A network consists of 2,56,000 hosts has organized into subnets at three levels.
ID: 3634122 • Letter: A
Question
A network consists of 2,56,000 hosts has organized into subnets at three levels. At level 1, all 2,56,000 hosts are connected, at level 2 , hosts are divided into four groups, and at level 3, each group at level 2 is further divided into four subgroups. Assume partitions at level 2 and level 3 are performed uniformly. Indicate first and last address of groups and sub-groups at each level. Derive the network masks at each level of routing to the desired destinations. The required IP address space is provided by ISP starting from 132.96.0.0.
Explanation / Answer
assuming 2,56,000 ip's are allocated then it needs 18 bits for host address so
network mask at level 1 is :255.252.0.0
using 2 bits less for host address as divided into 4 groups--->
network mask at level 2 is:255.255.0.0
using 2 bits less for host address as divided into 4 groups---->
network mask at level 3 is:255.255.192.0
given ip is 132.96.0.0,network mask not given assumed 255.255.252.0 as we need 2^18 ip addresses which is 2,56,000,so taking last 18 bits for host address.
for level 1:-range of ip's is 132.96.0.0 to 132.99.255.255 and network mask 255.252.0.0
for level 2:range of ip's for group 1 is :-132.96.0.0 to 132.96.255.255 and network mask 255.255.0.0
range of ip's for group 2 is :-132.98.0.0 to 132.96.255.255 and network mask 255.255.0.0
range of ip's for group 3 is :-132.97.0.0 to 132.96.255.255 and network mask 255.255.0.0
range of ip's for group 1 is :-132.99.0.0 to 132.96.255.255 and network mask 255.255.0.0
in level 3 we are using 2 more bits for network address so network mask is 255.255.192.0
as we are dividing into 4 groups,they have those bits as 00,01,10,11 respectively...from what i told its easy to calculate for level 3 the range of ip's like
for level 3, group 1 in level 2 is divided into 4 groups so group 11 will be having ip's from 132.96.0.0 to 132.96.63.255
for group 12 it will be 132.96.0.0 to 132.127.255 and so on like this you can calculate
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