Computer Network Classfull and Classless IP Address, Subnetting, CIDR, VLSM (ET: 1.5 Hours)
IP Class, Subnetting, VLSM
    • 1
      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
      ClassFirst Octet RangeDefault Subnet MaskNetwork BitsHost BitsMain Use
      A0 – 127255.0.0.0824Large networks
      B128 – 191255.255.0.01616Medium-sized networks
      C192 – 223255.255.255.0248Small networks
      D224 – 239Not applicableMulticastMulticasting
      E240 – 255Not applicableExperimentalExperimental / 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.
      ClassPrivate IP Range
      Class A10.0.0.0 – 10.255.255.255
      Class B172.16.0.0 – 172.31.255.255
      Class C192.168.0.0 – 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 = 2h
      Usable host addresses = 2h − 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 OctetClass
      0–127Class A
      128–191Class B
      192–223Class C
      224–239Class D
      240–255Class 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.

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: 2h - 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: 2h - 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: 2h - 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: 2h - 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: 2s ≥ 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 = 2h = 211 = 2,048 addresses.
      Step 4: Calculate Usable Hosts
      • Formula: Usable Hosts = 2h - 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 = 2s = 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:
      DepartmentRequired Hosts
      Teacher Department100
      O/S Lab90
      Network (NW) Lab50
      Staff10
      Find the following for each subnet:
      1. Network Address
      2. Prefix Length / Subnet Mask
      3. Total IP Addresses
      4. Usable Host Addresses
      5. First Usable IP Address
      6. Last Usable IP Address
      7. 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

Topic Wise Question Bank
All Topics
    • ☆
      1
      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

      SubnetNetwork AddressCIDRSubnet MaskBroadcast AddressUsable Host Range
      Subnet 1192.168.1.0/26255.255.255.192192.168.1.63192.168.1.1 – 192.168.1.62
      Subnet 2192.168.1.64/26255.255.255.192192.168.1.127192.168.1.65 – 192.168.1.126
      Subnet 3192.168.1.128/26255.255.255.192192.168.1.191192.168.1.129 – 192.168.1.190
      Subnet 4192.168.1.192/26255.255.255.192192.168.1.255192.168.1.193 – 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



    • ☆
      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 Network_mathSubnetting
      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_math | IT Mathematical Question

      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



    • ☆
      4
      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.


    • ☆
      5
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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.


    • ☆
      6
      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

      ItemValue
      Network Address14.24.74.0/25
      Broadcast Address14.24.74.127

      Subnet B

      Required Addresses = 60
      Allocated Block Size = 64 addresses
      CIDR Mask = /26

      ItemValue
      Network Address14.24.74.128/26
      Broadcast Address14.24.74.191

      Final Answer

      SubnetNetwork AddressBroadcast Address
      Subnet A14.24.74.0/2514.24.74.127
      Subnet B14.24.74.128/2614.24.74.191

    • ☆
      7
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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

      ItemValue
      Network Address14.24.74.0/25
      Broadcast Address14.24.74.127

      Subnet B

      Required Addresses = 60
      Allocated Block Size = 64 addresses
      CIDR Mask = /26

      ItemValue
      Network Address14.24.74.128/26
      Broadcast Address14.24.74.191

      Final Answer

      SubnetNetwork AddressBroadcast Address
      Subnet A14.24.74.0/2514.24.74.127
      Subnet B14.24.74.128/2614.24.74.191

    • ☆
      8
      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

    • ☆
      9
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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

    • ☆
      10
      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).



    • ☆
      11
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      (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).



    • ☆
      12
      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.

    • ☆
      13
      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


    • ☆
      14
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      (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


    • ☆
      15
      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:

      QuestionAnswer
      Subnet Mask255.255.255.128
      Hosts per subnet126 usable hosts
      Broadcast Address10.20.30.127


    • ☆
      16
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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:

      QuestionAnswer
      Subnet Mask255.255.255.128
      Hosts per subnet126 usable hosts
      Broadcast Address10.20.30.127


    • ☆
      17
      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



    • ☆
      18
      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



    • ☆
      19
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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



    • ☆
      20
      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


    • ☆
      21
      Computer Network_mathSubnetting
      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
      Combined Bank, O(IT-23), 26_math | IT Mathematical Question

      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


    • ☆
      22
      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

    • ☆
      23
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question
      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

    • ☆
      24
      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

      NetworkBinary (last octet)
      192.168.0.00000 0000
      192.168.1.00000 0001
      192.168.2.00000 0010
      192.168.3.00000 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.


    • ☆
      25
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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

      NetworkBinary (last octet)
      192.168.0.00000 0000
      192.168.1.00000 0001
      192.168.2.00000 0010
      192.168.3.00000 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.


    • ☆
      26
      Computer Network_mathSubnetting
      A core router receives a packet with destination IP address 192.168.20.45. The router's forwarding table contains the following overlapping routes:
      • 192.168.0.0/16
      • 192.168.16.0/20
      • 192.168.20.0/22
      • 192.168.20.32/25
      Identify the exact routing interface the router will use to forward this packet
      RAKUB, ANSE, 26_math | IT Mathematical Question

      Given:
      Destination IP: 192.168.20.45

      Routing Table Entries:
      192.168.0.0/16
      192.168.16.0/20
      192.168.20.0/22
      192.168.20.32/25

      Step 1: Check Each Route for Match
      A packet matches a route when the destination IP is ANDed with the subnet mask and the result equals the network address.

      RouteMatch?
      192.168.0.0/16Yes
      192.168.16.0/20Yes
      192.168.20.0/22Yes
      192.168.20.32/25Yes

      Step 2: Apply Longest Prefix Match Rule
      When multiple routes match, the router selects the route with the longest subnet prefix (the most specific network).

      RoutePrefix Length
      192.168.0.0/1616
      192.168.16.0/2020
      192.168.20.0/2222
      192.168.20.32/2525 ← Longest

      Final Answer:
      The router will forward the packet using the route 192.168.20.32/25 because it has the longest prefix match (25 bits), making it the most specific route available.

      Why This Rule Exists:
      Longest Prefix Match ensures that traffic follows the most precise route. A /25 network covers a smaller and more specific range of addresses than a /16 network. Therefore, when both routes match, the router chooses the more specific route to deliver the packet accurately.


    • ☆
      27
      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


    • ☆
      28
      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



    • ☆
      29
      Computer Network_mathSubnetting
      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_math | IT Mathematical Question

      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


    • ☆
      30
      Computer Network_mathSubnetting
      Convert the decimal IP address 192.168.101.5 into binary IP address. Fill-up the following in tabular form:
      Rupali, ANE, 23_math | IT Mathematical Question

      Conversion of IP Address (192.168.101.5) into Binary:

      Decimal IP: 192.168.101.5

      Binary IP: 11000000.10101000.01100101.00000101



    • ☆
      31
      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: 232-20 = 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:

      DepartmentRequired SizeNetworkPrefixAddress RangeBroadcast
      Manager2048245.248.128.0/21245.248.128.0 – 245.248.135.255245.248.135.255
      HR1024245.248.136.0/22245.248.136.0 – 245.248.139.255245.248.139.255
      Admin1024245.248.140.0/22245.248.140.0 – 245.248.143.255245.248.143.255

      Final Answer:

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



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
1. What is Classful IP Addressing?
2. Which of the following is a class of IPv4 address in traditional classful addressing?
3. How many classes are defined in traditional Classful IPv4 addressing?
4. Which classful IPv4 class is primarily designed for very large networks?
5. Which classful IPv4 class is traditionally used for medium-sized networks?
6. Which classful IPv4 class is traditionally used for smaller networks?
7. What is Classless Addressing also known as?
8. What is the primary motivation for using Classless Addressing?
9. Which of the following is an advantage of Classless Addressing?
10. What identifies the network and host portions in Classless Addressing?
11. What does VLSM stand for?
12. What is the main purpose of VLSM?
13. Which subnetting method is more flexible for networks with different host requirements?
14. What is subnetting?
15. What is the first step in a typical subnetting calculation?
16. Which formula is used to determine the number of usable host addresses in a traditional IPv4 subnet?
17. How many usable host addresses are available in a /24 IPv4 subnet?
18. How many total IP addresses are available in a /26 subnet?
19. How many usable host addresses are available in a /26 subnet?
20. How many bits must be borrowed from a /24 network to create 4 equal subnets?
21. What is the new prefix length when a /24 network is divided into 4 equal subnets?
22. What is the dotted-decimal subnet mask for /26?
23. How many equal subnets are created by borrowing 3 bits?
24. If 4 bits are borrowed for subnetting, how many equal subnets can be created?
25. A network 216.21.5.0 is required to support 30 hosts per subnet. How many host bits are required?
26. For a subnet requiring 52 usable hosts, how many host bits are required?
27. What subnet prefix provides 62 usable hosts?
28. What subnet prefix provides 126 usable hosts?
29. What subnet prefix provides 254 usable hosts?
30. What is a network address?
31. What is the broadcast address of an IPv4 subnet?
32. What is the first usable host address of a subnet?
33. What is the last usable host address of a subnet?
34. For 192.168.10.0/24 divided into 4 equal subnets, what is the network address of the second subnet?
35. For 192.168.10.0/24 divided into 4 equal subnets, what is the broadcast address of the second subnet?
36. For 192.168.10.0/24 divided into 4 equal subnets, what is the first usable IP of the second subnet?
37. What is the block size for a subnet mask of 255.255.255.192?
38. What is CIDR route aggregation also called?
39. What is the main benefit of supernetting?
40. Four /24 networks 192.168.0.0/24 through 192.168.3.0/24 can be aggregated into which supernet?
41. In VLSM, which subnet should generally be allocated first?
42. A department requires 100 usable hosts. Which subnet is the smallest suitable choice?
43. A department requires 90 usable hosts. Which subnet is the smallest suitable choice?
44. A department requires 50 usable hosts. Which subnet is the smallest suitable choice?
45. A department requires 10 usable hosts. Which subnet is the smallest suitable choice?

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