Understanding Class B Subnetting

This question asks us to determine the maximum number of subnets and hosts within each subnet when a Class B IP address space is subnetted using a 6-bit subnet number. Subnetting involves dividing a larger network into smaller, more manageable sub-networks.

Class B IP Addresses Explained

A Class B IP address is identified by the first two bits being 10. The default allocation for a Class B network is:

• The first 16 bits (the first two octets) represent the Network portion.
• The last 16 bits (the last two octets) represent the Host portion.

The default subnet mask for a Class B network is 255.255.0.0.

Subnetting a Class B Network

Subnetting a network involves "borrowing" bits from the host portion of the IP address to create a subnet portion. The number of bits borrowed determines the number of possible subnets and reduces the number of available host bits per subnet.

In this problem, we are told that 6 bits are used for the subnet number. These 6 bits are borrowed from the original 16 host bits of the Class B address.

• Original Host Bits: 16
• Bits Borrowed for Subnet: 6
• Remaining Host Bits: $16 - 6 = 10$

Now, let's calculate the maximum number of subnets and the maximum number of hosts per subnet using these values.

Calculating Maximum Number of Subnets

The number of possible subnets is determined by the number of bits allocated to the subnet portion. If 'n' is the number of subnet bits, the total number of possible subnet combinations is $2^n$.

Here, the number of subnet bits is 6. So, total possible subnets = $2^6 = 64$.

However, in older subnetting practices, the first and last subnets were often reserved for specific purposes (the subnet where all subnet bits are '0' and the subnet where all subnet bits are '1'). When these are reserved, the number of usable subnets becomes $2^n - 2$.

Usable subnets (older practice) = $2^6 - 2 = 64 - 2 = 62$.

Given the options, it appears the question refers to the older practice where the first and last subnets are not used for assigning hosts, leading to 62 usable subnets.

Calculating Maximum Number of Hosts per Subnet

The number of possible host addresses within each subnet is determined by the number of remaining host bits. If 'm' is the number of host bits, the total number of possible host addresses is $2^m$.

Here, the number of remaining host bits is 10. So, total possible host addresses = $2^{10} = 1024$.

Within any given network or subnet, two addresses are reserved and cannot be assigned to individual hosts:

• The network address (where all host bits are 0). This identifies the subnet itself.
• The broadcast address (where all host bits are 1). This is used to send data to all devices on the subnet.

Therefore, the number of usable host addresses per subnet is $2^m - 2$.

Usable hosts per subnet = $2^{10} - 2 = 1024 - 2 = 1022$.

Summary of Results

Based on a 6-bit subnet number for a Class B network, assuming the older practice of reserving the first and last subnets and the network/broadcast addresses within each subnet:

• Maximum number of usable subnets: 62
• Maximum number of usable hosts per subnet: 1022

The maximum number of subnets is 62, and the maximum number of hosts in each subnet is 1022.

Revision Table: Class B Subnetting Key Concepts

Additional Information: Modern Subnetting and CIDR

While the calculation $2^n - 2$ for usable subnets was standard practice historically (due to reserving the all-zeros and all-ones subnets), modern networking using CIDR (Classless Inter-Domain Routing) typically allows the use of all subnets ($2^n$). However, the $2^m - 2$ rule for usable hosts per subnet (reserving the network and broadcast addresses) still universally applies.

If the question implied modern CIDR practices, the number of subnets would be $2^6 = 64$. But since '62 subnets' is an option and matches with '1022 hosts' (which implies the $2^m-2$ rule is used for hosts), it strongly suggests the question is based on the older subnetting convention where the first and last subnets are unusable.

The custom subnet mask for this scenario would be derived from the 16 default network bits plus the 6 borrowed subnet bits, totaling $16 + 6 = 22$ network/subnet bits. The subnet mask would have 22 bits set to '1'. In binary, this is:

11111111.11111111.11111100.00000000

Converting this to decimal: 255.255.252.0. This is the subnet mask for a Class B network subnetted with 6 subnet bits.