Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 9)
9.
What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
Answer: Option
Explanation:
A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.
Discussion:
14 comments Page 1 of 2.
Racharla Chandra Kanth said:
1 decade ago
There are 21 N/w bits and the given IP is a Class B IP so by default 16 bits are for n/w. Here 5 bits are used for subnetting from the third octet.
66: 01000010
Subbnet 11111000 Anding
-------------
01000000 = 64.
Therefore n/w id: 172.16.64.0.
66: 01000010
Subbnet 11111000 Anding
-------------
01000000 = 64.
Therefore n/w id: 172.16.64.0.
Abhijeet said:
1 decade ago
That 66 came from where?
Stranger said:
10 years ago
Who next subnet be 72? If we increased by 8.
Stranger said:
10 years ago
How next subnet be 72?
Manu Ratheesh said:
10 years ago
Given IP address is 172.16.66.0/21 => NID is 21 bits and HID is 11 bits.
=> Subnet Mask is 255.255.248.0 (SM is obtained by placing 1 in all NID bits and 0 in all HID bits).
In order to get the sub network number or network address or subnet id, perform bitwise AND between the IP address and subnet Mask.
=> (172.16.66.0) AND (255.255.248.0) = 172.16.64.0. Option (C).
=> Subnet Mask is 255.255.248.0 (SM is obtained by placing 1 in all NID bits and 0 in all HID bits).
In order to get the sub network number or network address or subnet id, perform bitwise AND between the IP address and subnet Mask.
=> (172.16.66.0) AND (255.255.248.0) = 172.16.64.0. Option (C).
Yogesh said:
9 years ago
Why 172.16.0.0 is not the subnet network?
Nehh said:
9 years ago
172.16.0.0 cannot be the Network address because according to classful addressing we take class B address directly 255.255.0.0.Here in classless addressing CIDR notation /21 is given. So subnet mask will be on that basis and after AND IP and SMask, you will get network address 172.16.64.0.
Sechonge said:
9 years ago
There are 21 N/w bits and the given IP is a Class B IP so by default 16 bits are for n/w. Here 5 bits are used for subnetting from the third octet.
66: 01000010
Subnet 11111000 Adding
-------------
01000000 = 64.
Therefore n/w id: 172.16.64.0.
66: 01000010
Subnet 11111000 Adding
-------------
01000000 = 64.
Therefore n/w id: 172.16.64.0.
Munna said:
9 years ago
There are 21 N/w bits and the given IP is a Class B IP so by default 16 bits are for n/w.
Here 5 bits are used for subnetting from the third octet.
66: 01000010
Subnet 11111000 Adding
-------------
01000000 = 64.
Therefore n/w id: 172.16.64.0.
Here 5 bits are used for subnetting from the third octet.
66: 01000010
Subnet 11111000 Adding
-------------
01000000 = 64.
Therefore n/w id: 172.16.64.0.
Ehsan Faghih said:
9 years ago
Consider it we have 1 bit for subnet. Then how many of sub-network we could have? answer:2.
Why: because one bit only can determine 2 states (ways) for us! (0 or 1).
In this case, when we determine the subnet mask we found that we have 5bits for subnets.
Now tell me " how many states provide for us by 5bits? 32 or 64?
YES! 64 states is true answer!
Then we will have 64 sub-network by 5bits subnets.
Why: because one bit only can determine 2 states (ways) for us! (0 or 1).
In this case, when we determine the subnet mask we found that we have 5bits for subnets.
Now tell me " how many states provide for us by 5bits? 32 or 64?
YES! 64 states is true answer!
Then we will have 64 sub-network by 5bits subnets.
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