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 1 of 2.
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)
Damy mgimba said:
4 years ago
I'm not understanding, anyone, can explain more?
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.
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)
Naomi Nash said:
6 years ago
Then what's the no of networks and host per network?
Being 1 bit in the last bit is affected, the number of networks (which is the same as subnets) is;
Subnet= 2^n.
=2^1,
=2.
A number of Hosts= 2^k-2.
Number of Host bits= 32-25=7 bits,
=2^7-2,
=128-2,
=126 Hosts per subnet.
Being 1 bit in the last bit is affected, the number of networks (which is the same as subnets) is;
Subnet= 2^n.
=2^1,
=2.
A number of Hosts= 2^k-2.
Number of Host bits= 32-25=7 bits,
=2^7-2,
=128-2,
=126 Hosts per subnet.
(2)
Amala said:
8 years ago
Then what's the no of networks and host per network?
Mani said:
8 years ago
172.16.112.1/25 question simply 24 bit octal +1 another eight 0, so ip address 172.16.112.0 mention 1;.
(1)
Jimbo jones said:
9 years ago
I think the answer is 172.16.0.0 network. It's still a class b network so you can only touch the 3rd octet.
Gurpreet singh said:
9 years ago
Sometimes it happens our IP series belongs to class b IP but subnet mask of class. So when ever its happen you have to see IP address behave on that you can understand series, in this question. IP address starts with 172 that mean class b IP.
Lucas Mathew said:
9 years ago
Why use class C subnet mask if the IP is in class B?
(1)
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