Question

If an ISP has a 10.0.0.0/8 you are required to give 125, 30, 500, 6, 2, 2, 2 and host addresses. Supply the needed information in table given.Solve and show solution.

Hosts	n	Prefix Length(32-n)	Max of hosts(2n – 2)	Inc 2n	Network IP	Subnet Mask	Broadcast IP

Answer 1

Arrange the hosts requirements in descending order and assign the highest IP to the highest host requirement.

500

125

30

6

2

2

2

For 500 hosts addresses, 9(2^9=512) bits are required for the host part to uniquely identify each host. The remaining 32-9-8=15 bits are used for subnet part. Thus the subnet mask is /23(8+15).

For 125 hosts addresses, 7(2^7=126) bits are required for host part to uniquely identify each host. The remaining 32-7-8=17 bits are used for subnet part. Thus the subnet mask is/25(8+17).

For 30 hosts addresses, 5(2^5=32) bits are required for host part to uniquely identify each host. The remaining 32-5-8=19 bits are used for subnet part. Thus the subnet mask is /27(8+19).

For 6 hosts addresses, 3(2^3=8) bits are required for host part to uniquely identify each host. The remaining 32-3-8=21 bits are used for subnet part. Thus the subnet mask is /29(8+21).

For 2 hosts addresses, 2(2^2=4) bits are required for host part to uniquely identify each host. The remaining 32-2-8=22 bits are used for subnet part. Thus the subnet mask is/30(8+22).

Hosts	n	Prefix Length	Max of hosts	Inc 2n	Network IP	Subnet Mask	Broadcast
500	9	32-9=23	512-2=510	512	10.0.0.0	/23	10.0.1.255
125	7	32-7=25	128-2=126	128	10.0.2.0	/25	10.0.2.127
30	5	32-5=27	32-2=30	32	10.0.2.128	/27	10.0.2 159
6	3	32-3=29	8-2=6	8	10.0.2.160	/29	10.0.2.167
2	2	32-2=30	4-2=2	4	10.0.2.168	/30	10.0.2.171
2	2	32-2=30	4-2=2	4	10.0.2.172	/30	10.0.2.175
2	2	32-2=30	4-2=2	4	10.0.2.176	/30	10.0.2.179