Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 7)
7.
If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?
Answer: Option
Explanation:
A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
Discussion:
14 comments Page 2 of 2.
Darshil Soni said:
6 years ago
This answer is incorrect because, this is clasless inter domain routing. And /25 represents this:
||||||||.||||||||.||||||||.|0000000 this means that the subnet mask is 255.255.255.128.
There will be 2 Networks with 510 valid hosts in it. And the block size will be 128.
Network IP - First Host - Last Host - Broadcast IP
172.16.0.0 - 172.16.0.1 - 172.16.127.254 - 172.16.127.255
172.16.128.0 - 172.16.128.1 - 172.16.255.254 - 172.16.255.255
172.16.256.0
||||||||.||||||||.||||||||.|0000000 this means that the subnet mask is 255.255.255.128.
There will be 2 Networks with 510 valid hosts in it. And the block size will be 128.
Network IP - First Host - Last Host - Broadcast IP
172.16.0.0 - 172.16.0.1 - 172.16.127.254 - 172.16.127.255
172.16.128.0 - 172.16.128.1 - 172.16.255.254 - 172.16.255.255
172.16.256.0
(5)
Arnold Subhashnagar said:
5 years ago
@All.
It is said that it is a 1/25 so the valid IP address will fall only in this domain of having 24-bit octet containing the one extra bit as per the question.
It is said that it is a 1/25 so the valid IP address will fall only in this domain of having 24-bit octet containing the one extra bit as per the question.
Damy mgimba said:
5 years ago
I'm not understanding, anyone, can explain more?
Shotande Kehinde said:
2 years ago
172.16.112.1/25 will have 2subnets.i.e I bit is borrowed from the host bit x=1, therefore x=2^1 =2.
Expectedly the block size is 128 resulting to the 2subnets given as 172.16.112.0.
And 172.16.112.128.
Note; 172.16.112.1 is the first usable host of 172.16.112.0.
And is 172.16.112.0.
Expectedly the block size is 128 resulting to the 2subnets given as 172.16.112.0.
And 172.16.112.128.
Note; 172.16.112.1 is the first usable host of 172.16.112.0.
And is 172.16.112.0.
(1)
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