Networking Task Assessment Questions

Question 2.

1. Router A subnet:
   Subnet Mask: 255.255.255.192
   Network ID: 192.168.10.0/26 (First address of the subnet block)
   Broadcast ID: 192.168.10.63 (Last address of the subnet block)
   Host Address Range: 192.168.10.1 to 192.168.10.126
2. For Router B subnet:
   Subnet Mask: 255.255.255.192
   Network ID: 192.168.19.64/26 (First address of the subnet block)
   Broadcast ID: 192.168.10.127 (Last address of the subnet block)
   Host Address Range: 192.168.10.65 to 192.168.10.126

Explanation:
Given IP address space= 192.168.10.0/25 hence given network address is 192.168.10.0 and subnet mask 255.255.255.128
Can also put 126 hosts in the network because total 128 address are available in which first address is 192.168.10.0 and last address 192.168.10.128 is the broadcast address
As given that two routers A and B are connected with 50 PCs and 20 PCs respectively, we divide the given range into two subnets for connecting PCs in two routers. The formula to accommodate maximum hosts in the minimum subnet size is 2^n-2 (n is the number of bits used to represent a host and the -2 means to subtract the two unused address (network and broadcast address).
Applying the formula 2^6-2= 64 – 2 = 62.
Now if we divide our network into two subnets with a network mask 26, 62 hosts can be connected maximum in each subnet.