Final answer:
To form subnets for six departments in an organization, we need to calculate the number of subnet bits required based on the number of hosts in each department. Using a /24 subnet mask, the possible arrangement of subnets with their subnet number, subnet mask, and IP address ranges are provided.
Step-by-step explanation:
Subnetting for Six Departments
To form subnets for six departments in an organization with a class C network 192.168.10, we need to determine the number of subnets required to accommodate the number of hosts in each department.
The number of hosts required for each department is: 25, 20, 93, 12, 8, and 22.
Using the formula 2^n - 2, where 'n' is the number of host bits, we can calculate the number of subnet bits required:
Department 1: 2^5-2 = 30 hosts
Department 2: 2^5-2 = 30 hosts
Department 3: 2^7-2 = 126 hosts
Department 4: 2^4-2 = 14 hosts
Department 5: 2^3-2 = 6 hosts
Department 6: 2^5-2 = 30 hosts
Since the total of all hosts is 180, which is less than 256 (the total number of hosts in a class C network), we can use a /24 subnet mask for this organization.
The possible arrangement of subnets with their subnet number, subnet mask, and range of IP addresses are as follows:
Department 1: Subnet number: 192.168.10.0/27, Subnet mask: 255.255.255.224, Range: 192.168.10.1 - 192.168.10.30
Department 2: Subnet number: 192.168.10.32/27, Subnet mask: 255.255.255.224, Range: 192.168.10.33 - 192.168.10.62
Department 3: Subnet number: 192.168.10.64/25, Subnet mask: 255.255.255.128, Range: 192.168.10.65 - 192.168.10.126
Department 4: Subnet number: 192.168.10.128/28, Subnet mask: 255.255.255.240, Range: 192.168.10.129 - 192.168.10.142
Department 5: Subnet number: 192.168.10.144/29, Subnet mask: 255.255.255.248, Range: 192.168.10.145 - 192.168.10.150
Department 6: Subnet number: 192.168.10.152/27, Subnet mask: 255.255.255.224, Range: 192.168.10.153 - 192.168.10.182