Question

Part 1: Develop the IPV4 Addressing Scheme Given an IP address and mask of 192.168.1.0 /24 (address / mask), design an IP addressing scheme that satisfies the following requirements Subnet Number of Hosts Subnet A 20 Subnet B 60 Use subnet zero for the first subnet. No subnet calculators are allowed Host computers will use the last IP address in the subnet. The network router will use the first usable host address. The switch will use the second usable host address. Develop addressing scheme.

Answer 1

List of subnets possible:

Subnet mask(in slash notation)	No of possible hosts
/24	256
/25	128
/26	64
/27	32
/28	16
/29	8
/30	4

Subnet A requires 20 hosts and subnet B requires 60 hosts.

Subnet with higher number of hosts is assigned first. Hence subnet B is assigned the IP address 192.168.1.0/26. As there are 64 possible IP addresses with /26 mask, one is used for network router, second IP address is used for switch and last IP address is used for broadcast by all hosts in that subnet. Thus available number of hosts are 61 hosts. Hence subnet B is given the IP address 192.168.1,0/26.

Next subnet to be assigned is subnet A. Subnet A requires 20 hosts. Thus /27 is assigned. The network IP for subnet A is 192.168.1.64/27. With /27 the number of hosts possible is 32. One is assigned for network router and other for switch and last address for broadcast. Hence 29 host addresses are available. Hence subnet A is given the IP address 192.168.1.64/27.