The number and grouping of hosts are as follows:
Student LAN
Student computer: 460
Router (LAN Gateway): 1
Switch (Management): 20
Total student subnets: 481
Teacher LAN
Teacher Computer: 64
Router (LAN Gateway): 1
Switch (Management): 4
Total Teacher Subnets: 69
Administrator LAN
Administrator computer: 20
Servers: 1
Router (LAN Gateway): 1
Switch (Management): 1
Total administrator Subnets: 23
WAN
The Wan:2 between routers
WAN Total: 2
Calculate and assign addresses using VLSM
Use the VLSM allocation method to assign a smaller block of addresses to each network as needed.
Address Block 172.16.0.0/22 (subnet mask 255.255.252.0) has been assigned to the entire internetwork. 10 bits will be used to define the host address and subnet. The result is a 1024 IPv4 local address, ranging from 172.16.0.0 to 172.16.3.0.
Student LAN
The largest subnet is a student LAN that requires 460 addresses.
Use the formula available host number = 2^n-2, borrowing 9 bits as the host part, and 512-2 = 510 available host addresses. This quantity is in line with current requirements and there is little room for future development.
Using 9 master seats also leaves 1 bits available for locally defined subnet addresses. We use the smallest available address to derive the subnet address 172.16.0.0/23.
The mask for student subnets is calculated as follows:
Address: 172.16.0.0
Binary representation:
10101100.00010000.00000000.00000000
Mask: 255.255.254.0
Represents 23 bits in binary:
11111111.11111111.11111110.00000000
In the student network, the IPV4 host address range is:
172.16.0.1 to 172.16.1.254, broadcast address is 172.16.1.255.
Because these addresses are already assigned to student LANs, they can no longer be assigned to the rest of the subnets: Teacher LAN, Administrator LAN, and WAN. The 172.16.2.0 address is the address within the range of 172.16.3.255.
Teacher LAN
The second largest network is the teacher LAN. This network requires a minimum of 66 addresses. If you use 6 as an exponent in the power formula of 2, 2^6-2 can only provide 62 available addresses. Therefore, you must use an address block that has 7 master seats. The calculation 2^7-2 can draw an address block with 126 addresses. This leaves 25 bits to be assigned to the network address. The next available address block of the same size is the 172.16.2.0/25 network.
Address: 172.16.2.0
Binary representation:
10101100.00010000.0000010.00000000
Mask: 255.255.255.128
Represents 25 bits in binary:
11111111.11111111.1111111.10000000
The IPV4 host address range provided is:
172.16.2.1 to 172.16.2.126, broadcast address is 172.16.2.127.
From the original address block 172.16.0.0/22, assign the address 172.16.0.0 to 172.16.2.127. The remaining assignable addresses are 172.16.2.128 to 172.16.3.255.
Administrator LAN
The administrator LAN needs to support 23 hosts. Therefore, you need 6 master seats and use the calculation: 2^6-2.
The next available address block that can support these hosts is the 172.16.2.128/26 address block.
Address: 172.16.2.128
Binary representation:
10101100.00010000.0000010.10000000
Mask: 255.255.255.192
Represents 26 bits in binary:
11111111.11111111.1111111.11000000
The IPV4 host address range provided is:
172.16.2.129 to 172.16.2.190, broadcast address is 172.16.2.191.
The result is a 62 unique IPV4 address for the administrator LAN.
WAN
The last network segment is a WAN connection and requires 2 host addresses. Support WAN links with just 2 primary seats. 2^2-2 = 2.
This leaves 8 bits to define the local subnet address. The next available address block is 172.16.2.192/30.
Address: 172.16.2.192
Binary representation:
10101100.00010000.0000010.11000000
Mask: 255.255.255.252
Represents 30 bits in binary:
11111111.11111111.1111111.11111100
The IPV4 host address range provided is:
172.16.2.193 to 172.16.2.194, broadcast address is 172.16.2.195.
CCNA Cisco subnetting (compute and assign addresses using VLSM)