice).

IPv6 address structure:

|<------ 64 bits ------>|<------ 64 bits ------>|
  Global routing prefix     Interface identifier
  (network portion)         (host portion)

Typical allocation:

• ISP gets /32 from RIR
• ISP assigns /48 to each customer
• Customer uses /48 to allocate /64 subnets
• Each /64 can hold 2^64 hosts (EUI-64 auto-configuration uses 64 host bits)

Key differences:

Feature	IPv4	IPv6
Address length	32 bits	128 bits
Notation	Dotted decimal	Colon-hexadecimal
No broadcast	Has broadcast	No broadcast (uses multicast)
Standard LAN prefix	/24 (common)	/64 (mandatory for SLAAC)
Loopback	127.0.0.1/8	::1/128
Private/ULA	RFC 1918	fc00::/7 (ULA)
Link-local	169.254.0.0/16	fe80::/10

IPv6 subnetting example: Given 2001:db8:abcd::/48, allocate /64 subnets:

• 2001:db8:abcd:0000::/64
• 2001:db8:abcd:0001::/64
• ...
• 2001:db8:abcd:ffff::/64

Total: 2^16 = 65,536 /64 subnets from one /48.

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Q19. A company is given 2001:0db8:0001::/48. How many /64 subnets can they create? Show one subnet's full address range. (Medium)

From /48 to /64: difference = 16 bits. Number of /64 subnets = 2^16 = 65,536.

Example subnet: 2001:0db8:0001:0042::/64

• Network: 2001:0db8:0001:0042:0000:0000:0000:0000
• First host: 2001:0db8:0001:0042:0000:0000:0000:0001
• Last host: 2001:0db8:0001:0042:ffff:ffff:ffff:ffff
• No broadcast (IPv6 uses multicast)
• Usable hosts: 2^64 - 1 (theoretical; excluding network address itself)

In practice, the all-zeros interface ID and the subnet-router anycast address (all-zeros for interface) are reserved, but with 2^64 addresses this is purely theoretical.

IPv6 address compression rules:

• Leading zeros in each group can be omitted: 0042 -> 42
• One consecutive run of all-zero groups can be replaced with ::
• Example: 2001:0db8:0001:0042:0000:0000:0000:0001 = 2001:db8:1:42::1

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Network Design and Practical Scenarios

Q20. How do you determine if two IP addresses are in the same subnet? (Easy)

Perform bitwise AND between each IP address and the subnet mask. If both results are equal, the addresses are in the same subnet.

Example:

• IP1: 192.168.1.100
• IP2: 192.168.1.200
• Mask: 255.255.255.0 (/24)

IP1 AND Mask:

• 192.168.1.100 AND 255.255.255.0 = 192.168.1.0

IP2 AND Mask:

• 192.168.1.200 AND 255.255.255.0 = 192.168.1.0

Same result = same subnet. They can communicate directly without a router.

Counterexample with /25:

• IP1: 192.168.1.100, IP2: 192.168.1.200, Mask: /25 (255.255.255.128)
• IP1 AND Mask = 192.168.1.0 (first half: .0-.127)
• IP2 AND Mask = 192.168.1.128 (second half: .128-.255)
• Different subnets: they need a router to communicate.

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Q21. What is a subnet calculator and when would you use one in production? (Easy)

A subnet calculator is a tool (online or CLI) that automates subnet math: given an IP and mask, it returns network address, broadcast, host range, usable hosts, and adjacent subnets.

CLI example (ipcalc):

ipcalc 192.168.10.0/26
# Network:   192.168.10.0/26
# Broadcast: 192.168.10.63
# HostMin:   192.168.10.1
# HostMax:   192.168.10.62
# Hosts/Net: 62

Production use cases:

• IP address management (IPAM) systems like NetBox automate subnet tracking
• Network engineers use it to verify configurations before pushing to devices
• Troubleshooting: confirm a host's subnet membership before checking routing
• Capacity planning: knowing remaining free blocks in an address plan

Important: In interviews, demonstrate you can do the math mentally or on paper. Subnet calculator reliance without understanding shows shallow knowledge.

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Q22. Explain the concept of a "host route" and a "default route." (Medium)

Host route: A route with /32 prefix length (IPv4) or /128 (IPv6) matching exactly one IP address. Used for:

• Loopback interfaces on routers (OSPF router-ID)
• Policy-based routing to specific hosts
• Overriding an aggregate route for one host

Default route: 0.0.0.0/0 (IPv4) or ::/0 (IPv6). Matches any destination not matched by a more specific route. Points to the gateway of last resort.

Longest prefix match rule: When multiple routes match a destination, the router selects the one with the longest (most specific) prefix.

Routing table:
192.168.0.0/16   -> Router A
192.168.1.0/24   -> Router B
192.168.1.5/32   -> Router C
0.0.0.0/0        -> ISP router

Packet to 192.168.1.5: goes to Router C (/32 wins)
Packet to 192.168.1.99: goes to Router B (/24 wins)
Packet to 192.168.5.1: goes to Router A (/16 wins)
Packet to 8.8.8.8: goes to ISP router (/0, default route)

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Q23. What is RFC 1918 and why does it matter for subnetting design? (Medium)

RFC 1918 defines private IPv4 address space that is not routable on the public internet:

• 10.0.0.0/8 (Class A private, 16.7 million addresses)
• 172.16.0.0/12 (Class B private range, 1.05 million addresses, covers 172.16.x.x through 172.31.x.x)
• 192.168.0.0/16 (Class C private, 65,536 addresses)

Why it matters for design:

1. NAT requirement: Private addresses require Network Address Translation to reach the internet. Routers at the edge translate private to public IP.

2. Address conservation: With IPv4 exhaustion, private addressing + NAT allowed the internet to grow far beyond 4.3 billion theoretical addresses.

3. Design choice:

   • Large enterprises: use 10.0.0.0/8 (massive space for VLSM)
   • Medium networks: use 172.16.0.0/12
   • Small offices/home: use 192.168.0.0/16

4. Overlapping VPNs problem: When merging two companies or connecting VPNs, overlapping RFC 1918 space creates routing conflicts. Careful address planning prevents this.

Overlap scenario:

• Company A: 10.0.0.0/8 internally
• Company B: 10.0.0.0/8 internally
• Merging: need NAT-on-a-stick or re-addressing. Expensive.

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Q24. How does DHCP interact with subnetting? (Easy)

DHCP (Dynamic Host Configuration Protocol) assigns IP addresses to hosts from a configured pool within a subnet.

DHCP server configuration ties to subnetting:

• Scope: defines the subnet (e.g., 192.168.1.0/24)
• Range: usable host addresses to hand out (e.g., .100 - .200)
• Exclusions: static reservations for servers, printers
• Gateway: one entry, typically the router interface IP
• Lease time: duration of the assignment

DHCP and subnets: A DHCP server can serve multiple subnets using either:

1. One server per subnet (simple but less scalable)
2. DHCP relay agent (ip helper-address on Cisco): routers forward DHCP broadcasts to a centralized DHCP server on another subnet as unicast. The server uses the relay agent's IP (giaddr field) to determine which scope to serve.

Important: DHCP leases are within the subnet. A host in 192.168.1.0/24 will only receive an address from the 192.168.1.x scope, not from 192.168.2.x.

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Q25. A network scan returns the address 10.0.0.255 in a /16 network. Is this the broadcast? (Medium)

No. In 10.0.0.0/16, the broadcast address is 10.0.255.255, not 10.0.0.255.

Explanation:

• /16 mask: 255.255.0.0
• Host bits: last 16 bits
• All host bits set to 1: 00000000.00000000.11111111.11111111
• That is: 10.0.255.255

10.0.0.255 is the broadcast of the /24 subnet 10.0.0.0/24, but in a /16 network it is just a regular host address.

Common mistake: Assuming the last 255 in any position is the broadcast. Broadcast depends on the subnet mask, not just the last octet.

Test:

• 10.0.0.0/16: broadcast = 10.0.255.255
• 10.0.0.0/24: broadcast = 10.0.0.255
• 10.0.0.0/25: broadcast = 10.0.0.127

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Q26. What is a "discontiguous network" and what problem does it cause? (Hard)

A discontiguous network occurs when subnets of the same major network (classful boundary) are separated by subnets belonging to a different major network. This commonly happened in classful routing protocols.

Visual:

Site A           Core Router         Site B
10.1.1.0/24 --- [172.16.1.0/24] --- 10.2.2.0/24

Both 10.x networks are part of major network 10.0.0.0/8 (Class A)
but separated by the 172.16.x.x network.

Problem with classful routing protocols (RIPv1, IGRP): These protocols summarize routes at classful boundaries. Both sites advertise "10.0.0.0/8" to the core router. The core sees two paths to 10.0.0.0/8 and load-balances, sending some packets to the wrong site.

Solutions:

1. Use classless routing protocols (OSPF, EIGRP, RIPv2, BGP) that carry prefix length in updates
2. Disable auto-summarization: no auto-summary in EIGRP/RIPv2
3. Redesign address plan to eliminate discontiguous subnets

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Q27. In binary, what is the subnet mask for /19? Calculate an example subnet. (Hard)

/19 in binary: 11111111.11111111.11100000.00000000 = 255.255.224.0

Block size in third octet: 256 - 224 = 32. Valid third-octet values: 0, 32, 64, 96, 128, 160, 192, 224.

Example: is 172.17.68.0/19 valid? Third octet = 68. Is 68 a multiple of 32? 68/32 = 2 remainder 4. No, 68 is not on a /19 boundary. Closest valid boundary below 68: 64. Subnet: 172.17.64.0/19.

172.17.64.0/19:

• Network: 172.17.64.0
• Broadcast: 172.17.95.255 (64 + 32 - 1 = 95 in third octet)
• Host range: 172.17.64.1 - 172.17.95.254
• Usable hosts: 2^13 - 2 = 8,190

Host count check: 32-19 = 13 host bits. 2^13 = 8,192. Minus 2 = 8,190.

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Q28. A router receives two routes: 10.5.4.0/22 and 10.5.6.0/24. A packet arrives for 10.5.6.150. Which route wins and why? (Hard)

The /24 route wins by the longest prefix match rule.

Verify that both routes match 10.5.6.150:

10.5.4.0/22 check:

• Mask: 255.255.252.0
• 10.5.6.150 AND 255.255.252.0: third octet 6 AND 252 = 4 (00000110 AND 11111100 = 00000100)
• Result: 10.5.4.0. This matches the route 10.5.4.0/22. Yes, 10.5.6.150 is within the /22.

10.5.6.0/24 check:

• Mask: 255.255.255.0
• 10.5.6.150 AND 255.255.255.0 = 10.5.6.0. Matches exactly.

Both routes match. Longest prefix (/24) is more specific than /22. Packet forwarded via the /24 route.

Intuition: /24 covers only 256 addresses, while /22 covers 1,024 addresses. The /24 "knows more" about how to reach 10.5.6.x specifically.

Longest prefix match is why supernetting works without black-holing: Specific routes override the aggregate, while the aggregate catches traffic for gaps.

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Frequently Asked Questions

Q: What is the fastest mental subnetting method for interviews? Practice the magic number method: identify the interesting octet, compute 256 minus that octet value, and count multiples to find subnets quickly without binary conversion.

Q: Why can't I use 0 and 255 as host addresses? In the same octet, yes, but all-zeros host bits = network address and all-ones host bits = broadcast address for that subnet. So 192.168.1.0 and 192.168.1.255 in a /24 are reserved. However, 10.0.0.0 and 10.255.255.255 are still valid hosts in a /16 subnet.

Q: What is the difference between a subnet and a VLAN? Subnetting is Layer 3 (IP division). VLANs are Layer 2 (logical switch port groupings). Typically one subnet corresponds to one VLAN, but they are separate concepts. Inter-VLAN routing requires a Layer 3 device (router or Layer 3 switch with SVIs).

Q: How many bits of a /30 are host bits? 32 - 30 = 2 host bits. This gives 2^2 = 4 total addresses, with 2 usable hosts.

Q: Is VLSM required for modern routing protocols? VLSM requires classless routing protocols (OSPF, EIGRP, RIPv2, BGP). All modern protocols support it. Only legacy protocols (RIPv1, IGRP) do not; they are obsolete.

Q: Where does subnetting appear in GATE and competitive exams? Subnetting problems appear regularly in GATE CS networking sections, typically as calculation questions requiring network address, broadcast, and host count determination. VLSM allocation problems also appear.

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Internal Links

• Computer Networks Interview Questions
• OSI Model Interview Questions
• TCP/IP Interview Questions
• HTTP, HTTPS and DNS Interview Questions
• TCS Interview Questions
• Infosys Interview Questions