Subnet question?

Question

All of your corporate servers have static IP addresses. Some users on your network are complaining that they cannot contact a server named Server 2, which was recently added to the network. You check the TCP/IP configuration on Server 2 and notice the following:

Connection-specific DNS suffix:
IP Address 192.168.64.100
Subnet Mask 255.255.255.0
Default Gateway 192.168.79.1

You notice that the subnet mask value is incorrect. Which of the following mask values would be correct?

A) 255.255.252.0
B) 255.255.254.0
C) 255.255.248.0
D) 255.255.224.0

I am not sure where to start on this. I have several other questions similar to this. What would be the answer and why?

Answer 1

It's a Variable Length Subnet Masking (VLSM) question using RFC1918 classful addressing.

The key here is that they are asking for the correct subnet mask. That indicates the other information given is probably correct.

The easiest way to do this particular problem (but this doesn't always apply) is to look at the addresses supplied - 192.168.64.100 and 192.168.79.1. It's obvious from these two addresses (first two octets are identical) that the subnetting happens in the third octet.

By simply subtracting 64 (third octet from the first address, and also the lowest of the two numbers) from 79 (third octet from the second address, largest of the two numbers), we get a number of 15.

We must round this up to the next highest power of two, which is 16 or 2^4.

We can do the next part a couple of ways. Since we know that the octet is by definition 8 bits wide, we can subtract the 4 least significant bits (from the right), or turn them to zeros, which will give us a binary number of 11110000. Converting this to decimal should yeild the value 240. This is the absolute largest number the third octect of the mask can be and the two addresses be in the same IP network.

An alternative to this is to simple subtract 16 from 256 ( there are 256 possible values of the third octet 0-255), which will also yeild ---- 240.

Now we compare this to the list of possible answers and simple find the mask that comes closest to that number, without exceeding it.

In this case, the correct answer is 255.255.224.0 also known as a /19 in CIDR notation.

Answer 2

Im trying to figure this out also. With the IP given your subnet is correct, but I do believe your Default Gateway is incorrect.

Your third octet should be the same as the IP's.

Seriously this class C subnet is very common. 192.168.x.x will always have a Mask of 255.255.255.0 with the wild card of 0.0.0.255.

Sorry I couldn't help

Answer 3

What error are they getting? Have you tried to PING using IP and the server_name?? If so does one work and one not work? If so its not a subnet issue its a DNS issue. What subnets do you have on your network? If they are all the same then go look at another server and set the SM to that same. Depending on your subnet scope any of those SM could work(that is how you subnetted it)k.

Answer 4

Go through the links...!