M2: Computer Network (Subnetting, CIDR, VLSM)

Computer Network Classfull and Classless IP Address, Subnetting, CIDR, VLSM (ET: 1.5 Hours)

Video Tutorial: Theory-- Classfull and Classless IP Address, Subnetting, CIDR, VLSM

IP Class, Subnetting, VLSM

Classfull IP Addressing

Classful IP Addressing: Classful IP addressing is a method of dividing IPv4 addresses into different classes based on the value of the first few bits of the address. IPv4 addresses are divided into five classes: Class A, Class B, Class C, Class D, and Class E.
1. Classes of IPv4 Address

Class	First Octet Range	Default Subnet Mask	Network Bits	Host Bits	Main Use
A	0 – 127	255.0.0.0	8	24	Large networks
B	128 – 191	255.255.0.0	16	16	Medium-sized networks
C	192 – 223	255.255.255.0	24	8	Small networks
D	224 – 239	Not applicable	Multicast		Multicasting
E	240 – 255	Not applicable	Experimental		Experimental / Future use

Note: Class D and Class E do not use the normal network/host division or default subnet masks like Classes A, B, and C.
2. Class A
Class A addresses have a first-octet range of 0–127. The first octet represents the network portion, while the remaining three octets represent the host portion.
Default Subnet Mask: 255.0.0.0
Example: 55.123.145.67
Here, 55 lies between 0 and 127, so it is a Class A address.
Network = 55
Host = 123.145.67
3. Class B
Class B addresses have a first-octet range of 128–191. The first two octets represent the network portion, while the last two octets represent the host portion.
Default Subnet Mask: 255.255.0.0
Example: 172.16.25.10
Here, 172 lies between 128 and 191, so it is a Class B address.
Network = 172.16
Host = 25.10
4. Class C
Class C addresses have a first-octet range of 192–223. The first three octets represent the network portion, while the last octet represents the host portion.
Default Subnet Mask: 255.255.255.0
Example: 192.168.1.254
Here, 192 lies between 192 and 223, so it is a Class C address.
Network = 192.168.1
Host = 254
5. Class D
Class D addresses have a first-octet range of 224–239. They are mainly used for multicasting, where data is sent to a group of devices.
Example: Live video streaming, video conferencing, and other multicast applications.
6. Class E
Class E addresses have a first-octet range of 240–255. They are reserved mainly for experimental or research purposes and are not normally used for ordinary host addressing.
7. Private IP Address Ranges
Some IPv4 address ranges are reserved for private networks such as home Wi-Fi, LANs, and internal organizational networks.

Class	Private IP Range
Class A	10.0.0.0 – 10.255.255.255
Class B	172.16.0.0 – 172.31.255.255
Class C	192.168.0.0 – 192.168.255.255

8. Important Formulas for Classful Addressing
Class A:
Number of host bits = 24
Total addresses per network = 2²⁴
Usable hosts per network = 2²⁴ − 2
Class B:
Number of host bits = 16
Total addresses per network = 2¹⁶
Usable hosts per network = 2¹⁶ − 2
Class C:
Number of host bits = 8
Total addresses per network = 2⁸
Usable hosts per network = 2⁸ − 2 = 254
General Formula:
Total addresses = 2^h
Usable host addresses = 2^h − 2
Here, h = number of host bits.
9. How to Identify the Class of an IP Address
Look at the first octet of the IPv4 address.

First Octet	Class
0–127	Class A
128–191	Class B
192–223	Class C
224–239	Class D
240–255	Class E

Examples:
55.123.145.67 → Class A
172.16.25.10 → Class B
192.168.1.254 → Class C
224.x.x.x → Class D
240.x.x.x → Class E
Exam Tip: For Class A, B, and C, remember the default masks: A = 255.0.0.0, B = 255.255.0.0, C = 255.255.255.0.

2
Classless Addressing
What is Classless Addressing?
Classless Addressing is an IP addressing method in which IP addresses are not divided into fixed classes such as Class A, Class B, Class C, Class D, and Class E.
In classless addressing, the network portion of an IP address is identified using a prefix length, represented by /n. The value of n indicates the number of bits used to identify the network or block.
For example, 200.10.0.0/28 means that the first 28 bits are used to identify the network, while the remaining 4 bits are used for hosts.
Classless Addressing was introduced to overcome the limitations of classful addressing and provides more efficient, flexible, and scalable IP address allocation.
Why Classless Addressing is Needed?
In classful addressing, IP addresses were divided into fixed classes and each class had a fixed default subnet mask. This often resulted in inefficient use of IP addresses.
For example, a Class A or Class B network can provide a very large number of host addresses, but many organizations may not actually require that many hosts. As a result, a large number of IP addresses may be wasted.
3
Advantages of Classless Addressing
1. Efficient IP Address Allocation — IP addresses can be allocated according to actual requirements, reducing address wastage.
2. Flexible — There is no fixed requirement to use Class A, B, or C network sizes.
3. Scalable — It can support the continuous growth of networks and the Internet.
4. Simplified Routing — Classless addressing can reduce the size and complexity of routing tables.
5. Supports CIDR — Classless addressing supports Classless Inter-Domain Routing (CIDR).
6. Supports VLSM — It also supports Variable Length Subnet Masking (VLSM), allowing different subnet sizes according to requirements.
7. Route Aggregation — Multiple network routes can be combined into a single summarized route, helping reduce routing table size.
4
Variable Length Subnet Masking (VLSM)
Variable Length Subnet Masking (VLSM) is a subnetting technique that allows a Class A, Class B, or Class C network to be divided into subnets of different sizes.
In traditional subnetting, a network is divided into equal-sized subnets. For example, a network can be divided into 2, 4, or 8 equal subnets. Each subnet has the same number of IP addresses.
In VLSM, the network can be divided into subnets according to the actual host requirements. Therefore, different subnets can have different numbers of IP addresses.
For example, one subnet may be created for 100 hosts, another for 50 hosts, and another for 20 hosts. This provides more efficient utilization of IP addresses.
Why is VLSM Needed?
In normal subnetting, all subnets are of equal size. This can cause IP address wastage when different parts of a network require different numbers of hosts.
For example, if one department requires 100 hosts and another department requires only 20 hosts, allocating the same number of IP addresses to both departments would waste many IP addresses.
VLSM solves this problem by allowing each subnet to have a size based on its actual requirement.

Variable Length Subnet Masking (VLSM) হলো এমন একটি subnetting technique যার মাধ্যমে একটি Class A, Class B অথবা Class C network-কে ভিন্ন ভিন্ন আকারের subnet-এ ভাগ করা যায়।
সাধারণ subnetting-এ একটি network-কে সমান আকারের subnet-এ ভাগ করা হয়। যেমন একটি network-কে 2, 4 অথবা 8টি সমান subnet-এ ভাগ করা যায়। ফলে প্রতিটি subnet-এ সমান সংখ্যক IP address থাকে।
কিন্তু VLSM-এ subnet-গুলো সমান আকারের হওয়া বাধ্যতামূলক নয়। Network-এর প্রকৃত host requirement অনুযায়ী প্রতিটি subnet-এর size নির্ধারণ করা যায়।
উদাহরণস্বরূপ, একটি subnet-এ 100টি host, অন্য একটি subnet-এ 50টি host এবং আরেকটি subnet-এ 20টি host রাখা যেতে পারে। এর ফলে IP address-এর অপচয় কম হয়।
VLSM কেন প্রয়োজন?
সাধারণ subnetting-এ সব subnet-এর size সমান হয়। কিন্তু বাস্তবে বিভিন্ন network বা department-এর host requirement সমান নাও হতে পারে। তাই একই size-এর subnet ব্যবহার করলে অনেক IP address অব্যবহৃত থেকে যায়।
যেমন, একটি department-এর জন্য 100টি host-এর প্রয়োজন হলেও অন্য একটি department-এর জন্য মাত্র 20টি host-এর প্রয়োজন হতে পারে। উভয় department-কে একই size-এর subnet দিলে অতিরিক্ত IP address নষ্ট হবে।
VLSM এই সমস্যার সমাধান করে। এটি host requirement অনুযায়ী subnet-এর size নির্ধারণ করতে দেয়।

Class	First Octet Range	Default Subnet Mask	Network Bits	Host Bits	ব্যবহার
A	0 – 127	255.0.0.0	8	24	বড় Network
B	128 – 191	255.255.0.0	16	16	মাঝারি Network
C	192 – 223	255.255.255.0	24	8	ছোট Network
D	224 – 239	প্রযোজ্য নয়	Multicast	Multicasting
E	240 – 255	প্রযোজ্য নয়	Experimental	পরীক্ষামূলক ব্যবহার

Subnetting Mathematical Problem

- 1
  What is Subnetting?Steps of Subnetting (মুখস্ত করার দরকার নেই স্টেপ গুলো ম্যাথ করতে কাজে লাগবে)
  Subnetting is the process of dividing a single IP network into two or more smaller logical networks, called subnets.
  A subnet is a logical subdivision of an IP network. It allows a large network to be divided into smaller networks so that devices can be organized into separate logical networks.
  1. Identify the class of the IP address — Determine whether the given IP address belongs to Class A, Class B, or Class C and note its default subnet mask.
  2. Convert the default subnet mask into binary — Convert the subnet mask into its binary representation to understand the network and host portions.
  3. Determine the required hosts per subnet — Identify how many hosts are required in each subnet and determine the appropriate subnet generator and octet position.
  4. Generate the new subnet mask — Create a new subnet mask by borrowing appropriate bits from the host portion of the IP address.
  5. Determine the subnet ranges — Use the subnet generator to calculate the network ranges, network addresses, usable host addresses, and broadcast addresses for each subnet.
- 2
  Subnet the IP address 216.21.5.0 into 30 hosts in each subnet. (With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 1
  Subnetting Solution: 216.21.5.0
  Step 1: Host Bits Needed
  Formula: 2^h - 2 ≥ 30 hosts (where h is the number of host bits).
  For h = 5: 2⁵ - 2 = 32 - 2 = 30 usable hosts per subnet.
  Therefore, 5 host bits are allocated.
  Step 2: Subnet Mask & CIDR
  Total bits in IPv4 address = 32 bits.
  Network Prefix (CIDR) = 32 - 5 host bits = /27.
  Subnet Mask in Binary: 11111111.11111111.11111111.11100000.
  Subnet Mask in Decimal: 255.255.255.224.
  Step 3: Block Size (Magic Number)
  Block Size = 256 - 224 = 32 (or 2⁵ = 32).
  Each subnet increments by 32 in the 4th octet.
  Step 4: Total Subnets Generated
  Class C default prefix = /24. Subnet bits borrowed = 27 - 24 = 3 bits.
  Total Subnets = 2³ = 8 subnets created.
  Step 5: Generated Subnet Ranges
  Subnet 1: 216.21.5.0 - 216.21.5.31
  Subnet 2: 216.21.5.32 - 216.21.5.63
  Subnet 3: 216.21.5.64 - 216.21.5.95
  Subnet 4: 216.21.5.96 - 216.21.5.127
  Subnet 5: 216.21.5.128 - 216.21.5.159
  Subnet 6: 216.21.5.160 - 216.21.5.191
  Subnet 7: 216.21.5.192 - 216.21.5.223
  Subnet 8: 216.21.5.224 - 216.21.5.255
  Step 6: First Subnet Details
  Network Address: 216.21.5.0
  First Usable IP: 216.21.5.1
  Last Usable IP: 216.21.5.30
  Broadcast Address: 216.21.5.31
- 3
  Subnet the IP address 196.10.20.0 into 52 hosts in each subnet.(With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 2
  Subnetting Solution: 196.10.20.0
  Step 1: Host Bits Needed
  Formula: 2^h - 2 ≥ 52 hosts (where h is the number of host bits).
  For h = 6: 2⁶ - 2 = 64 - 2 = 62 usable hosts per subnet.
  Therefore, 6 host bits are allocated.
  Step 2: Subnet Mask & CIDR
  Total bits in IPv4 address = 32 bits.
  Network Prefix (CIDR) = 32 - 6 host bits = /26.
  Subnet Mask in Binary: 11111111.11111111.11111111.11000000.
  Subnet Mask in Decimal: 255.255.255.192.
  Step 3: Block Size (Magic Number)
  Block Size = 256 - 192 = 64 (or 2⁶ = 64).
  Each subnet increments by 64 in the 4th octet.
  Step 4: Total Subnets Generated
  Class C default prefix = /24. Subnet bits borrowed = 26 - 24 = 2 bits.
  Total Subnets = 2² = 4 subnets created.
  Step 5: Generated Subnet Ranges
  Subnet 1: 196.10.20.0 - 196.10.20.63
  Subnet 2: 196.10.20.64 - 196.10.20.127
  Subnet 3: 196.10.20.128 - 196.10.20.191
  Subnet 4: 196.10.20.192 - 196.10.20.255
  Step 6: First Subnet Details
  Network Address: 196.10.20.0
  First Usable IP: 196.10.20.1
  Last Usable IP: 196.10.20.62
  Broadcast Address: 196.10.20.63
- 4
  Subnet the IP address 150.15.0.0 into 500 hosts in each subnet.(With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 3
  Subnetting Solution: 150.15.0.0
  Step 1: Host Bits Needed
  Formula: 2^h - 2 ≥ 500 hosts (where h is the number of host bits).
  For h = 9: 2⁹ - 2 = 512 - 2 = 510 usable hosts per subnet.
  Therefore, 9 host bits are allocated.
  Step 2: Subnet Mask & CIDR
  Total bits in IPv4 address = 32 bits.
  Network Prefix (CIDR) = 32 - 9 host bits = /23.
  Subnet Mask in Binary: 11111111.11111111.11111110.00000000.
  Subnet Mask in Decimal: 255.255.254.0.
  Step 3: Block Size (Magic Number)
  Block Size = 256 - 254 = 2 in the 3rd octet.
  Each subnet increments by 2 in the 3rd octet.
  Step 4: Total Subnets Generated
  Class B default prefix = /16. Subnet bits borrowed = 23 - 16 = 7 bits.
  Total Subnets = 2⁷ = 128 subnets created.
  Step 5: First Few Subnet Ranges
  Subnet 1: 150.15.0.0 - 150.15.1.255
  Subnet 2: 150.15.2.0 - 150.15.3.255
  Subnet 3: 150.15.4.0 - 150.15.5.255
  Subnet 4: 150.15.6.0 - 150.15.7.255
  Subnet 128 (Last): 150.15.254.0 - 150.15.255.255
  Step 6: First Subnet Details
  Network Address: 150.15.0.0
  First Usable IP: 150.15.0.1
  Last Usable IP: 150.15.1.254
  Broadcast Address: 150.15.1.255
- 5
  Subnet the IP address 10.0.0.0 into 100 hosts in each subnet.(With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 4
  Subnetting Solution: 10.0.0.0
  Step 1: Host Bits Needed
  Formula: 2^h - 2 ≥ 100 hosts (where h is the number of host bits).
  For h = 7: 2⁷ - 2 = 128 - 2 = 126 usable hosts per subnet.
  Therefore, 7 host bits are allocated.
  Step 2: Subnet Mask & CIDR
  Total bits in IPv4 address = 32 bits.
  Network Prefix (CIDR) = 32 - 7 host bits = /25.
  Subnet Mask in Binary: 11111111.11111111.11111111.10000000.
  Subnet Mask in Decimal: 255.255.255.128.
  Step 3: Block Size (Magic Number)
  Block Size = 256 - 128 = 128 (or 2⁷ = 128).
  Each subnet increments by 128 in the 4th octet (rolling over into the 3rd octet).
  Step 4: Total Subnets Generated
  Class A default prefix = /8. Subnet bits borrowed = 25 - 8 = 17 bits.
  Total Subnets = 2¹⁷ = 131,072 subnets created.
  Step 5: First Few Subnet Ranges
  Subnet 1: 10.0.0.0 - 10.0.0.127
  Subnet 2: 10.0.0.128 - 10.0.0.255
  Subnet 3: 10.0.1.0 - 10.0.1.127
  Subnet 4: 10.0.1.128 - 10.0.1.255
  Subnet 131072 (Last): 10.255.255.128 - 10.255.255.255
  Step 6: First Subnet Details
  Network Address: 10.0.0.0
  First Usable IP: 10.0.0.1
  Last Usable IP: 10.0.0.126
  Broadcast Address: 10.0.0.127
- 6
  In a block of addresses, we know the IP address of one host is 25.34.12.56/16. What is the first address (network address) in this block?(With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 5
  Network Address Solution: 25.34.12.56/16
  Step 1: Identify CIDR Prefix
  Prefix = /16 (the first 16 bits represent the network portion).
  Host Bits = 32 - 16 = 16 bits (the remaining 16 bits represent the host portion).
  Step 2: Determine Subnet Mask
  Subnet Mask in Binary: 11111111.11111111.00000000.00000000
  Subnet Mask in Decimal: 255.255.0.0
  Step 3: Bitwise AND Operation
  Keep the first 2 octets (16 bits) unchanged: 25.34
  Set all host bits (last 2 octets) to zero: 0.0
  Step 4: Final Result
  First Address (Network Address): 25.34.0.0
  Full Notation: 25.34.0.0/16
- 7
  In a block of addresses, we know the IP address of one host is 182.44.82.16/26. What is the first address (network address) in this block?
  Network Address Solution: 182.44.82.16/26
  Step 1: Identify CIDR Prefix
  Prefix = /26 (the first 26 bits represent the network portion).
  Host Bits = 32 - 26 = 6 bits (the remaining 6 bits represent the host portion).
  Step 2: Determine Subnet Mask & Block Size
  Subnet Mask in Binary: 11111111.11111111.11111111.11000000
  Subnet Mask in Decimal: 255.255.255.192
  Block Size (Magic Number) = 256 - 192 = 64 (or 2⁶ = 64).
  Step 3: Calculate Subnet Interval
  Subnets in the 4th octet increment by 64: 0, 64, 128, 192.
  The host IP 4th octet is 16, which falls in the range 0 to 63.
  Step 4: Final Result
  First Address (Network Address): 182.44.82.0
  Full Notation: 182.44.82.0/26
- 8
  In a block of addresses, we know the IP address of one host is 25.34.12.56/16. What is the last address (limited broadcast address) in this block? (With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 6
  Broadcast Address Solution: 25.34.12.56/16
  Step 1: Identify CIDR Prefix
  Prefix = /16 (the first 16 bits represent the network portion).
  Host Bits = 32 - 16 = 16 bits (the remaining 16 bits represent the host portion).
  Step 2: Determine Network Address
  Keep the first 2 octets (16 bits) unchanged: 25.34
  Set all 16 host bits to 0: 25.34.0.0 (Network Address).
  Step 3: Calculate Broadcast Address
  To find the last address in the block, set all 16 host bits to 1 (binary: 11111111.11111111).
  The last 2 octets become 255.255.
  Step 4: Final Result
  Last Address (Broadcast Address): 25.34.255.255
  Full Notation: 25.34.255.255/16
- 9
  In a block of addresses, we know the IP address of one host is 182.44.82.16/26. What is the last address (limited broadcast address) in this block?
  Broadcast Address Solution: 182.44.82.16/26
  Step 1: Identify CIDR Prefix
  Prefix = /26 (the first 26 bits represent the network portion).
  Host Bits = 32 - 26 = 6 bits (the remaining 6 bits represent the host portion).
  Step 2: Determine Network Address
  Subnet Mask in Decimal: 255.255.255.192 (Block size = 256 - 192 = 64).
  Host 4th octet is 16, placing it in the range 0 to 63.
  Network Address = 182.44.82.0.
  Step 3: Calculate Broadcast Address
  To find the last address, set all 6 host bits to 1 (Network + Block Size - 1).
  4th Octet = 0 + 64 - 1 = 63.
  Step 4: Final Result
  Last Address (Broadcast Address): 182.44.82.63
  Full Notation: 182.44.82.63/26
- 10
  An organization has a class B network and wishes to form subnets for 64 departments. Find The Subnet Mask(With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 7
  Subnet Mask Solution: Class B (64 Departments)
  Step 1: Class B Default Prefix
  Default Class B Network Prefix = /16 (16 network bits).
  Default Subnet Mask = 255.255.0.0.
  Step 2: Subnet Bits Needed
  Formula: 2^s ≥ 64 subnets (where s is the borrowed subnet bits).
  For s = 6: 2⁶ = 64 subnets.
  Therefore, 6 subnet bits are borrowed.
  Step 3: New CIDR Prefix
  New Prefix Length = Default Prefix + Borrowed Bits = 16 + 6 = /22.
  Remaining Host Bits = 32 - 22 = 10 host bits per subnet.
  Step 4: Calculate Subnet Mask
  Binary Mask: 11111111.11111111.11111100.00000000
  3rd Octet Calculation: 128 + 64 + 32 + 16 + 8 + 4 = 252.
  Dotted Decimal Mask: 255.255.252.0.
  Step 5: Subnet Capacity Details
  Usable Hosts per Subnet = 2¹⁰ - 2 = 1024 - 2 = 1,022 hosts.
  Block Size (Magic Number) = 256 - 252 = 4 in the 3rd octet.
  Step 6: Final Answer
  Subnet Mask: 255.255.252.0
  CIDR Notation: /22
- 11
  If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?(With Video Learn Every Step Carefully)
  🎥 Video Solution: Subnetting Problem 8
  Maximum Hosts Solution: Mask 255.255.248.0
  Step 1: Convert Subnet Mask to Binary
  Dotted Decimal: 255.255.248.0
  Binary Representation: 11111111.11111111.11111000.00000000
  Prefix Length (CIDR): /21 (21 1-bits in mask).
  Step 2: Determine Host Bits
  Total Bits in IPv4 Address = 32 bits.
  Host Bits (h) = 32 - Network Bits = 32 - 21 = 11 bits (0-bits in mask).
  Step 3: Calculate Total Addresses
  Total IP Addresses per Subnet = 2^h = 2¹¹ = 2,048 addresses.
  Step 4: Calculate Usable Hosts
  Formula: Usable Hosts = 2^h - 2
  Subtract 1 address for the Network Address.
  Subtract 1 address for the Broadcast Address.
  Usable Hosts = 2048 - 2 = 2,046 hosts.
  Step 5: Final Result
  Maximum Usable Hosts per Subnet: 2,046
  Total IP Addresses per Subnet: 2,048
- 12
  The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?
  🎥 Video Solution: Subnetting Problem 9
  Class B Subnetting Solution: 6-Bit Subnet Number
  Step 1: Identify Default Class B Bits
  Network Bits = 16 bits (Default Class B prefix is /16).
  Default Host Bits = 32 - 16 = 16 bits.
  Step 2: Calculate Maximum Subnets
  Subnet Bits (s) = 6 bits.
  Formula: Maximum Subnets = 2^s = 2⁶ = 64 subnets.
  Usabe Subnet =64-2=62
  Step 3: Calculate Remaining Host Bits
  Remaining Host Bits (h) = 16 - 6 = 10 bits.
  New Network Prefix = 16 + 6 = /22.
  Step 4: Calculate Hosts per Subnet
  Total IP Addresses per Subnet = 2¹⁰ = 1,024 addresses.
  Usable Hosts per Subnet = 2¹⁰ - 2 = 1024 - 2 = 1,022 hosts.
  Step 5: Final Result
  Maximum Number of Subnets: 64
  Maximum Usable Hosts per Subnet: 1,022 (Total 1,024 IP addresses)
  Subnet Mask: 255.255.252.0 (/22)
- 13
  Given the IP network 172.11.0.0/16, use VLSM (Variable Length Subnet Masking) to allocate IP addresses to the following departments according to their host requirements:
  Department Required Hosts
  Teacher Department 100
  O/S Lab 90
  Network (NW) Lab 50
  Staff 10
  Find the following for each subnet:
  Network Address
  Prefix Length / Subnet Mask
  Total IP Addresses
  Usable Host Addresses
  First Usable IP Address
  Last Usable IP Address
  Broadcast Address
  VLSM Solution: Network 172.11.0.0/16
  1. Teacher Department
  Required Hosts: 100 (Needs 7 host bits: 2⁷ = 128 ≥ 100 + 2)
  Prefix / Subnet Mask: /25 (255.255.255.128)
  Total IPs / Usable Hosts: 128 / 126
  Network Address: 172.11.0.0
  First Usable IP: 172.11.0.1
  Last Usable IP: 172.11.0.126
  Broadcast Address: 172.11.0.127
  2. O/S Lab
  Required Hosts: 90 (Needs 7 host bits: 2⁷ = 128 ≥ 90 + 2)
  Prefix / Subnet Mask: /25 (255.255.255.128)
  Total IPs / Usable Hosts: 128 / 126
  Network Address: 172.11.0.128
  First Usable IP: 172.11.0.129
  Last Usable IP: 172.11.0.254
  Broadcast Address: 172.11.0.255
  3. Network (NW) Lab
  Required Hosts: 50 (Needs 6 host bits: 2⁶ = 64 ≥ 50 + 2)
  Prefix / Subnet Mask: /26 (255.255.255.192)
  Total IPs / Usable Hosts: 64 / 62
  Network Address: 172.11.1.0
  First Usable IP: 172.11.1.1
  Last Usable IP: 172.11.1.62
  Broadcast Address: 172.11.1.63
  4. Staff Department
  Required Hosts: 10 (Needs 4 host bits: 2⁴ = 16 ≥ 10 + 2)
  Prefix / Subnet Mask: /28 (255.255.255.240)
  Total IPs / Usable Hosts: 16 / 14
  Network Address: 172.11.1.64
  First Usable IP: 172.11.1.65
  Last Usable IP: 172.11.1.78
  Broadcast Address: 172.11.1.79
  Allocation Summary
  Teacher: 172.11.0.0/25 (172.11.0.1 - 172.11.0.126)
  O/S Lab: 172.11.0.128/25 (172.11.0.129 - 172.11.0.254)
  NW Lab: 172.11.1.0/26 (172.11.1.1 - 172.11.1.62)
  Staff: 172.11.1.64/28 (172.11.1.65 - 172.11.1.78)
- 14
  If you are assigned an IP address 200.160.93.0/19 and asked to deploy VLSM, there are some requirements as follows:
  Amberkhana: 800 hosts
  Zindabazar: 90 hosts
  Lamabazar: 1000 hosts
  Subidbazar: 150 hosts
  Now assign network address for each area, also find broadcast address of those areas.
  🎥 Video Solution: VLSM Solution
  VLSM Solution: Network 200.160.93.0/19
  Base Network Address Calculation:
  The host IP 200.160.93.0/19 has a 3rd octet of 93. With a /19 prefix, the block size in the 3rd octet is 32 (range increments: 0, 32, 64, 96...). Thus, 93 falls into the 200.160.64.0/19 network block.
  1. Lamabazar
  Required Hosts: 1000 (Needs 10 host bits: 2¹⁰ = 1024 ≥ 1000 + 2)
  Prefix / Subnet Mask: /22 (255.255.252.0)
  Total IPs / Usable Hosts: 1024 / 1022
  Network Address: 200.160.64.0
  First Usable IP: 200.160.64.1
  Last Usable IP: 200.160.67.254
  Broadcast Address: 200.160.67.255
  2. Amberkhana
  Required Hosts: 800 (Needs 10 host bits: 2¹⁰ = 1024 ≥ 800 + 2)
  Prefix / Subnet Mask: /22 (255.255.252.0)
  Total IPs / Usable Hosts: 1024 / 1022
  Network Address: 200.160.68.0
  First Usable IP: 200.160.68.1
  Last Usable IP: 200.160.71.254
  Broadcast Address: 200.160.71.255
  3. Subidbazar
  Required Hosts: 150 (Needs 8 host bits: 2⁸ = 256 ≥ 150 + 2)
  Prefix / Subnet Mask: /24 (255.255.255.0)
  Total IPs / Usable Hosts: 256 / 254
  Network Address: 200.160.72.0
  First Usable IP: 200.160.72.1
  Last Usable IP: 200.160.72.254
  Broadcast Address: 200.160.72.255
  4. Zindabazar
  Required Hosts: 90 (Needs 7 host bits: 2⁷ = 128 ≥ 90 + 2)
  Prefix / Subnet Mask: /25 (255.255.255.128)
  Total IPs / Usable Hosts: 128 / 126
  Network Address: 200.160.73.0
  First Usable IP: 200.160.73.1
  Last Usable IP: 200.160.73.126
  Broadcast Address: 200.160.73.127
  Allocation Summary
  Lamabazar: Network: 200.160.64.0/22 | Broadcast: 200.160.67.255
  Amberkhana: Network: 200.160.68.0/22 | Broadcast: 200.160.71.255
  Subidbazar: Network: 200.160.72.0/24 | Broadcast: 200.160.72.255
  Zindabazar: Network: 200.160.73.0/25 | Broadcast: 200.160.73.127

Department	Required Hosts
Teacher Department	100
O/S Lab	90
Network (NW) Lab	50
Staff	10

Previous Job Question: Subnetting

Topic Wise Question Bank

All Topics

☆

Computer NetworkSubnetting

An organization has been assigned the IPv4 network address 192.168.1.0/24. As part of the network deployment, the network administrator is required to divide the address space into four equal-sized subnets to support different departments. Determine the Network Address, Subnet Mask (both CIDR and dotted-decimal notation)

CB, O(IT-24), 26 | Officer (IT)

Subnetting of 192.168.1.0/24 into Four Equal-Sized Subnets

To divide the network 192.168.1.0/24 into 4 equal-sized subnets:

• Original Prefix = /24
• Number of Required Subnets = 4 = 2²
• Borrowed Bits = 2
• New Prefix = /26
• Subnet Mask = 255.255.255.192

Subnet Details

Subnet	Network Address	CIDR	Subnet Mask	Broadcast Address	Usable Host Range
Subnet 1	192.168.1.0	/26	255.255.255.192	192.168.1.63	192.168.1.1 – 192.168.1.62
Subnet 2	192.168.1.64	/26	255.255.255.192	192.168.1.127	192.168.1.65 – 192.168.1.126
Subnet 3	192.168.1.128	/26	255.255.255.192	192.168.1.191	192.168.1.129 – 192.168.1.190
Subnet 4	192.168.1.192	/26	255.255.255.192	192.168.1.255	192.168.1.193 – 192.168.1.254

Final Answer

Network Addresses: 192.168.1.0, 192.168.1.64, 192.168.1.128, 192.168.1.192
Subnet Mask (CIDR): /26
Subnet Mask (Dotted Decimal): 255.255.255.192

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2
Computer NetworkSubnetting
Using this IP 172.16.16.137/22 find the answer of the following question.

i. Subnet Mask
ii. Block Size.
iii. Network Address
iv. Broadcast Address
v. Total valid Host.
CB, SO(IT), 22 | Senior Officer (IT)
Subnet Mask: /22 corresponds to the subnet mask 255.255.252.0.
Block Size: $2^(32−22)=2^10=1024$ .
Network Address: 172.16.16.0/22.
Broadcast Address: 172.16.19.255/22.
Total Valid Hosts: $2^(32−22)−2=1024−2=1022$
Explanation:
Given IP Address: 172.16.16.137/22
Step 1: /22 বোঝা
/22 মানে Network bits = 22
Host bits = 32 − 22 = 10 bits
Step 2: Subnet Mask লেখা
/22 এর Binary Subnet Mask:
11111111.11111111.11111100.00000000
Decimal Subnet Mask = 255.255.252.0
Step 3: IP Address কে Binary তে রূপান্তর
172 = 10101100
16 = 00010000
16 = 00010000
137 = 10001001
IP (Binary):
10101100.00010000.00010000.10001001
Step 4: AND Operation (Network Address)
IP Address AND Subnet Mask করা হয়।
যেখানে Subnet Mask = 0 → Host bits 0 হয়ে যায়。
```
IP:     10101100.00010000.00010000.10001001
Mask:   11111111.11111111.11111100.00000000
------------------------------------------------
Result: 10101100.00010000.00010000.00000000
```
Binary Result কে Decimal এ রূপান্তর করলে:
Network Address = 172.16.16.0
Step 5: Broadcast Address বের করা
Broadcast address পেতে সব Host bits = 1 করা হয়。
```
Network (Binary):
10101100.00010000.00010000.00000000

Host bits (10 bits) = 1111111111

Broadcast (Binary):
10101100.00010000.00010011.11111111
```
Binary থেকে Decimal করলে:
Broadcast Address = 172.16.19.255
Step 6: Valid Host Range
First Host = Network + 1 → 172.16.16.1
Last Host = Broadcast − 1 → 172.16.19.254
Step 7: Total Valid Hosts
Host bits = 10
Total = 2¹⁰ = 1024
Valid Hosts = 1024 − 2 = 1022
☆
3
Computer NetworkSubnetting
A company named ICB has been allotted the IP address 192.198.10.0/22. Using subnetting, allocate subnets for Finance, Admin, IT, Loan, and HR requiring 250, 125, 120, 250, and 60 hosts respectively. Find the subnet mask, valid host addresses, and network address for each subnet.
ICB, AP, 26 | Assistant Programmer
The given network is 192.198.10.0/22, which provides 1024 IP addresses (from 192.198.8.0 to 192.198.11.255). VLSM is used to allocate different subnet sizes based on host requirements.
Step 1: Sort requirements (descending)
250, 250, 125, 120, 60 hosts
Step 2: Required subnet sizes
- 250 hosts → /24 (256 addresses)
- 250 hosts → /24
- 125 hosts → /25 (128 addresses)
- 120 hosts → /25
- 60 hosts → /26 (64 addresses)
Step 3: Subnet allocation
Finance: 192.198.8.0/24
Mask: 255.255.255.0
Valid hosts: 192.198.8.1 – 192.198.8.254
Loan: 192.198.9.0/24
Mask: 255.255.255.0
Valid hosts: 192.198.9.1 – 192.198.9.254
Admin: 192.198.10.0/25
Mask: 255.255.255.128
Valid hosts: 192.198.10.1 – 192.198.10.126
IT: 192.198.10.128/25
Mask: 255.255.255.128
Valid hosts: 192.198.10.129 – 192.198.10.254
HR: 192.198.11.0/26
Mask: 255.255.255.192
Valid hosts: 192.198.11.1 – 192.198.11.62
Conclusion: VLSM minimizes wastage by allocating IPs according to requirement.
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4
Computer NetworkSubnetting
An organization is granted the IPv4 network block 14.24.74.0/24 and needs to segment it into two subnets: Subnet A (requires 120 addresses) and Subnet B (requires 60 addresses). Allocating sequentially from the requirement first to maximize remaining address space, state only the Network Address (with its CIDR mask) and the Broadcast Address for both subnets.
Combined Bank, SO(IT-24), 26 | Senior Officer (IT)
Given:
Network Block = 14.24.74.0/24
Subnet A requires = 120 addresses
Subnet B requires = 60 addresses
Allocate the largest subnet first.
Subnet A
Required Addresses = 120
Allocated Block Size = 128 addresses
CIDR Mask = /25
Item Value
Network Address 14.24.74.0/25
Broadcast Address 14.24.74.127
Subnet B
Required Addresses = 60
Allocated Block Size = 64 addresses
CIDR Mask = /26
Item Value
Network Address 14.24.74.128/26
Broadcast Address 14.24.74.191
Final Answer
Subnet Network Address Broadcast Address
Subnet A 14.24.74.0/25 14.24.74.127
Subnet B 14.24.74.128/26 14.24.74.191
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5
Computer NetworkSubnetting
Differentiate between IPv4 and IPv6. Given the IP address 192.168.10.0/24, calculate. Number of subnets if borrowed bits = 3 and hosts per subnet.
Combined Bank, AE(IT)/AME, 26 | AME/ANE/AE
Subnet Calculation:
Given IP: 192.168.10.0/24
Borrowed bits = 3
- Number of Subnets = 2³ = 8
- New subnet mask = /24 + 3 = /27
- Host bits = 32 - 27 = 5
- Hosts per subnet = 2⁵ - 2 = 32 - 2 = 30
Final Answer:
- Subnets = 8
- Hosts per subnet = 30
Total Time in a Month
- 30 days = 30 × 24 = 720 hours
- Downtime = 4 hours
Availability Calculation
- Availability = [(Total Time − Downtime) / Total Time] × 100
- = [(720 − 4) / 720] × 100
- = (716 / 720) × 100 ≈ 99.44%
Revenue Loss
- Transactions per hour = 500
- Fee per transaction = 0.50 Tk
- Loss per hour = 500 × 0.50 = 250 Tk
- Total loss = 250 × 4 = 1000 Tk
Subnet Calculation:
Given IP: 192.168.10.0/24
Borrowed bits = 3
- Number of Subnets = 2³ = 8
- New subnet mask = /24 + 3 = /27
- Host bits = 32 - 27 = 5
- Hosts per subnet = 2⁵ - 2 = 32 - 2 = 30
Final Answer:
- Subnets = 8
- Hosts per subnet = 30
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6
Computer NetworkSubnetting
A Classless IP Address is: 105.38.89.230/20. Find out the answer of the following question.
(i) What is Net id and Host id?
(ii) What is network address and broadcast address?
(ii) What is network size?
(iv) If this classless IP address is used to classfull IP address what will be the class?
CB, AE(IT)/AHME/SO, 21 | AME/ANE/AE
(i) Net ID and Host ID:
The CIDR notation /20 means the first 20 bits are for the Net ID and the remaining 12 bits are for the Host ID.
Net ID (first 20 bits): 105.38.80.0/20
Host ID (remaining 12 bits) represents the specific host in the network.
(ii) Network Address and Broadcast Address:
Network Address: Set all host bits to 0.
=>105.38.80.0 (first 20 bits).
Broadcast Address: Set all host bits to 1.
=>105.38.95.255 (last 12 bits as 1).
(iii) Network Size:
12 host bits: 2^12 = 4096 total addresses.
Usable addresses: 4096 - 2 = 4094 (since network and broadcast are reserved).
(iv) Classful IP Address:
The first octet 105 in binary is 01101001, which starts with 0.
This indicates the address is Class A (1.0.0.0 to 127.255.255.255).
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7
Computer NetworkSubnetting
What is the primary motivation for transitioning from classful IP addressing to classless IP addressing?
Sonali Bank, ADA, 24 | ADA
The primary motivation for moving from classful IP addressing to classless IP addressing (CIDR) was to use IP addresses more efficiently
- Address Wastage: Classful addressing allocated fixed-size blocks, leading to large wastage of IP addresses.
- Flexibility: CIDR allows flexible allocation of IP addresses based on actual need.
- Routing Efficiency: CIDR reduces the size of routing tables by using route aggregation.
- Scalability: Helps support the rapid growth of the internet.
- Better Utilization: Ensures optimal use of available IP address space.
Classful IP addressing থেকে CIDR (classless)-এ যাওয়ার মূল উদ্দেশ্য ছিল IP address-এর সঠিক ও efficient ব্যবহার নিশ্চিত করা
- Address Wastage: Classful পদ্ধতিতে fixed block দেওয়ায় অনেক IP address অপচয় হতো।
- Flexibility: CIDR প্রয়োজন অনুযায়ী IP address বরাদ্দ করতে দেয়।
- Routing Efficiency: Route aggregation-এর মাধ্যমে routing table ছোট হয়।
- Scalability: Internet দ্রুত বড় হওয়ার সাথে মানিয়ে নিতে সাহায্য করে।
- Better Utilization: IP address-এর সর্বোত্তম ব্যবহার নিশ্চিত করে।
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8
Computer NetworkSubnetting
A bank has the network block 192.168.10.0/24. The IT manager wants to divide this into 4 equal subnets.
(a) How many bits do you need to borrow to make 4 subnets?
(b) What is the new Subnet Mask in dotted-decimal format?
(c) Write down the Network Address, the First Usable IP, and the Broadcast Address for the second subnet created. Show your calculation.
Combined Bank, SO(IT-23), 26 | Senior Officer (IT)
(a) Number of bits to borrow
To create 4 subnets, we use the formula: 2ⁿ = 4
So, n = 2 bits must be borrowed from the host part.
(b) New Subnet Mask
Original network: /24
Borrowed bits: 2
New prefix: /26
Subnet mask = 255.255.255.192
(c) Second Subnet Details
Block size = 256 - 192 = 64
Subnets are:
1st: 192.168.10.0 - 192.168.10.63
2nd: 192.168.10.64 - 192.168.10.127
So for the second subnet:
Network Address: 192.168.10.64
First Usable IP: 192.168.10.65
Broadcast Address: 192.168.10.127
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9
Computer NetworkSubnetting
Consider the IP address 10.20.30.0/25 now answer the below question:
a. What is the subnet mask of the above IP address?
b. How many host per subnet have?
c. What is the Broadcast address of this 10.20.30.0/3 IP address?
Janata, ASA, 21 | AME/ANE/AE
Given:
IP Address = 10.20.30.0/25
a) Subnet Mask
A /25 prefix means 25 bits are used for the network.
Subnet Mask = 255.255.255.128
Binary form:
11111111.11111111.11111111.10000000
b) Number of hosts per subnet
Host bits = 32 − 25 = 7 bits
Total IP addresses per subnet = 2⁷ = 128
Usable hosts = 128 − 2 = 126 hosts
(2 addresses are reserved for network and broadcast)
c) Broadcast address
With /25, each subnet block size = 128.
Subnet range:
10.20.30.0 → 10.20.30.127
Network address = 10.20.30.0
Broadcast address = 10.20.30.127
Usable host range:
10.20.30.1 – 10.20.30.126
Final Answers:
Question Answer
Subnet Mask 255.255.255.128
Hosts per subnet 126 usable hosts
Broadcast Address 10.20.30.127
☆
10
Computer NetworkSubnetting
Using VLSM, subnet the block 10.100.0.0/22 to allocate address space for the following departments. Write the Subnet Address and CIDR mask for each: Dept A: 250 hosts, Dept B: 120 hosts, Dept C: 50 hosts. (Allocate efficiently and show the order of allocation.)
CB, ANE, 2026 | Assistant Network Engineer
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11
Computer NetworkSubnetting
Divide the network 192.168.10.0/24 into four equal-sized subnets and determine the network address, subnet mask, usable host range, and broadcast address of each subnet.
CB, ANE, 2026 | Assistant Network Engineer
☆
12
Computer NetworkSubnetting
What is Subnet mask and Localhost IP address?
SPCBL, SAME, 22 | AME/ANE/AE

Subnet Mask
A subnet mask is a 32-bit number used to divide an IP address into network and host portions. It helps determine which part of the IP address identifies the network and which part identifies the device within that network.
Example: 255.255.255.0
Localhost IP Address
The localhost IP address refers to the local machine itself. It is used to test network applications on the same device.
Localhost IPv4: 127.0.0.1

Subnet Mask
Subnet mask হলো একটি 32-bit number যা একটি IP address-কে network অংশ এবং host অংশে ভাগ করতে ব্যবহৃত হয়। এটি নির্ধারণ করে কোন অংশ network নির্দেশ করে এবং কোন অংশ device নির্দেশ করে।
উদাহরণ: 255.255.255.0
Localhost IP Address
Localhost IP address হলো নিজের computer বা device-কে নির্দেশ করার জন্য ব্যবহৃত একটি special IP address। এটি মূলত testing-এর কাজে ব্যবহৃত হয়।
Localhost IPv4: 127.0.0.1
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13
Computer NetworkSubnetting
VLSM Subnetting . Given an IP address 172.16.0.0/20 For creating 4 subnets department of A,B,C,D with 4000,2000,6000 and 8000 hosts, find out every department first and last IP address. Also write the subnet mask of q.x.y.z/notation.
Rupali, ANE, 21 |
First, calculate the number of required bits for each subnet:
- For 4000 hosts: Needs at least 12 bits for host portion (2¹² = 4096).
- For 2000 hosts: Needs at least 11 bits for host portion (2¹¹ = 2048).
- For 6000 hosts: Needs at least 13 bits for host portion (2¹³ = 8192).
- For 8000 hosts: Needs at least 13 bits for host portion (2¹³ = 8192).
Subnet Calculations:
Department C (6000 hosts)
Subnet Mask: 172.16.0.0/19 (because 32 - 13 = 19 bits for network portion).
Subnet Range: 172.16.0.0 - 172.16.31.255
First IP: 172.16.0.1
Last IP: 172.16.31.254
Department D (8000 hosts)
Subnet Mask: 172.16.32.0/19 (same as above).
Subnet Range: 172.16.32.0 - 172.16.63.255
First IP: 172.16.32.1
Last IP: 172.16.63.254
Department A (4000 hosts)
Subnet Mask: 172.16.64.0/20 (because 32 - 12 = 20 bits for network portion).
Subnet Range: 172.16.64.0 - 172.16.79.255
First IP: 172.16.64.1
Last IP: 172.16.79.254
Department B (2000 hosts)
Subnet Mask: 172.16.80.0/21 (because 32 - 11 = 21 bits for network portion).
Subnet Range: 172.16.80.0 - 172.16.95.255
First IP: 172.16.80.1 Last IP: 172.16.95.254
☆
14
Computer NetworkSubnetting
Given the base network address 192.168.10.0/24, perform Variable Length Subnet Masking (VLSM) to accommodate the following host requirements. Determine the network address and subnet mask for each subnet.
Subnet Required Hosts Network Address Subnet Mask
A 25
B 22
C 8
D 2
E 2
BPDB, AE(CSE),2026 | Power Sector
☆
15
Computer NetworkSubnetting
Subnetting logic requires precise binary calculation. A network engineer is tasked with dividing the internal network 192.168.10.0/24 into exactly 4 equal subnets for four different bank branches.
Show the mathematical calculation to determine how many bits must be borrowed to create 4 subnets, and state the New Subnet Mask in Both CIDR notation and decimal format
Combined Bank, O(IT-23), 26 | Officer (IT)

Given:
Network Address = 192.168.10.0/24
Required Subnets = 4
Step 1: Determine Borrowed Bits
To create subnets, host bits are borrowed from the host portion.
Formula:
Number of subnets = 2ⁿ
Here, required subnets = 4
So,
2ⁿ = 4
2² = 4
Therefore, 2 bits must be borrowed.
Step 2: Find New CIDR Prefix
Original prefix = /24
Borrowed bits = 2
So,
New prefix = 24 + 2 = /26
Step 3: Find New Subnet Mask
A /26 subnet mask means first 26 bits are network bits.
Binary subnet mask:
11111111.11111111.11111111.11000000
Convert to decimal:
11111111 = 255
11111111 = 255
11111111 = 255
11000000 = 192
Final Answer:
Borrowed Bits = 2
New CIDR Notation = 192.168.10.0/26
New Subnet Mask = 255.255.255.192

প্রশ্ন: 192.168.10.0/24 Network-কে 4টি Equal Subnet-এ ভাগ করে New Subnet Mask নির্ণয় কর
প্রদত্ত:
Network Address = 192.168.10.0/24
Required Subnets = 4
ধাপ ১: Borrowed Bit নির্ণয়
Subnet তৈরি করার জন্য host portion থেকে bit borrow করা হয়।
Formula:
Number of subnets = 2ⁿ
এখানে required subnets = 4
সুতরাং,
2ⁿ = 4
2² = 4
তাই 2 bits borrow করতে হবে।
ধাপ ২: New CIDR Prefix নির্ণয়
Original prefix = /24
Borrowed bits = 2
সুতরাং,
New prefix = 24 + 2 = /26
ধাপ ৩: New Subnet Mask নির্ণয়
/26 subnet mask মানে প্রথম 26 bit network bit।
Binary subnet mask:
11111111.11111111.11111111.11000000
Decimal-এ রূপান্তর:
11111111 = 255
11111111 = 255
11111111 = 255
11000000 = 192
Final Answer:
Borrowed Bits = 2
New CIDR Notation = 192.168.10.0/26
New Subnet Mask = 255.255.255.192
☆
16
Computer NetworkSubnetting
Given IP address 192.168.1.0 and divided four subnets in equal size: A) find the new subnet mask (CIDR) B) find out first usable host address of subnetting
Sonali Bank, ADA, 26 | ADA
Given:
- IP Address = 192.168.1.0
- Default mask = /24
- Required subnets = 4
A) New Subnet Mask (CIDR)
- Required subnets = 4 = 2²
- So, borrowed bits = 2
- New CIDR = /24 + 2 = /26
- Subnet Mask = 255.255.255.192
B) First Usable Host Address of Each Subnet
- Subnet 1: 192.168.1.0/26 → First usable host = 192.168.1.1
- Subnet 2: 192.168.1.64/26 → First usable host = 192.168.1.65
- Subnet 3: 192.168.1.128/26 → First usable host = 192.168.1.129
- Subnet 4: 192.168.1.192/26 → First usable host = 192.168.1.193
☆
17
Computer NetworkSubnetting
An ISP has been assigned the networks 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, and 192.168.3.0/24. To reduce routing table entries, the network engineer decides to use CIDR-based route aggregation. Determine the supernet address and explain how supernetting improves routing efficiency.[assume]
Combined Bank, ADA-23, 26 | ADA
Given Networks:
- 192.168.0.0/24
- 192.168.1.0/24
- 192.168.2.0/24
- 192.168.3.0/24
Step 1: Convert to Binary
Network Binary (last octet)
192.168.0.0 0000 0000
192.168.1.0 0000 0001
192.168.2.0 0000 0010
192.168.3.0 0000 0011
Step 2: Find Common Prefix
Looking at the last octet in binary:
0 = 0000 0000
1 = 0000 0001
2 = 0000 0010
3 = 0000 0011
The first 22 bits are identical across all four networks. The last 2 bits vary (00, 01, 10, 11).
Step 3: Determine Supernet Address
Supernet address = 192.168.0.0/22
Mask: 255.255.252.0
Range: 192.168.0.0 to 192.168.3.255
Total hosts: 1024 (4 × 256)
How Supernetting Improves Routing Efficiency
Reduced table size: Instead of 4 separate /24 entries, the router stores 1 /22 entry.
Faster lookups: Smaller routing tables mean faster longest-prefix-match searches.
Less update traffic: Changes in one subnet do not require re-advertising all routes.
Hierarchical routing: ISPs can advertise a single aggregated route instead of multiple subnets.
Given Networks:
- 192.168.0.0/24
- 192.168.1.0/24
- 192.168.2.0/24
- 192.168.3.0/24
Step 1: Binary-তে Convert করা
Network Binary (last octet)
192.168.0.0 0000 0000
192.168.1.0 0000 0001
192.168.2.0 0000 0010
192.168.3.0 0000 0011
Step 2: Common Prefix খোঁজা
Last octet binary-তে দেখলে:
0 = 0000 0000
1 = 0000 0001
2 = 0000 0010
3 = 0000 0011
চারটা network-এর প্রথম 22 bits identical। শেষ 2 bits vary করে (00, 01, 10, 11)।
Step 3: Supernet Address নির্ধারণ
Supernet address = 192.168.0.0/22
Mask: 255.255.252.0
Range: 192.168.0.0 থেকে 192.168.3.255
Total hosts: 1024 (4 × 256)
Supernetting Routing Efficiency কীভাবে Improve করে
Reduced table size: 4টা আলাদা /24 entry-এর পরিবর্তে router 1টা /22 entry store করে।
Faster lookups: ছোট routing tables মানে দ্রুত longest-prefix-match search।
Less update traffic: এক subnet change হলেও full advertisement লাগে না।
Hierarchical routing: ISP একটাই aggregated route advertise করতে পারে।
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18
Computer NetworkSubnetting
What do you mean by Subnet and Subnet Mask? The network address 172.16.0.0/19 provides how many subnets and hosts? What is the function of OSPF?
Rupali, ANE, 23 | AME/ANE/AE
1. Subnet and Subnet Mask
Subnet: A subnet is a smaller logical division of a large IP network created to improve network management, performance, and security.
Subnet Mask: A subnet mask is a 32-bit number used to identify the network portion and the host portion of an IP address.
2. Subnet Calculation for Network 172.16.0.0/19
The IP address 172.16.0.0 belongs to Class B, whose default subnet mask is /16.
Given subnet mask is /19, so additional bits used for subnetting = 19 − 16 = 3 bits.
Number of Subnets:
Number of subnets = 2³ = 8 subnets.
Number of Hosts per Subnet:
Host bits = 32 − 19 = 13 bits.
Number of hosts per subnet = 2¹³ − 2 = 8190 hosts.
3. Function of OSPF
OSPF (Open Shortest Path First) is a link-state routing protocol used to find the shortest path for data transmission within an IP network.
- Uses cost-based shortest path algorithm (Dijkstra)
- Supports fast convergence
- Scales well for large networks
- Uses area-based routing for efficiency
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19
Computer NetworkSubnetting
Convert the decimal IP address 192.168.101.5 into binary IP address. Fill-up the following in tabular form:
Rupali, ANE, 23 | AME/ANE/AE
Conversion of IP Address (192.168.101.5) into Binary:
Decimal IP: 192.168.101.5
Binary IP: 11000000.10101000.01100101.00000101

Item	Value
Network Address	14.24.74.0/25
Broadcast Address	14.24.74.127

Item	Value
Network Address	14.24.74.128/26
Broadcast Address	14.24.74.191

Subnet	Network Address	Broadcast Address
Subnet A	14.24.74.0/25	14.24.74.127
Subnet B	14.24.74.128/26	14.24.74.191

Question	Answer
Subnet Mask	255.255.255.128
Hosts per subnet	126 usable hosts
Broadcast Address	10.20.30.127

Subnet	Required Hosts	Network Address	Subnet Mask
A	25
B	22
C	8
D	2
E	2

Network	Binary (last octet)
192.168.0.0	0000 0000
192.168.1.0	0000 0001
192.168.2.0	0000 0010
192.168.3.0	0000 0011

Network	Binary (last octet)
192.168.0.0	0000 0000
192.168.1.0	0000 0001
192.168.2.0	0000 0010
192.168.3.0	0000 0011

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Computer NetworkSubnetting

Given the network 245.248.128.0/20, divide the address space among three departments as follows:
(a) Manager: half of the address space
(b) HR: one-quarter of the address space
(c) Admin: the remaining one-quarter

Dwasa,AME,25 | Other

Subnetting: 245.248.128.0/20

Given Network:

Network Address: 245.248.128.0/20
Subnet Mask: 255.255.240.0
Total Addresses: 2^32-20 = 2¹² = 4096

The address space needs to be divided among three departments:

Manager: Half of the address space = 50% = 2048 addresses
HR: One-quarter of the address space = 25% = 1024 addresses
Admin: Remaining one-quarter = 25% = 1024 addresses

Step 1: Manager — Half of the Address Space

Manager needs 2048 addresses.

2048 = 2¹¹

Therefore, the new prefix is:

32 − 11 = /21

Network: 245.248.128.0/21
Subnet Mask: 255.255.248.0
Address Range: 245.248.128.0 – 245.248.135.255
Usable Host Range: 245.248.128.1 – 245.248.135.254
Broadcast: 245.248.135.255

Step 2: HR — One-Quarter of the Address Space

HR needs 1024 addresses.

1024 = 2¹⁰

Therefore, the new prefix is:

32 − 10 = /22

Network: 245.248.136.0/22
Subnet Mask: 255.255.252.0
Address Range: 245.248.136.0 – 245.248.139.255
Usable Host Range: 245.248.136.1 – 245.248.139.254
Broadcast: 245.248.139.255

Step 3: Admin — Remaining One-Quarter

Admin also needs 1024 addresses.

1024 = 2¹⁰

Therefore, the new prefix is /22.

Network: 245.248.140.0/22
Subnet Mask: 255.255.252.0
Address Range: 245.248.140.0 – 245.248.143.255
Usable Host Range: 245.248.140.1 – 245.248.143.254
Broadcast: 245.248.143.255

Final Subnet Allocation:

Department	Required Size	Network	Prefix	Address Range	Broadcast
Manager	2048	245.248.128.0	/21	245.248.128.0 – 245.248.135.255	245.248.135.255
HR	1024	245.248.136.0	/22	245.248.136.0 – 245.248.139.255	245.248.139.255
Admin	1024	245.248.140.0	/22	245.248.140.0 – 245.248.143.255	245.248.143.255

Final Answer:

Manager gets 245.248.128.0/21, HR gets 245.248.136.0/22, and Admin gets 245.248.140.0/22.

Assigment: Classfull and Classles IP , VLSM, Subnetting

1. What is Classful IP Addressing and Classless IP Addressing? Explain the major advantages of Classless Addressing.

2. What is Subnetting? Explain the basic steps involved in subnetting an IP network.

3. Subnet 216.21.5.0 into subnets supporting 30 hosts in each subnet. Show the complete calculation.

4. If a Class B network has a subnet mask of 255.255.248.0, determine the maximum number of usable hosts per subnet.

5. Given 172.16.16.137/22, find the Subnet Mask, Block Size, Network Address, Broadcast Address, and Total Usable Hosts.

6. Given 105.38.89.230/20, find the Network ID, Host ID, Network Address, Broadcast Address, and Network Size.

7. A bank has 192.168.10.0/24 and wants 4 equal subnets. Find the borrowed bits, new subnet mask, and the Network Address, First Usable IP, and Broadcast Address of the second subnet.

8. An ISP has 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, and 192.168.3.0/24. Determine the supernet address using CIDR route aggregation and explain its benefit.

9. Given 172.11.0.0/16, use VLSM to allocate subnets for Teacher Department (100 hosts), O/S Lab (90 hosts), Network Lab (50 hosts), and Staff (10 hosts). Find the Network Address, Prefix Length, Usable Host Range, and Broadcast Address for each subnet.

10. A company has been assigned 192.198.10.0/22. Using VLSM, allocate subnets for Finance (250 hosts), Admin (125 hosts), IT (120 hosts), Loan (250 hosts), and HR (60 hosts). Find the subnet mask, network address, and valid host address range for each department.

Classfull and Classles IP, VLSM, Subnetting 45 questions