ip address

logistic_guy

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If an Ethernet port on a router were assigned an IP address of \(\displaystyle 172.16.112.1/25\), what would be the valid subnet address of this interface?

\(\displaystyle \bold{A.} \ 172.16.112.0\)
\(\displaystyle \bold{B.} \ 172.16.0.0\)
\(\displaystyle \bold{C.} \ 172.16.96.0\)
\(\displaystyle \bold{D.} \ 172.16.255.0\)
\(\displaystyle \bold{E.} \ 172.16.128.0\)
 
If an Ethernet port on a router were assigned an IP address of \(\displaystyle 172.16.112.1/25\), what would be the valid subnet address of this interface?

\(\displaystyle \bold{A.} \ 172.16.112.0\)
\(\displaystyle \bold{B.} \ 172.16.0.0\)
\(\displaystyle \bold{C.} \ 172.16.96.0\)
\(\displaystyle \bold{D.} \ 172.16.255.0\)
\(\displaystyle \bold{E.} \ 172.16.128.0\)
Is this an assignment from an instructional class?

Please provide a copy of your attempts to solve it - and indicate where you are stuck.
 
Thank you a lot professor khan for passing by. The IP address is my game. It is just that I was touring the United States in the last two months and had not practised anything. I am not a Genius like you, so I forgot some stuff. This problem can be solved easily if my \(\displaystyle 1000\) posts were not deleted as they were my references.

Is this an assignment from an instructional class?
I don't think so.

Please provide a copy of your attempts to solve it - and indicate where you are stuck.
Give me back my references, and you will be surprised by my attempt. I am sure that you know it is very boring to reinvent the wheel. I just need to take a look at my references for \(\displaystyle 1\) second to remember the strategy of how to attack this problem. However, my intuition tells me that the answer is \(\displaystyle \bold{A}\).
 
@khansaheb

Suppose you have a network consists of three computers and they have the following IPs:

\(\displaystyle 172.16.112.1\)
\(\displaystyle 172.16.112.2\)
\(\displaystyle 172.16.112.3\)

It makes sense that the network address is \(\displaystyle 172.16.112.0\). The multiple choice answers are tricky because each one of them is a valid network address in class \(\displaystyle \text{B}\) but the ip address \(\displaystyle 172.16.112.1\) can lie only within the network address \(\displaystyle 172.16.112.0\).
 
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