An organization is granted the block 211.17.180.0/24. The administrator wants to create 32subnets. i) find the first and last address in subnet 1 ii) find the first and last address insubnet 32

To find the first and last address in subnet 1, we need to determine the subnet mask for the given block. The block 211.17.180.0/24 has a subnet mask of 24 bits, which means the first 24 bits are fixed and the remaining 8 bits can be used for subnetting.

To create 32 subnets, we need to borrow enough bits from the host portion to accommodate 32 subnets. Since 2^5 = 32, we need to borrow 5 bits. This leaves us with 3 bits for the host portion.

To calculate the subnet mask for the subnets, we add the borrowed bits to the original subnet mask. So the new subnet mask will be 24 + 5 = 29 bits.

Now, let's find the first and last address in subnet 1. Since we borrowed 5 bits for subnetting, the first subnet will have a network address of 211.17.180.0. To find the first address in subnet 1, we increment the host portion by 1, which gives us 211.17.180.1.

To find the last address in subnet 1, we need to determine the number of hosts in each subnet. Since we have 3 bits for the host portion, we have 2^3 - 2 = 6 usable host addresses per subnet. Therefore, the last address in subnet 1 will be 211.17.180.6.

To find the first and last address in subnet 32, we follow the same process. The network address for subnet 32 will be 211.17.180.224. The first address in subnet 32 will be 211.17.180.225, and the last address will be 211.17.180.230.

So, the first and last addresses in subnet 1 are 211.17.180.1 and 211.17.180.6, respectively. The first and last addresses in subnet 32 are 211.17.180.225 and 211.17.180.230, respectively.