6.8.1: Skills Integration Challenge-Planning Subnets and Configuring IP Addresses

December 1, 2023 Alice AUSTIN

Subnet 1(existing): (Consists of R2,1A,1B)
First subnet, existing student LAN (off of router R2 Central), up to 60 hosts

We can give 64 hosts here to make it proper

nsub = n + log2 (256/64) = 24 + 2 = 26

So the subnet mask is 255.255.255.192
Invert of the subnet mask 0. 0. 0. 63
Network address is 192.168.23.0/26
Last address(OR Operation) 192.168.23.63/26

First Usable address 192.168.23.1/26
Last Usable address 192.168.23.62/26

Subnet 2(future):(Consist of R2,1A,1B)
Second subnet, future student LAN, up to 28 hosts

We can give 32 hosts here to make it proper

nsub = n + log2 (256/32) = 24 + 3 = 27

So the subnet mask is 255.255.255.224
Invert of the subnet mask 0. 0. 0. 31
Network address is 192.168.23.64/27
Last address (OR Operation) 192.168.23.95/27

First Usable address 192.168.23.65/27
Last Usable address 192.168.23.94/27

Subnet 3(existing):(Consist of R1,Eagle_Server)
Third subnet, existing ISP LAN, up to 12 hosts

We can give 16 hosts here to make it proper

nsub = n + log2 (256/16) = 24 + 4 = 28

So the subnet mask is 255.255.255.240
Invert of the subnet mask 0. 0. 0. 15
Network address is 192.168.23.96/28
Last address (OR Operation) 192.168.23.111/28

First Usable address 192.168.23.95/28
Last Usable address 192.168.23.110/28

Subnet 4(future):( Consist of R1,Eagle_Server)
Fourth subnet, future ISP LAN, up to 6 hosts

We can give 8 hosts here to make it proper

nsub = n + log2 (256/8) = 24 + 5 = 29

So the subnet mask is 255.255.255.248
Invert of the subnet mask 0. 0. 0. 7
Network address is 192.168.23.112/29
Last address (OR Operation) 192.168.23.119/29

First Usable address 192.168.23.113/29
Last Usable address 192.168.23.118/29

Subnet 5 :(Consist of existing WAN)
Fifth subnet, point to point link , up to 4 hosts (2 for connection and 2 for net and broadcast address)
We can give 8 hosts here to make it proper
nsub = n + log2 (256/4) = 24 + 6 = 30

So the subnet mask is 255.255.255.252
Invert of the subnet mask 0. 0. 0. 3
Network address is 192.168.23.120/30
Last address (OR Operation) 192.168.23.123/30

First Usable address 192.168.23.121/30
Last Usable address 192.168.23.122/30

Subnet 6 :(Consist of future WAN)
6th subnet, point to point link, up to 4 hosts (2 for connection and 2 for net and broadcast address)
We can give 4 hosts here to make it proper
nsub = n + log2 (256/4) = 24 + 6 = 30

So the subnet mask is 255.255.255.252
Invert of the subnet mask 0. 0. 0. 3
Network address is 192.168.23.124/30
Last address (OR Operation) 192.168.23.127/30

First Usable address 192.168.23.125/30
Last Usable address 192.168.23.126/30

Subnet 7 :(Consist of future WAN)
6th subnet, point to point link, up to 4 hosts (2 for connection and 2 for net and broadcast address)
We can give 4 hosts here to make it proper
nsub = n + log2 (256/4) = 24 + 6 = 30

So the subnet mask is 255.255.255.252
Invert of the subnet mask 0. 0. 0. 3
Network address is 192.168.23.128/30
Last address (OR Operation) 192.168.23.131/30

First Usable address 192.168.23.129/30
Last Usable address 192.168.23.130/30

source

ip address