Computer Networks - Computer Networks Section 3
- A
204.204.204.128/255.255.255.192
204.204.204.0/255.255.255.128
204.204.204.64/250.255.255.128
- B
204.204.204.0/255.255.255.192
204.204.204.192/255.255.255.128
204.204.204.64/255.255.255.128
- C
204.204.204.128/255.255.255.128
204.204.204.192/255.255.255.192
204.204.204.224/255.255.255.192
- D
204.204.204.128/255.255.255.128
204.204.204.64/255.255.255.192
204.204.204.0/255.255.255.192
- Option : D
- Explanation : MSB in last 8 bits helps us to get two subnets
10000000 ⟶ subnet1
00000000 ⟶subnet2
subnet2 is divided into 2 more subnets using 7th bit
00000000 ⟶ subnet2(0)
01000000 ⟶ subnet2(1)
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- Option : C
- Explanation : Subnet address of interface = 144. 16 . 68 .0
Subnet address of incoming IP
= 144. 16. 68. 117 BITWISE AND 255.255.255.0
= 144. 16. 68.0
Both subnetwork addresses are same, so packet is forwarded to interface eth2
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18. A subnetted Class B network has the following broadcast address : 144.16.95.255. Its subnet mask
- Option : D
- Explanation : For broadcast address of a subnet, all the host
bits are set to 1.
Broadcast address for subnet is
.95.255 ---> .0101 1111. 1111 1111
So, Possible host bits can be 13 to 0
Hence could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
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- Option : C
- Explanation : The two computer C1 and C2 are configured as
follows :
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- Option : C
- Explanation : Maximum number of subnets
= 26 - 2 = 62
Maximum number of host in each subnet
= 210 – 2 = 1022
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