Answer :
To divide the [tex]$172.16.45.0/24$[/tex] network into 9 subnets with the maximum number of hosts possible on each subnet, follow these steps:
1. Determine the number of additional bits needed:
To create 9 subnets, we need to determine how many additional bits are required to represent at least 9 subnets. Since subnetting involves binary representation, we need at least [tex]$2^x \geq 9$[/tex] subnets. The smallest power of 2 greater than or equal to 9 is [tex]$2^4 = 16$[/tex].
Therefore, we need an additional 4 bits to create the required subnets.
So, the number of additional bits needed is [tex]\( 4 \)[/tex].
2. Calculate the new subnet mask in CIDR notation:
The original prefix length is 24 (as in [tex]$172.16.45.0/24$[/tex]). By adding the 4 additional bits, the new prefix length becomes [tex]\( 24 + 4 = 28 \)[/tex].
So, the CIDR value for the new subnet mask is [tex]\( /28 \)[/tex].
3. Convert the new subnet mask to dotted decimal format:
A /28 CIDR notation corresponds to the following binary representation of the subnet mask:
```
11111111.11111111.11111111.11110000
```
Converting this binary representation to dotted decimal format, we get [tex]\( 255.255.255.240 \)[/tex].
4. Determine the number of subnets created:
Given the additional 4 bits, the total number of subnets created will be [tex]\( 2^4 = 16 \)[/tex].
5. Calculate the number of hosts per subnet:
With a /28 subnet mask, each subnet has [tex]\( 32 - 28 = 4 \)[/tex] bits for host addresses. The total number of possible addresses is [tex]\( 2^4 = 16 \)[/tex]. Subtracting the network and broadcast addresses, the number of usable hosts per subnet is [tex]\( 14 \)[/tex].
6. Fill in the Network ID, First Host, Last Host, and Broadcast Address for the first 4 subnets:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Subnet} & \text{Network ID} & \text{First Host} & \text{Last Host} & \text{Broadcast} \\ \hline 1 & 172.16.45.0 & 172.16.45.1 & 172.16.45.14 & 172.16.45.15 \\ \hline 2 & 172.16.45.16 & 172.16.45.17 & 172.16.45.30 & 172.16.45.31 \\ \hline 3 & 172.16.45.32 & 172.16.45.33 & 172.16.45.46 & 172.16.45.47 \\ \hline 4 & 172.16.45.48 & 172.16.45.49 & 172.16.45.62 & 172.16.45.63 \\ \hline \end{array} \][/tex]
So, the answers are:
- Additional bits needed: [tex]\( 4 \)[/tex]
- CIDR value for the new subnet mask: [tex]\( /28 \)[/tex]
- Dotted decimal value for the new subnet mask: [tex]\( 255.255.255.240 \)[/tex]
- Number of subnets created: [tex]\( 16 \)[/tex]
- Number of hosts per subnet: [tex]\( 14 \)[/tex]
Delimited table with detailed subnet information for the first four subnets:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Subnet} & \text{Network ID} & \text{First Host} & \text{Last Host} & \text{Broadcast} \\ \hline 1 & 172.16.45.0 & 172.16.45.1 & 172.16.45.14 & 172.16.45.15 \\ \hline 2 & 172.16.45.16 & 172.16.45.17 & 172.16.45.30 & 172.16.45.31 \\ \hline 3 & 172.16.45.32 & 172.16.45.33 & 172.16.45.46 & 172.16.45.47 \\ \hline 4 & 172.16.45.48 & 172.16.45.49 & 172.16.45.62 & 172.16.45.63 \\ \hline \end{array} \][/tex]
1. Determine the number of additional bits needed:
To create 9 subnets, we need to determine how many additional bits are required to represent at least 9 subnets. Since subnetting involves binary representation, we need at least [tex]$2^x \geq 9$[/tex] subnets. The smallest power of 2 greater than or equal to 9 is [tex]$2^4 = 16$[/tex].
Therefore, we need an additional 4 bits to create the required subnets.
So, the number of additional bits needed is [tex]\( 4 \)[/tex].
2. Calculate the new subnet mask in CIDR notation:
The original prefix length is 24 (as in [tex]$172.16.45.0/24$[/tex]). By adding the 4 additional bits, the new prefix length becomes [tex]\( 24 + 4 = 28 \)[/tex].
So, the CIDR value for the new subnet mask is [tex]\( /28 \)[/tex].
3. Convert the new subnet mask to dotted decimal format:
A /28 CIDR notation corresponds to the following binary representation of the subnet mask:
```
11111111.11111111.11111111.11110000
```
Converting this binary representation to dotted decimal format, we get [tex]\( 255.255.255.240 \)[/tex].
4. Determine the number of subnets created:
Given the additional 4 bits, the total number of subnets created will be [tex]\( 2^4 = 16 \)[/tex].
5. Calculate the number of hosts per subnet:
With a /28 subnet mask, each subnet has [tex]\( 32 - 28 = 4 \)[/tex] bits for host addresses. The total number of possible addresses is [tex]\( 2^4 = 16 \)[/tex]. Subtracting the network and broadcast addresses, the number of usable hosts per subnet is [tex]\( 14 \)[/tex].
6. Fill in the Network ID, First Host, Last Host, and Broadcast Address for the first 4 subnets:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Subnet} & \text{Network ID} & \text{First Host} & \text{Last Host} & \text{Broadcast} \\ \hline 1 & 172.16.45.0 & 172.16.45.1 & 172.16.45.14 & 172.16.45.15 \\ \hline 2 & 172.16.45.16 & 172.16.45.17 & 172.16.45.30 & 172.16.45.31 \\ \hline 3 & 172.16.45.32 & 172.16.45.33 & 172.16.45.46 & 172.16.45.47 \\ \hline 4 & 172.16.45.48 & 172.16.45.49 & 172.16.45.62 & 172.16.45.63 \\ \hline \end{array} \][/tex]
So, the answers are:
- Additional bits needed: [tex]\( 4 \)[/tex]
- CIDR value for the new subnet mask: [tex]\( /28 \)[/tex]
- Dotted decimal value for the new subnet mask: [tex]\( 255.255.255.240 \)[/tex]
- Number of subnets created: [tex]\( 16 \)[/tex]
- Number of hosts per subnet: [tex]\( 14 \)[/tex]
Delimited table with detailed subnet information for the first four subnets:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Subnet} & \text{Network ID} & \text{First Host} & \text{Last Host} & \text{Broadcast} \\ \hline 1 & 172.16.45.0 & 172.16.45.1 & 172.16.45.14 & 172.16.45.15 \\ \hline 2 & 172.16.45.16 & 172.16.45.17 & 172.16.45.30 & 172.16.45.31 \\ \hline 3 & 172.16.45.32 & 172.16.45.33 & 172.16.45.46 & 172.16.45.47 \\ \hline 4 & 172.16.45.48 & 172.16.45.49 & 172.16.45.62 & 172.16.45.63 \\ \hline \end{array} \][/tex]
