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.
Simham said:
9 years ago
Why are taking Class B? This is a classless domain. Can any one please tell me.
(1)
Mehedi hasan dipu said:
10 years ago
I think this answer is incorrect what will be the right answer. If this answer is correct so then please describe the correct answer.
Manu Ratheesh said:
10 years ago
Given IP is 172.16.112.1/25 => NID is 25 bits.
We know that subnet mask is obtained by replacing all NID bits with 1 and HID bits with 0.
So SM is 255.255.255.10000000 (First 25 bits is NID and the last 7 bit is HID).
i.e, SM is 255.255.255.128.
Now the subnet address is obtained by (SM) AND (IP).
=> (255.255.255.128) AND (172.16.112.1) = 172.16.112.0.
Option (A).
We know that subnet mask is obtained by replacing all NID bits with 1 and HID bits with 0.
So SM is 255.255.255.10000000 (First 25 bits is NID and the last 7 bit is HID).
i.e, SM is 255.255.255.128.
Now the subnet address is obtained by (SM) AND (IP).
=> (255.255.255.128) AND (172.16.112.1) = 172.16.112.0.
Option (A).
Stranger said:
10 years ago
I don't understand this please help.
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