Provide a detailed answer to each question. Consider a network with the IP addre

Question

Provide a detailed answer to each question.

Consider a network with the IP address 192.168.1.1/16

Part I

What is the network prefix for this network, in the form a.b.c.d/x?

Assuming that this network is to be broken down into two subnets of equal size, write down the network prefix of each subnetwork.

Assume that the 1st subnet is divided into 5 equal subnetworks, and write down the network prefix of each subnet.

Of the 5 subnets, select the 1st one and write down the first usable address, the last usable address, and the broadcast address.

Part II

For the other subnet of the original address range above, assume that 4 subnetworks need to be created with the following details A - 300 hosts, B - 500 hosts, C - 88 hosts and D - 60 hosts.

Using the concept of Variable Length Subnetting, derive and write down the network prefixed of A, B, C and D.

Assuming that network A is further divided into two subnetworks (A1 and A2) of size 30 and 200, write down the network prefixed of these subnets.

Explanation / Answer

Consider a network with the IP address 192.168.1.1/16
Part I
What is the network prefix for this network, in the form a.b.c.d/x?
192.168.1.1/16
Assuming that this network is to be broken down into two subnets of equal size, write down the network prefix of each subnetwork.
here we have to divide into 2 subnet so that 1 but will be increase in mask.
192.168.1.1/17
192.168.3.1/17

Assume that the 1st subnet is divided into 5 equal subnetworks, and write down the network prefix of each subnet.
network is divided into 5 subnet so that atleast 3 bit will be used in subnet mask
192.168.1.1/20
192.168.2.1/20
192.168.3.1/20
192.168.4.1/20
192.168.5.1/20
  
Of the 5 subnets, select the 1st one and write down the first usable address, the last usable address, and the broadcast address.
1. 192.168.1.1/20
broadcast address : 192.168.15.255

192.168.2.1/20
broadcast address : 192.168.15.255

192.168.3.1/20
broadcast address : 192.168.15.255

192.168.4.1/20
broadcast address : 192.168.15.255

192.168.5.1/20
broadcast address : 192.168.15.255

Part II
For the other subnet of the original address range above, assume that 4 subnetworks need to be created with the following details A - 300 hosts, B - 500 hosts, C - 88 hosts and D - 60 hosts.
Using the concept of Variable Length Subnetting, derive and write down the network prefixed of A, B, C and D.
Assuming that network A is further divided into two subnetworks (A1 and A2) of size 30 and 200, write down the network prefixed of these subnets.
sol:
A - 300 hosts, B - 500 hosts, C - 88 hosts and D - 60 hosts.
A need 300 host so that it need 9 bit address from host whose range is max 2^9 - 2 = 510 host
so that subnet mask will be
A : 192.168.1.1/18
now A will be divided into 2 subnet of A1:30 , A2:200
A1 : 192.168.2.1/19
A1 : 192.168.6.1/19
B : 192.168.2.1/18
C : 192.168.3.1/18
D : 192.168.4.1/18

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