Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 10)
10.
The network address of 172.16.0.0/19 provides how many subnets and hosts?
Answer: Option
Explanation:
A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.
Discussion:
20 comments Page 1 of 2.
Pragati said:
1 decade ago
How to calculate number of hosts?
Virendra singh said:
1 decade ago
19-16 = 3 (16 bit network address for class B).
2^3 = 8 subnet.
Total 32 bits.
32-19 = 13 bits available for host so.
2^13 = 80192 hosts for each subnet.
But 2 addresses are reserved for network and broadcast address therefore 80192-2 = 80190 hosts for each subnet are available.
2^3 = 8 subnet.
Total 32 bits.
32-19 = 13 bits available for host so.
2^13 = 80192 hosts for each subnet.
But 2 addresses are reserved for network and broadcast address therefore 80192-2 = 80190 hosts for each subnet are available.
Rambabu said:
1 decade ago
It is wrong answer because.
Number of hosts per subnet is ok, but 2^3 = 8 is wrong answer,
2^3 = 8-2 => 6.
Only 6 subnet are useful.
172.16.0.0---it is used for Network id.
And 172.16.255.255 is used for broadcast address, so we can't use 8 subnet id's even we have 3 bit's for subting.
Number of hosts per subnet is ok, but 2^3 = 8 is wrong answer,
2^3 = 8-2 => 6.
Only 6 subnet are useful.
172.16.0.0---it is used for Network id.
And 172.16.255.255 is used for broadcast address, so we can't use 8 subnet id's even we have 3 bit's for subting.
Vasil said:
1 decade ago
Everything is correct, 3-bits are 8-subnets 2^3=8.
And then we have left 13 bits to work with which is:
2^13 = 80192-2 = 80190.
11111111.11111111.11100000.00000000.
1- = 19.
0 = 13---5+8 = 13.
I hope this is clear.
And then we have left 13 bits to work with which is:
2^13 = 80192-2 = 80190.
11111111.11111111.11100000.00000000.
1- = 19.
0 = 13---5+8 = 13.
I hope this is clear.
SATHISH.C said:
1 decade ago
No actually I accept the answer of Rambabu because 2^3 = 8-2 = 6.
1 is for n/w and 1 for broadcast.
Host is right 2^13 = 8192-2 = 8190.
1 is for n/w and 1 for broadcast.
Host is right 2^13 = 8192-2 = 8190.
Vikas beniwal said:
1 decade ago
No. of network doesn't require to subtract anything.
Its only applicable for number of hosts that one is used for Broadcast and for the network.
So the formula for host is 2^n-2.
No. of subnets/networks is 2^n.
Its only applicable for number of hosts that one is used for Broadcast and for the network.
So the formula for host is 2^n-2.
No. of subnets/networks is 2^n.
Kalimullah Ahmadzai said:
1 decade ago
@Vikas Beniwal and @Vasil are right. we don't need to subtract from the network portion.
Davil said:
10 years ago
Wrong answer. 2 power 5 is number of host per subnet.
Gajulapalli naveen said:
10 years ago
SubnetMask CIDR NetBits HostBits #of Nets TotalHosts Hosts/Net.
255.255.224.0/19 3 13 6(8*) 8190 49140 (65520*).
Host is right 2^13 = 8192-2 = 8190.
Its class B address.
11111111 11111111 11100000 00000000.
Class A Class B Class C.
We have 2^3 8-sub networks. For 1 sub network we are calculating the how many hosts we have.
8192 hosts we have from these hosts we are subtracting 2 IP address network id and broadcast address.
255.255.224.0/19 3 13 6(8*) 8190 49140 (65520*).
Host is right 2^13 = 8192-2 = 8190.
Its class B address.
11111111 11111111 11100000 00000000.
Class A Class B Class C.
We have 2^3 8-sub networks. For 1 sub network we are calculating the how many hosts we have.
8192 hosts we have from these hosts we are subtracting 2 IP address network id and broadcast address.
Mehedi hasan said:
10 years ago
Answer is wrong, the correct answer will be 16128 and useable host: 16126 but the subnet is ok.
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