Subnetting and supernetting

Subnetting & Supernetting – Complete Exam Guide 2025 | SSC, IBPS, UPSC, Railway
🌐 Networking · Competitive Exam 2025

Subnetting & Supernetting
Complete Exam Guide

IP Classes · Subnet Masks · CIDR · VLSM · Formulas · Worked Examples · 50 MCQs — everything you need for SSC, IBPS, UPSC, Railway, NDA, GATE & Bank exams

SSC CGL / CHSLIBPS PO / ClerkRBI Grade B UPSC GS-IIIRailway NTPCNDA / CDSGATE CSBank Internal
🌐

FoundationIP Address Basics

IPv4 Address Structure (32 bits = 4 octets)
192
Octet 1
8 bits
.
168
Octet 2
8 bits
.
10
Octet 3
8 bits
.
5
Octet 4
8 bits
Network Part
Identifies the network
+
Host Part
Identifies device in the network
PropertyIPv4IPv6
Address length32 bits128 bits
Format4 octets, decimal8 groups, hexadecimal
Total addresses~4.3 billion (2³²)~340 undecillion (2¹²⁸)
Notation example192.168.1.102001:0db8::1
Exam focus⭐⭐⭐⭐⭐⭐⭐⭐
⚡ Core Facts for Exam
  • IPv4 = 32-bit address divided into 4 octets (each 0–255)
  • Each octet = 8 bits; total = 4 × 8 = 32 bits
  • Subnet Mask decides which part is Network and which is Host
  • CIDR notation: the number after "/" = count of network bits
  • IPv6 uses 128 bits (not 32) — designed to replace IPv4
📊

Most Asked in Banking & Govt ExamsIPv4 Classes (A to E)

A
1 – 126
16,777,214 hosts
/8 · 255.0.0.0
Large orgs, ISPs
B
128 – 191
65,534 hosts
/16 · 255.255.0.0
Medium orgs
C
192 – 223
254 hosts
/24 · 255.255.255.0
Small networks
D
224 – 239
Multicast
No default mask
No hosts — multicast
E
240 – 255
Reserved
No default mask
Research / future use
ClassFirst Octet RangeDefault MaskCIDRNetworksHosts/NetworkUse
A1–126255.0.0.0/812616,777,214Very large orgs
B128–191255.255.0.0/1616,38465,534Medium orgs
C192–223255.255.255.0/242,097,152254Small networks
D224–239N/AN/AMulticast
E240–255N/AN/AExperimental
🎯 Exam Tricks — IP Classes
  • 127.x.x.x is skipped from Class A — reserved for Loopback (127.0.0.1 = localhost)
  • 0.x.x.x is also reserved — so Class A starts at 1
  • Classes D and E have NO subnet mask and NO hosts
  • Trick: A=1–126, B=128–191, C=192–223 — memorise the starting numbers
  • Classful vs Classless: Modern internet uses CIDR (classless) — class system is now legacy
🔒

Frequently Asked in All ExamsPrivate & Special IP Addresses

10.0.0.0/8

Class A Private
Range: 10.0.0.0 – 10.255.255.255
Hosts: ~16.7 million

172.16.0.0/12

Class B Private
Range: 172.16.0.0 – 172.31.255.255
Hosts: ~1 million

192.168.0.0/16

Class C Private
Range: 192.168.0.0 – 192.168.255.255
Hosts: ~65,000

127.0.0.0/8

Loopback
127.0.0.1 = localhost
Tests TCP/IP on own machine

169.254.0.0/16

APIPA (Link-Local)
Assigned when DHCP fails
Self-assigned by OS

0.0.0.0

Unspecified / Default Route
Used in routing tables
Represents "all networks"

✂️

Core TopicWhat is Subnetting?

Subnetting = dividing a large IP network into smaller networks called subnets. Achieved by borrowing bits from the host portion of the address and adding them to the network portion — making the subnet mask longer.

Reason for SubnettingExplanationExample
🔒 SecurityDifferent departments get separate subnets; breach in one doesn't spreadHR vs Finance vs IT subnet
📦 Reduce broadcast trafficSmaller subnet = smaller broadcast domain = less noiseLAN with 10 devices vs 254
💾 Efficient IP useAllocate only what is needed; avoid wasting addresses30-host subnet instead of 254
🏦 Better managementSeparate ATMs, branches, servers — easier troubleshootingBank ATM: 10.10.10.0/28
🚦 Traffic controlRouters between subnets control inter-subnet trafficInter-VLAN routing
📌 Key Terms
  • Network ID — first address of a subnet (all host bits = 0). Cannot be assigned to a host.
  • Broadcast Address — last address (all host bits = 1). Sends to ALL hosts in subnet. Cannot be assigned.
  • Usable Host Range — all addresses between Network ID and Broadcast address.
  • Block Size / Increment256 − subnet mask last octet (e.g., 256−192 = 64 for /26)
  • Borrowed Bits — extra bits moved from host to network to create subnets
  • CIDR = Classless Inter-Domain Routing — uses /24, /26 etc. instead of class system
🧮

Must MemoriseSubnetting Formulas

Usable Hosts per Subnet

2h − 2

h = host bits (32 − prefix). Subtract 2 for Network ID and Broadcast address.

Number of Subnets Created

2b

b = borrowed bits (new prefix − old prefix). Modern networks allow all subnets.

Block Size (Increment)

256 − mask_octet

The "jump" between subnets. E.g., /26 mask=192 → block = 256−192 = 64.

Host Bits from Prefix

32 − prefix

E.g., /26 → host bits = 32−26 = 6 → 2⁶−2 = 62 usable hosts.

Wildcard Mask

255.255.255.255 − mask

Inverse of subnet mask. Used in Cisco ACLs and routing protocols.

Broadcast Address

Network ID + block − 1

E.g., Network .64/26 (block 64) → Broadcast = .64 + 64 − 1 = .127

📋

Print & MemoriseComplete Subnet Mask Reference Table

Subnet MaskCIDRUsable HostsBlock SizeSubnets from /24Typical Use
255.255.255.0/242542561Standard LAN
255.255.255.128/251261282Medium subnet
255.255.255.192/2662644Dept subnet
255.255.255.224/2730328Small dept
255.255.255.240/28141616ATM / small
255.255.255.248/296832Point clusters
255.255.255.252/302464P2P links
255.255.255.254/310 (or 2)*2128P2P (RFC 3021)
255.255.255.255/321 host only1256Host route
255.255.0.0/1665,534Class B default
255.0.0.0/816,777,214Class A default
✅ Quick Memory Trick — Hosts Per Subnet
  • /24 = 254 | /25 = 126 | /26 = 62 | /27 = 30 | /28 = 14 | /29 = 6 | /30 = 2
  • Pattern: each step halves the hosts → 254 → 126 → 62 → 30 → 14 → 6 → 2
  • Block sizes double each step up: 256, 128, 64, 32, 16, 8, 4
  • /30 is used for point-to-point WAN links (only 2 usable hosts — one each end)
🔬

Step-by-StepWorked Subnetting Examples

📌 Example 1 — Find Subnet Info for 192.168.1.70/26

1
Find host bits

Prefix = /26 → host bits = 32 − 26 = 6 bits

2
Calculate usable hosts

2⁶ − 2 = 64 − 2 = 62 usable hosts

3
Find block size

Mask last octet = 192 → block = 256 − 192 = 64
Subnets start at: .0, .64, .128, .192

4
Identify the subnet

IP is .70 → falls in .64 block → Network ID = 192.168.1.64

5
Find broadcast & range

Broadcast = .64 + 64 − 1 = 192.168.1.127
Usable range = .65 to .126

✅ Result
192.168.1.64
Network ID
192.168.1.127
Broadcast
.65 – .126
Host Range
62
Usable Hosts

📌 Example 2 — Divide 192.168.1.0/24 into 4 Equal Subnets

1
Find borrowed bits needed

Need 4 subnets → 2ᵇ ≥ 4 → b = 2 bits borrowed

2
New prefix

/24 + 2 = /26 | Mask = 255.255.255.192

3
Block size = 64

Subnets: .0, .64, .128, .192

✅ 4 Subnets Created (/26 — 62 hosts each)
SubnetNetwork IDHost RangeBroadcast
1192.168.1.0.1 – .62192.168.1.63
2192.168.1.64.65 – .126192.168.1.127
3192.168.1.128.129 – .190192.168.1.191
4192.168.1.192.193 – .254192.168.1.255

📌 Example 3 — Network 192.168.100.130/25 — Which Subnet?

1
/25 block size

256 − 128 = block of 128 → subnets: .0 and .128

2
130 falls in?

130 > 128 → belongs to subnet starting at 192.168.100.128

✅ Result
192.168.100.128
Network ID
192.168.100.255
Broadcast
.129 – .254
Host Range
126
Usable Hosts
🎯

Advanced ConceptVLSM — Variable Length Subnet Masking

VLSM = using different subnet mask sizes within the same network. Unlike fixed subnetting (all subnets same size), VLSM lets you right-size each subnet — assign /25 to a large dept, /28 to a small one, /30 for a point-to-point link.

📌 VLSM Example — Allocate for: Dept A (100 hosts), Dept B (50 hosts), WAN link (2 hosts)

1
Dept A — 100 hosts

Need 2ʰ − 2 ≥ 100 → h=7 → 126 hosts → use /25 (255.255.255.128)

2
Dept B — 50 hosts

Need 2ʰ − 2 ≥ 50 → h=6 → 62 hosts → use /26 (255.255.255.192)

3
WAN link — 2 hosts

2ʰ − 2 ≥ 2 → h=2 → use /30 (255.255.255.252) — classic P2P link

✅ Best Answer: VLSM allocation /25 + /26 + /30
🔗

Opposite of SubnettingSupernetting / Route Aggregation

Supernetting = combining multiple smaller networks into one larger network. Also called Route Aggregation, Route Summarization, or Prefix Aggregation. The subnet mask becomes shorter (fewer network bits). Mainly used by ISPs to reduce routing table size.

✂️ Subnetting

  • Purpose: Divide large network into smaller ones
  • Mask: Becomes LONGER (/24 → /26)
  • Used by: LAN admins, banks, offices
  • Hosts: Per network decreases
  • Borrows: bits FROM host part
  • Result: More networks, fewer hosts each

🔗 Supernetting

  • Purpose: Combine small networks into one large
  • Mask: Becomes SHORTER (/24 → /22)
  • Used by: ISPs, large WAN networks
  • Hosts: Per network increases
  • Returns: bits BACK to host part
  • Result: Fewer routes, larger network each
📌 3 Rules for Supernetting (Must All Be Met)
  • Rule 1 — Consecutive: Networks must be adjacent (no gaps). 192.168.0.0, .1.0, .2.0 ✅ — but not .0.0, .2.0, .3.0 ❌
  • Rule 2 — Same Mask: All networks must have the same prefix length (/24 with /24 etc.)
  • Rule 3 — Boundary Aligned: First network must start at the correct boundary. For 4×/24 networks, must start at multiples of 4 (e.g., .0, .4, .8)
CIDR Aggregation Quick Table (combining /24 networks)
2
/23
512 IPs
4
/22
1,024 IPs
8
/21
2,048 IPs
16
/20
4,096 IPs
32
/19
8,192 IPs
64
/18
16,384 IPs

Practice TheseSupernetting Worked Examples

📌 Combine 192.168.0.0/24 + 192.168.1.0/24

1
Count networks

2 networks → need to reduce prefix by 1 bit (2¹ = 2)

2
New prefix

/24 − 1 = /23

✅ Supernet: 192.168.0.0/23 (covers .0.0 to .1.255 — 510 usable hosts)

📌 Combine 10.10.0.0/24 → 10.10.3.0/24 (4 networks)

1
Count networks

4 networks → reduce prefix by 2 bits (2² = 4)

2
New prefix

/24 − 2 = /22

✅ Supernet: 10.10.0.0/22 (covers 10.10.0.0 to 10.10.3.255)

📌 Combine 192.168.16.0/24 → 192.168.23.0/24 (8 networks)

1
8 networks

Reduce by 3 bits (2³ = 8)

2
New prefix

/24 − 3 = /21

✅ Supernet: 192.168.16.0/21 (covers .16.0 to .23.255)
📝

Tap Any Option to Reveal AnswerMCQ Practice — 50 Questions

Score: 0 / 0
Q.01Basics🔥 Most Asked
IPv4 address is how many bits long?
✔ Correct: B — 32 bits
IPv4 addresses are 32 bits written as 4 octets (4 × 8 = 32). IPv6 uses 128 bits. This is the single most asked networking fact in competitive exams.
Q.02Classes🔥 Most Asked
Which class of IP address has a default subnet mask of 255.255.255.0?
✔ Correct: C — Class C
Class C (192–223) has default mask 255.255.255.0 (/24), giving 254 usable hosts. Class A = 255.0.0.0 (/8). Class B = 255.255.0.0 (/16).
Q.03Formula🔥 Most Asked
The formula for usable hosts per subnet is:
✔ Correct: B — 2^(host bits) − 2
We subtract 2 because the first address (all host bits = 0) is the Network ID and the last address (all host bits = 1) is the Broadcast address — neither can be assigned to a host.
Q.04Calculation🔥 Most Asked
How many usable hosts are in a /26 subnet?
✔ Correct: B — 62
/26 → host bits = 32−26 = 6 → 2⁶ − 2 = 64 − 2 = 62 usable hosts. Mask = 255.255.255.192. Block size = 64.
Q.05Calculation
How many usable hosts does a /29 subnet provide?
✔ Correct: B — 6
/29 → host bits = 32−29 = 3 → 2³ − 2 = 8 − 2 = 6 usable hosts. This subnet is often used for small server clusters.
Q.06CIDR🔥 Most Asked
CIDR stands for:
✔ Correct: B — Classless Inter-Domain Routing
CIDR = Classless Inter-Domain Routing. It replaces the rigid classful system (A/B/C) by allowing variable-length prefixes like /22, /27. Both subnetting and supernetting use CIDR notation.
Q.07Subnet ID🔥 Most Asked
What is the broadcast address of subnet 192.168.1.64/26?
✔ Correct: A — 192.168.1.127
/26 block size = 64. Subnet starts at .64. Broadcast = .64 + 64 − 1 = .127. Host range: .65 – .126.
Q.08Block Size
What is the block size (increment) for a /27 subnet?
✔ Correct: C — 32
/27 mask = 255.255.255.224. Block = 256 − 224 = 32. Subnets start at .0, .32, .64, .96 etc. Usable hosts = 30.
Q.09Subnet ID
IP address 192.168.100.130/25 — which subnet does it belong to?
✔ Correct: B — 192.168.100.128/25
/25 splits at .0 (range 0–127) and .128 (range 128–255). IP .130 falls in the 128–255 range → subnet = 192.168.100.128/25.
Q.10Private IP🔥 Most Asked
Which of these is a PRIVATE IP address?
✔ Correct: B — 192.168.1.1
Private IP ranges: 10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16. 8.8.8.8 = Google DNS (public). 203.0.113.x = TEST-NET (documentation). 4.2.2.2 = public DNS.
Q.11Subnets🔥 Most Asked
How many subnets are created if you borrow 3 bits from the host portion?
✔ Correct: C — 8
Borrowed bits = 3 → subnets = 2³ = 8. For each additional borrowed bit, subnets double: 1 bit=2, 2 bits=4, 3 bits=8, 4 bits=16.
Q.12Mask
255.255.255.240 is equivalent to which CIDR prefix?
✔ Correct: C — /28
240 in binary = 11110000 (4 ones). Total network bits = 24+4 = 28 → /28. Block size = 256−240 = 16. Usable hosts = 14.
Q.13Special IP🔥 Most Asked
What is the loopback IP address range in IPv4?
✔ Correct: A — 127.0.0.0/8
127.0.0.1 (localhost) is the loopback address. The entire 127.x.x.x range is reserved. Used to test TCP/IP stack on the local machine. 169.254.x.x = APIPA (when DHCP fails).
Q.14APIPA
APIPA (Automatic Private IP Addressing) uses which IP range?
✔ Correct: B — 169.254.0.0/16
APIPA = 169.254.0.0/16. A computer auto-assigns an IP in this range when it cannot reach a DHCP server. 10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16 are the three standard private ranges.
Q.15Subnets from /24
How many /28 subnets can be created from a single /24 network?
✔ Correct: C — 16
Borrowed bits = 28 − 24 = 4 → subnets = 2⁴ = 16. Each /28 has 14 usable hosts and block size 16. 16 × 16 = 256 = full /24 space.
Q.16Network ID
What is the Network ID of 192.168.1.0/29 containing IP 192.168.1.10?
✔ Correct: A — 192.168.1.8
/29 block size = 8. Subnets: .0–.7, .8–.15, .16–.23… IP .10 falls in .8–.15 → Network ID = 192.168.1.8, Broadcast = .15, Usable = .9–.14.
Q.17VLSM🔥 Most Asked
VLSM stands for:
✔ Correct: B — Variable Length Subnet Masking
VLSM allows different subnet mask sizes within the same network — /25 for a large dept, /30 for a WAN link. Requires classless routing protocols (OSPF, RIPv2, EIGRP).
Q.18P2P
Which subnet mask is best for a point-to-point WAN link (only 2 hosts needed)?
✔ Correct: C — /30
/30 provides exactly 2 usable hosts (2² − 2 = 2) — perfect for router-to-router links where one IP goes to each end. Mask = 255.255.255.252.
Q.19Hosts Needed
A department needs 200 hosts in one subnet. What is the smallest (most efficient) subnet mask to use?
✔ Correct: A — /24 (254 usable hosts)
Need ≥ 200 hosts. /25 = only 126 (not enough). /24 = 254 (just enough). We use the smallest subnet that fits, which is /24 here.
Q.20Wildcard
What is the wildcard mask for subnet mask 255.255.255.0?
✔ Correct: A — 0.0.0.255
Wildcard mask = inverse of subnet mask = 255.255.255.255 − 255.255.255.0 = 0.0.0.255. Used in Cisco ACLs and routing protocols like OSPF.
Q.21Supernetting🔥 Most Asked
Supernetting is also known as:
✔ Correct: B — Route Aggregation / Route Summarization
Supernetting = Route Aggregation / Route Summarization / Prefix Aggregation. Combines multiple networks to reduce routing table entries. Used mainly by ISPs.
Q.22Supernetting🔥 Most Asked
Combining two /24 networks results in which supernet mask?
✔ Correct: B — /23
2 networks → reduce prefix by 1 bit (2¹=2) → /24 − 1 = /23. Covers 512 IPs total. Example: 192.168.0.0/24 + 192.168.1.0/24 = 192.168.0.0/23.
Q.23Supernetting
What is the supernet of 10.10.0.0/24 through 10.10.3.0/24?
✔ Correct: B — 10.10.0.0/22
4 networks (.0, .1, .2, .3) → reduce by 2 bits (2²=4) → /24 − 2 = /22. The supernet covers 10.10.0.0 to 10.10.3.255.
Q.24Supernetting
Combining 8 Class C (/24) networks creates a supernet with mask:
✔ Correct: B — /21
8 networks = 2³ → reduce by 3 bits → /24 − 3 = /21. Rule: combining N networks where N=2^n, reduce prefix by n bits.
Q.25Supernetting Rule
Which condition is NOT required for supernetting?
✔ Correct: C — same country (irrelevant!)
The 3 rules for supernetting: (1) Consecutive networks, (2) Same mask, (3) Boundary aligned. Geography has nothing to do with it.
Q.26Supernetting
In supernetting the subnet mask becomes:
✔ Correct: B — Shorter (fewer network bits)
Supernetting = shorter prefix (e.g., /24 → /22). Subnetting = longer prefix (e.g., /24 → /26). Shorter prefix = larger network = more hosts.
Q.27IP Class🔥 Most Asked
What class does IP 172.20.5.1 belong to?
✔ Correct: B — Class B
First octet = 172 → falls in Class B range (128–191). 172.16.0.0/12 is also a private Class B range. Default mask = 255.255.0.0 (/16).
Q.28Broadcast
The broadcast address sends data to:
✔ Correct: B — All hosts in the subnet
Broadcast address = last IP in the subnet (all host bits = 1). Sends to all devices within that subnet. Cannot be assigned to a single host.
Q.29Four Subnets
Company has 192.168.0.0/24 and needs exactly 4 equal subnets. New mask?
✔ Correct: C — /26
4 subnets = 2² → borrow 2 bits → /24 + 2 = /26. Each /26 subnet has 62 usable hosts and block size 64.
Q.30ISP
Supernetting is most commonly used by:
✔ Correct: C — ISPs
ISPs use supernetting to advertise fewer, larger routes to the internet, keeping routing tables small. Also used in MPLS/WAN networks and bank data centres.
Q.31Aggregation
16 networks of /24 combined gives supernet mask:
✔ Correct: A — /20
16 = 2⁴ → reduce by 4 bits → /24 − 4 = /20. Covers 4,096 IPs total (16 × 256).
Q.32Mask value
255.255.252.0 in CIDR notation is:
✔ Correct: B — /22
252 = 11111100 in binary = 6 ones. Network bits = 16 + 6 = 22 → /22. This is the supernet for 4 Class C (/24) networks.
Q.33Subnet reduce
What does subnetting reduce in a network?
✔ Correct: B — Broadcast domain size
Subnetting splits a big broadcast domain into smaller ones, reducing unnecessary broadcast traffic. This improves performance — fewer devices receive unwanted broadcast packets.
Q.34NAT
Which mechanism translates private IP addresses to public IP addresses?
✔ Correct: C — NAT (Network Address Translation)
NAT maps private IPs (192.168.x.x) to a public IP for internet access. Enables many devices to share one public IP. Standard feature on all home routers.
Q.35Network ID
IP: 203.0.113.5, Mask: 255.255.255.252 — Network ID is:
✔ Correct: B — 203.0.113.4
/30 block = 4. Subnets: .0–.3, .4–.7, .8–.11... IP .5 is in .4 block → Network ID = 203.0.113.4, Broadcast = .7.
Q.36Banking use
In banking, which is NOT a typical use of subnetting?
✔ Correct: C — Printing cheques
Banks use subnetting for ATM networks, branch LANs, server isolation, VPN segments, firewall zones, and VoIP networks. Cheque printing is a physical process unrelated to IP subnetting.
Q.37Hosts from /23
Network 172.16.0.0/23 has how many usable hosts?
✔ Correct: C — 510
/23 → host bits = 32 − 23 = 9 → 2⁹ − 2 = 512 − 2 = 510 usable hosts. A /23 covers two /24 networks combined.
Q.38Class D
Class D IP addresses (224–239) are used for:
✔ Correct: A — Multicast
Class D (224.0.0.0–239.255.255.255) is reserved for multicast — sending to a group of hosts simultaneously. Class E (240–255) is for experimental/research use.
Q.39Supernetting
192.168.0.0/24 to 192.168.3.0/24 combined → supernet is:
✔ Correct: A — 192.168.0.0/22
Networks .0, .1, .2, .3 = 4 consecutive /24s → reduce by 2 bits → /22.
Q.40Network ID
For IP 192.168.5.200/26, the Network ID is:
✔ Correct: A — 192.168.5.192
/26 blocks: .0, .64, .128, .192. IP .200 falls in .192–.255 range → Network ID = 192.168.5.192. Broadcast = .255. Hosts: .193–.254.
Q.41Supernet mask
255.255.252.0 is the supernet mask for how many combined /24 networks?
✔ Correct: C — 4 networks
255.255.252.0 = /22. /22 − /24 = 2 bits → 2² = 4 networks combined.
Q.42Combining /16
The combined supernet of 172.16.0.0/16 and 172.17.0.0/16 is:
✔ Correct: A — 172.16.0.0/15
2 × /16 → reduce by 1 bit → /15. Covers 172.16.0.0 to 172.17.255.255.
Q.438 subnets from /24
What mask creates exactly 8 subnets from a /24 network?
✔ Correct: A — /27
8 subnets = 2³ → borrow 3 bits → /24 + 3 = /27. Each /27 = 30 usable hosts, block size 32.
Q.44Which class?
IP address 10.0.0.1 belongs to which class?
✔ Correct: A — Class A
First octet = 10 → Class A range (1–126). 10.0.0.0/8 is the largest private IP range (Class A private). Default mask = 255.0.0.0 (/8).
Q.45500 hosts
You need a subnet for exactly 500 hosts from 10.0.0.0/16. Best mask?
✔ Correct: A — /23 (510 usable hosts)
/24 = only 254 (too few). /23 → 9 host bits → 2⁹ − 2 = 510 hosts (fits 500+). /22 = 1022 (wasteful). Use the smallest subnet that fits the requirement.
Q.46Supernetting reduces
Supernetting primarily reduces:
✔ Correct: B — Routing table entries
Supernetting's primary purpose: combine many small routes into one → fewer routing table entries → faster router lookups → better internet performance.
Q.47CIDR / classful
Supernetting works with which type of addressing?
✔ Correct: B — CIDR (Classless)
Supernetting requires CIDR — classful addressing is too rigid to support shorter-than-default prefixes. CIDR enables any prefix length (/8 to /30 etc.).
Q.48Boundary
When combining 4 × /24 networks via supernetting, the first network must start at a multiple of:
✔ Correct: B — 4
Combining 4 /24 networks requires the first to start at a boundary of 4 (e.g., .0, .4, .8, .12, .16…). 2 networks = multiple of 2; 8 networks = multiple of 8.
Q.49IPv4 exhaustion
The total number of IPv4 addresses available is approximately:
✔ Correct: B — ~4.3 billion (2³²)
IPv4 = 32 bits → 2³² = ~4.29 billion addresses. IPv6 = 128 bits → 2¹²⁸ ≈ 340 undecillion. IPv4 addresses were officially exhausted by IANA in 2011.
Q.50VLSM best answer🔥 Most Asked
Depts need 100, 50 and 12 hosts. Which approach is BEST?
✔ Correct: B — VLSM /25 + /26 + /28
VLSM lets you right-size: /25 (126 hosts → fits 100), /26 (62 hosts → fits 50), /28 (14 hosts → fits 12). Efficient, no wastage. This is the design question examiners love.

Last-Minute PrepQuick Revision Flash Cards

🌐 IPv4 Basics

  • IPv4 = 32 bits, 4 octets
  • IPv6 = 128 bits
  • Total IPv4 = 2³² ≈ 4.3 billion
  • Octet range: 0–255
  • CIDR = Classless Inter-Domain Routing

📊 IP Classes

  • A: 1–126 | /8 | 16.7M hosts
  • B: 128–191 | /16 | 65,534 hosts
  • C: 192–223 | /24 | 254 hosts
  • D: 224–239 | Multicast
  • E: 240–255 | Experimental
  • 127.x.x.x = Loopback (not A)

🔒 Private & Special IPs

  • 10.0.0.0/8 — Class A private
  • 172.16.0.0/12 — Class B private
  • 192.168.0.0/16 — Class C private
  • 127.0.0.1 — Loopback
  • 169.254.0.0/16 — APIPA

🧮 Key Formulas

  • Hosts = 2^h − 2 (h = host bits)
  • Subnets = 2^b (b = borrowed bits)
  • Host bits = 32 − prefix
  • Block = 256 − mask octet
  • Wildcard = 255.255.255.255 − mask

📋 Mask Reference

  • /24 → 254 hosts | block 256
  • /25 → 126 | block 128
  • /26 → 62 | block 64
  • /27 → 30 | block 32
  • /28 → 14 | block 16
  • /29 → 6 | /30 → 2

✂️ Subnetting

  • Divides large network → smaller
  • Mask becomes LONGER
  • Reduces broadcast domain
  • Better security & management
  • Banks use for ATM, branch, server nets

🔗 Supernetting

  • Combines small networks → one large
  • Mask becomes SHORTER
  • = Route Aggregation / Summarization
  • Reduces routing table entries
  • Used by ISPs and WAN networks
  • Requires CIDR (not classful)

📐 Aggregation Table

  • 2 × /24 → /23
  • 4 × /24 → /22
  • 8 × /24 → /21
  • 16 × /24 → /20
  • 32 × /24 → /19
  • 64 × /24 → /18

🎯 VLSM

  • = Variable Length Subnet Masking
  • Different mask sizes in same network
  • Most efficient IP allocation
  • Needs classless routing (OSPF, RIPv2)
  • P2P link → always /30

⭐ Exam Favourites

  • /24 = 254 hosts (most asked)
  • /30 = 2 hosts = P2P link
  • Loopback = 127.0.0.1
  • APIPA = 169.254.x.x
  • CIDR = Classless Inter-Domain Routing
  • 4 × /24 = /22
📌 Must-Know Full Forms
CIDRVLSMNAT APIPADHCPACL IPv4 / IPv6OSPFRIPv2 MPLSWANVPN ISPEIGRP
FeatureFixed SubnettingVLSM
Mask sizeSame for all subnetsDifferent for each subnet
IP efficiencyWastes addressesVery efficient
Routing protocolWorks with classfulRequires classless (OSPF, EIGRP, RIP v2)
Use caseSimple networksEnterprise / complex networks