Summary
Highlights
The video revisits the network topology, identifying Network B which requires 20 hosts. To accommodate this, a /27 subnet is chosen from the available /26 slice. The calculation involves subtracting the old slash (26) from the new (27), resulting in 2 subnets. The block size for this /27 subnet is determined to be 32 (256 - 224).
The two subnets generated from the previous calculation are 192.160.10.192/27 and 192.160.10.224/27. The first valid host for the 192.160.10.192/27 subnet is 192.160.10.193, and the last valid host is 192.160.10.222. The broadcast address is 192.160.10.223. This subnet is assigned to Network B.
For interconnecting networks (links), the video emphasizes using a /30 subnet mask as it is the most efficient and prevents IP address wastage. Breaking up the previous /27 subnet for a link, the calculation (30 - 27) yields 8 subnets. The block size for /30 is 4 (256 - 252).
The generated /30 subnets start from 192.160.10.224. One of these subnets, 192.160.10.224/30, is used for the link between Network A and Network B. Its first valid IP is 192.160.10.225, and the last is 192.160.10.226, with a broadcast of 192.160.10.227.
The video demonstrates how to assign IP addresses and default gateways. For LAN A, the default gateway is 192.160.10.190/26, and client PCs would receive addresses like 192.160.10.129, 192.160.10.130, up to 192.160.10.189. For the link, addresses like 192.160.10.225 and 192.160.10.226 would be used.
The video then addresses subnetting for networks requiring more than 256 hosts, using an example of 500 hosts. A /23 subnet is needed for 500 hosts. The calculation (23 - 22) yields 2 subnets. A new rule for block size is introduced: if the last byte of the mask is zero, subtract from the third byte (256 - 254), resulting in a block size of 2.
The two subnets derived from the /23 calculation are 12.83.160.0/23 and 12.83.162.0/23. The first valid host for the 12.83.160.0/23 subnet is 12.83.160.1, and the last valid host is 12.83.161.254, with a broadcast of 12.83.161.255. This is assigned to LAN A.
For Network B, requiring 100 hosts within the larger network, a /25 subnet is chosen. Calculating from the /23, (25 - 23) results in 4 subnets. The block size for /25 is 128 (256 - 128). The subnets generated are 12.83.162.0/25, 12.83.162.128/25, 12.83.163.0/25, and 12.83.163.128/25.