__- Network A 10.102.1.78 Network B 10.102.0.1 10.102.1.255 10.102.0.2 11 10.102
ID: 3605818 • Letter: #
Question
__- Network A 10.102.1.78 Network B 10.102.0.1 10.102.1.255 10.102.0.2 11 10.102.0.15 10.102.0.3 10.102.0.2 17.102.1.78 Network C 17.85.17.64 17.19.57.9 i) Assume that the three networks indicated in the diagram use the smallest possible subnets (i«e smallest number of addresses). Give the network and broadcast addresses for those subnets as well as the number of bits in the host part of addresses. [4 marks Network Network address Broadcast address Host bits Network Network E etwor ii) Something is wrong with the collection of networks above. What is it? 2 marks)Explanation / Answer
i)
Consider Ip addresses of Network A 10.102.1.78 , 10.102.1.255
and 10.102.0.15 , 10.102.0.2 which can be written as
10.102.00000001.78
10.102.00000001.255
10.102.00000000.15
10.102.00000000.2
Here the 1st 23 bits are common for all the 4 systems and hence
Network Address is 10.102.0.0
Broadcast Address is 10.102.00000001.11111111 which is 10.102.1.255
Network bits is 23 and hence host bits is 32 - 23 = 9.
Consider Ip addresses of Network B 10.102.0.1 , 10.102.0.2
and 10.102.0.3 which can be written as
10.102.0.00000001
10.102.0.00000010
10.102.0.00000011
Here the 1st 30 bits are common for all the 3 systems and hence
Network Address is 10.102.0.0
Broadcast Address is 10.102.0.00000011 which is 10.102.0.3
Network bits is 30 and hence host bits is 32 - 30 = 2.
Consider Ip addresses of Network C 17.102.1.78 , 17.85.17.64
and 17.19.57.9 Here 17 is common in all and hence
Network Address is 17.0.0.0
Broadcast Address is 17.255.255.255
Network bits is 8 and hence host bits is 32 - 8 = 24.
ii)
1) We are getting same network ID for both the networks A and B
but they are seperated above.
2) In Network A the IP 10.102.1.255 contains all 1's in its host address
which is not allowed.
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