Given the network 192.168.100.0/24, we are going to segment this network into two subnets, Subnet A and Subnet B. The hosts requirement for Subnet A is 100 and for Subnet B is 42. Calculate belowSubnet A:Subnet MaskNumber of Useable Hosts in this SubnetIP Network AddressFirst valid IP AddressLast valid IP AddressSubnet Broadcast IP AddressSubnet B:Subnet MaskNumber of Useable Hosts in this SubnetIP Network AddressFirst valid IP AddressLast valid IP AddressSubnet Broadcast IP Address

Answer and Explanation:

Each subnet can only be power of 2s.

Thus first subnest would have 2^7 = 128 IPs (100 were required)and second subnet would have 2^6 = 64 IPs (42 were required).

Subnet A with 100 Hosts required for 192.168.100.0/25 network:

Subnet mask -> 255:255:255:128

(Mask for 11111111:11111111:11111111:1000000)

Number of usable Host : 2^7 - 1 - 1 = 126

(Since we cannot use IP addresses ending in 0 and 127)

IP Network Address:

192.168.100.0/25

First Valid IP address:

192.168.100.1

Last Valid IP address:

192.168.100.126

Subnet Broadcast IP:

192.168.100.127

Subnet B with 42 Hosts required for 192.168.100.128/26 network:

Subnet mask -> 255:255:255:192

(Mask for 11111111:11111111:11111111:1100000)

Number of usable Host : 2^6 - 1 - 1 = 64-2 = 62

(Since we cannot use IP addresses ending in 0 and 63)

IP Network Address:

192.168.100.128/26

First Valid IP address:

192.168.100.129

Last Valid IP address:

192.168.100.190

Subnet Broadcast IP:

192.168.100.191

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