An organization is assigned an IPv6 address block of 2001:db8:0:ca00::/56. How many subnets can be created without using bits in the interface ID space?

An organization is assigned an IPv6 address block of 2001:db8:0:ca00::/56. How many subnets can be created without using bits in the interface ID space?

• A. 128
• B. 256
• C. 512
• D. 1024

Answer: (B)

The provided IPv6 address block is 2001:db8:0:ca00::/56. This notation indicates that the first 56 bits of the address are fixed as defined by the prefix, leaving the remaining bits to be used for subnetting and interface identification. In an IPv6 address, the first 64 bits are typically designated for the network and subnetting portion, while the last 64 bits are used for the interface ID. Since this address block is /56, the first 56 bits are used for the network portion, and the remaining 72 bits (from the total of 128 bits in an IPv6 address) can be utilized for creating subnets and interface identification. To determine how many subnets can be created without using bits from the interface ID space, we focus on the remaining bits that can be utilized for subnetting. Since the subnetting starts from the 57th bit to the 64th bit, there are 8 bits available for subnetting (from /56 to /64). To calculate the number of subnets possible with these 8 bits, we use the formula: \[ 2^{number\ of\ subnet\ bits} = 2^8 = 256 \

The provided IPv6 address block is 2001:db8:0:ca00::/56. This notation indicates that the first 56 bits of the address are fixed as defined by the prefix, leaving the remaining bits to be used for subnetting and interface identification.

In an IPv6 address, the first 64 bits are typically designated for the network and subnetting portion, while the last 64 bits are used for the interface ID. Since this address block is /56, the first 56 bits are used for the network portion, and the remaining 72 bits (from the total of 128 bits in an IPv6 address) can be utilized for creating subnets and interface identification.

To determine how many subnets can be created without using bits from the interface ID space, we focus on the remaining bits that can be utilized for subnetting. Since the subnetting starts from the 57th bit to the 64th bit, there are 8 bits available for subnetting (from /56 to /64).

To calculate the number of subnets possible with these 8 bits, we use the formula:

[

2^{number\ of\ subnet\ bits} = 2^8 = 256

\