Calculate the IPv4 subnet details given a host IP, original subnet mask, and new subnet mask for Problem 2 and Problem 3.

$$\text{Number of Subnet Bits} = \text{New CIDR} - \text{Original CIDR} = 17 - 8 = 9$$

• Number of Subnet Bits: 9

$$\text{Number of Subnets Created} = 2^9 = 512$$

• Number of Subnets Created: 512

$$\text{Number of Host Bits per Subnet} = 32 - 17 = 15$$

• Number of Host Bits per Subnet: 15

$$\text{Number of Hosts per Subnet} = 2^{15} - 2 = 32768 - 2 = 32766$$

• Number of Hosts per Subnet: 32766

Host IP: 10.101.99.228
New Subnet Mask: 255.255.128.0
Bitwise AND:

$$\begin{array}{c} 00001010.01100101.01100011.11100100 \\ \text{AND } 11111111.11111111.10000000.00000000 \\ \hline 00001010.01100101.00000000.00000000 \end{array}$$

In decimal: 10.101.0.0

• Network Address of this Subnet: 10.101.0.0

$$\text{First Host Address} = 10.101.0.0 + 1 = 10.101.0.1$$

• IPv4 Address of First Host on this Subnet: 10.101.0.1

The block size for the third octet is $256 - 128 = 128$.
The subnets in the third octet are multiples of 128: 0, 128.
Since the host IP's third octet (99) falls into the 0 block, the next network address is 10.101.128.0.

$$\text{Broadcast Address} = 10.101.128.0 - 1 = 10.101.127.255$$

• IPv4 Broadcast Address on this Subnet: 10.101.127.255

$$\text{Last Host Address} = 10.101.127.255 - 1 = 10.101.127.254$$

• IPv4 Address of Last Host on this Subnet: 10.101.127.254

:

Given:
Host IP Address: 172.22.32.12
Original Subnet Mask: 255.255.0.0
New Subnet Mask: 255.255.224.0

First, let's convert the new subnet mask to binary and determine its CIDR notation.
255.255.224.0
In binary: 11111111.11111111.11100000.00000000
The number of '1's is $8 + 8 + 3 + 0 = 19$. So, the new subnet mask is /19.
The original subnet mask 255.255.0.0 is /16.

$$\text{Number of Subnet Bits} = \text{New CIDR} - \text{Original CIDR} = 19 - 16 = 3$$

• Number of Subnet Bits: 3

$$\text{Number of Subnets Created} = 2^3 = 8$$

• Number of Subnets Created: 8

$$\text{Number of Host Bits per Subnet} = 32 - 19 = 13$$

• Number of Host Bits per Subnet: 13

$$\text{Number of Hosts per Subnet} = 2^{13} - 2 = 8192 - 2 = 8190$$

• Number of Hosts per Subnet: 8190

Host IP: 172.22.32.12
New Subnet Mask: 255.255.224.0
Bitwise AND:

$$\begin{array}{c} 10101100.00010110.00100000.00001100 \\ \text{AND } 11111111.11111111.11100000.00000000 \\ \hline 10101100.00010110.00100000.00000000 \end{array}$$

In decimal: 172.22.32.0

• Network Address of this Subnet: 172.22.32.0

$$\text{First Host Address} = 172.22.32.0 + 1 = 172.22.32.1$$

• IPv4 Address of First Host on this Subnet: 172.22.32.1

The block size for the third octet is $256 - 224 = 32$.
The subnets in the third octet are multiples of 32: 0, 32, 64, 96, 128, 160, 192, 224.
Since the host IP's third octet (32) falls into the 32 block, the next network address is 172.22.64.0.

$$\text{Broadcast Address} = 172.22.64.0 - 1 = 172.22.63.255$$

• IPv4 Broadcast Address on this Subnet: 172.22.63.255

$$\text{Last Host Address} = 172.22.63.255 - 1 = 172.22.63.254$$

• IPv4 Address of Last Host on this Subnet: 172.22.63.254