Calculate the IPv4 subnet details for a given Host IP, Original Subnet Mask, and New Subnet Mask, specifically for Problems 4 and 5.

$$\text{Number of Subnet Bits} = \text{New CIDR} - \text{Original CIDR} = 30 - 24 = 6$$

• Number of Subnet Bits: 6

$$\text{Number of Subnets Created} = 2^6 = 64$$

• Number of Subnets Created: 64

$$\text{Number of Host Bits per Subnet} = 32 - 30 = 2$$

• Number of Host Bits per Subnet: 2

$$\text{Number of Hosts per Subnet} = 2^2 - 2 = 4 - 2 = 2$$

• Number of Hosts per Subnet: 2

Host IP: 192.168.1.245
New Subnet Mask: 255.255.255.252
For the fourth octet:
Host IP fourth octet (245): $11110101_2$
Subnet Mask fourth octet (252): $11111100_2$
Bitwise AND:

$$\begin{array}{c} 11110101_2 \\ \text{AND } 11111100_2 \\ \hline 11110100_2 \end{array}$$

$11110100_2$ is 244 in decimal.
So, the Network Address is 192.168.1.244.

• Network Address of this Subnet: 192.168.1.244

$$\text{First Host Address} = 192.168.1.244 + 1 = 192.168.1.245$$

• IPv4 Address of First Host on this Subnet: 192.168.1.245

The block size for the fourth octet is $256 - 252 = 4$.
The subnets in the fourth octet are multiples of 4: ..., 240, 244, 248, ...
Since the host IP 245 falls into the 244 block, the next network address is 248.
The broadcast address for the 192.168.1.244 subnet is one less than the next network address.

$$\text{Broadcast Address} = 192.168.1.248 - 1 = 192.168.1.247$$

• IPv4 Broadcast Address on this Subnet: 192.168.1.247

$$\text{Last Host Address} = 192.168.1.247 - 1 = 192.168.1.246$$

• IPv4 Address of Last Host on this Subnet: 192.168.1.246

:

Given:
Host IP Address: 128.107.0.55
Original Subnet Mask: 255.255.0.0
New Subnet Mask: 255.255.255.0

First, let's convert the new subnet mask to binary and determine its CIDR notation.
255.255.255.0
In binary: 11111111.11111111.11111111.00000000
The number of '1's is $8 + 8 + 8 + 0 = 24$. So, the new subnet mask is /24.
The original subnet mask 255.255.0.0 is /16.

$$\text{Number of Subnet Bits} = \text{New CIDR} - \text{Original CIDR} = 24 - 16 = 8$$

• Number of Subnet Bits: 8

$$\text{Number of Subnets Created} = 2^8 = 256$$

• Number of Subnets Created: 256

$$\text{Number of Host Bits per Subnet} = 32 - 24 = 8$$

• Number of Host Bits per Subnet: 8

$$\text{Number of Hosts per Subnet} = 2^8 - 2 = 256 - 2 = 254$$

• Number of Hosts per Subnet: 254

Host IP: 128.107.0.55
New Subnet Mask: 255.255.255.0
Bitwise AND:

$$\begin{array}{c} 10000000.01101011.00000000.00110111 \\ \text{AND } 11111111.11111111.11111111.00000000 \\ \hline 10000000.01101011.00000000.00000000 \end{array}$$

In decimal: 128.107.0.0

• Network Address of this Subnet: 128.107.0.0

$$\text{First Host Address} = 128.107.0.0 + 1 = 128.107.0.1$$

• IPv4 Address of First Host on this Subnet: 128.107.0.1

For a /24 subnet, the network address is 128.107.0.0. The broadcast address is the highest address in that range.

$$\text{Broadcast Address} = 128.107.0.255$$

• IPv4 Broadcast Address on this Subnet: 128.107.0.255

$$\text{Last Host Address} = 128.107.0.255 - 1 = 128.107.0.254$$

• IPv4 Address of Last Host on this Subnet: 128.107.0.254