Member Posts: 241
I came across this question on a practice exam and was wondering what some methods were for solving it.

I understand how to determine the gateway address, but is there a simple way to determine the IP address and subnet mask?

Next up: 100-101 ICND1 :study:

Keep the Son in your eyes!

• Member Posts: 6 ■□□□□□□□□□
Wow, this is a hard question, can take some time to solve, but I believe I have the answer.

You know the default gateway will be 136.28.90.34 because this is the only router address, and as you mentioned, that part is easy.

The rest is more difficult. The first thing you will want to do is determine the subnet mask. You can do this by looking at the only information you have, which is the two addresses you know are on the subnet, 136.28.86.63 and 136.28.90.34. Since we know these have to be on the same subnet, one could make the assumption that the subnet mask will use the greatest amount of bits possible using these two addresses:

136.28.86.63 = xxxxxxxx.xxxxxxxx.01010110.xxxxxxxx
136.28.90.34 = xxxxxxxx.xxxxxxxx.01011010.xxxxxxxx

Therefore the subnet mask is 255.255.240.0. Now that we know that, we can eliminate all other subnet masks from the possible answers.

Next we will determine what the possible host range is for this network.

Range = 136.28.80.0 - 136.28.95.255

Using this range you can eliminate all remaining answers but two:

136.28.90.34 and 136.28.86.63.

Well we know that the latter of those two is already in use by Server1, so, the only available address not in use, thats on the same subnet, is 136.28.90.34.

Of course in real life if you had to use this method to determine which address to use, you probably did not plan out your network very well
MCSA 2003 (70-270, 70-290, 70-293), Security+
MCSA 2008 (70-640... Next up is 70-642, 70-646)
• Banned Posts: 12,314 ■■■■■■■■□□
Nice response, Nick.

Yes, as demonstrated, the technique is to determine the subnet mask from the two known addresses, and the result should leave you with the only other usable IP address in that range. Glad I missed this the first time around, so I didn't have to type all that
• Member Posts: 8 ■□□□□□□□□□
Using this range you can eliminate all remaining answers but two:
136.28.90.34 and 136.28.86.63.
Well we know that the latter of those two is already in use by Server1, so, the only available address not in use, thats on the same subnet, is 136.28.90.34.
Oops. The problem with that answer is that BOTH of those two adddresses are already in use: the first by the router, and the second by the existing Server1. So that answer cannot be correct. We're on the right track, but since we now know that this isn't quite right, it's obvious that our subnet MUST use at least one less subnet bit. So let's modify the original assumption from:
Since we know these have to be on the same subnet, one could make the assumption that the subnet mask will use the greatest amount of bits possible using these two addresses:
136.28.86.63 = xxxxxxxx.xxxxxxxx.01010110.xxxxxxxx
136.28.90.34 = xxxxxxxx.xxxxxxxx.01011010.xxxxxxxx

To this (adding another host bit):

Subnet Mask = = xxxxxxxx.xxxxxxxx.11100000.xxxxxxxx (255.255.224.0)

That gives us the subnet range of 136.28.64.0 to 136.28.96.255, which still covers the requirements, still includes the SERVER1 and ROUTER addresses, but ALSO now includes a third address from the options: 136.28.70.55.

Since that third option is the ONLY one that meets the criteria (and that isn't already accounted for and being used by another device), that's the actual answer. :^)

IP: 136.28.70.55
NM: 255.255.224.0
GW: 136.28.90.34
• Member Posts: 4,212 ■■■■■■■■□□
The default gateway is easy. It has to be a router, and it has to be on the same subnet. Only 1 address fits both of those, and that is the 136.28.90.34.

225.255.255.0
255.255.224.0 (255.255.11100000.0)
225.255.240.0 (255.255.11110000.0)

There are 2 known addresses on that subnet, 136.28.86.63 and 136.28.90.34. I'm no subnetting guru, but I can tell you hands down that it isn't a class C subnet mask (if it were the 3rd octet would need to be the same and 86 is not 90), so now we have it down to a 50-50.

We'll start listing out the network addresses for the other 2 masks. We will either have a counter of 32 or 16.
136.28.0.0
136.28.32.0
136.28.64.0
136.28.96.0

Since 86 and 90 both fall in that same gap, this is possible. Lets try with a counter of 16.

136.28.0.0
136.28.16.0
136.28.32.0
136.28.48.0
136.28.64.0
136.28.80.0
136.28.96.0

Both 86 and 60 still fall within the same range. Interesting...

I'm going to put that on hold for a second and move on to potential addresses. (Just listing anything with a 136.28.x.x address)
136.28.201.23
136.28.90.34
136.28.97.158
136.28.138.84
136.28.86.63
136.28.70.55

Remember that the range we are looking for is either 64-96 or 80-96. So you can now remove the 1st, 3rd, and 4th.
This leaves
136.28.90.34 (in both ranges, but also in use by the router)
136.28.86.63 (in both ranges, but also in use by server 1)
136.28.70.55 (1st range only)

So my answer would be the following. I only see only 1 valid IP and it is only in 1 subnet.
IP: 136.28.70.55