Danny, the subnet ID is 192.168.15.28, not 192.168.15.228.
A shortcut to derive the subnet ID
192.168.15.30 and subnet mask 255.255.255.252
Value in the Interesting octet = 30
Bit value of the last ON bit = 4 ( Last octet binary - 11111100 = 4)
Divide them - 30/4 = Round down 7.5 = 7
Multiply the result with the bit value 7*4 = 28 --> That's the subnet ID. 192.168.15.28 - 192.168.15.31
Thanks for your speedy response.
My bad. I had a typo and the IP address is 192.168.15.230 and not 192.168.15.30.
However, your explanation would still apply (and would save a lot of time).
So, for 192.168.15.230, you could divide 230/4 = 57.5, which rounded down would be 57
Multiplying the result then with the bit value (or increment): 57 * 4 = 228.
So, the subnet would be 192.168.15.228 - 192.168.15.231.
I think I will apply this when the subnet cannot be calculated straight away (by counting up multiples of the increment).
On a separate note, my concern comes from what I have read that you should be able to calculate subnet ids in less than 30 seconds for the icnd1 exam. Hence, I am looking for any possible way to minimise my calculation times.
Thanks very much Sarah again.
I find the method that is fastest, is the method that your most comfortable with. For me, i use the multiples of 2 to get a baseline, so 2, 4, 8, 16, 32, 64, 128, 256 etc etc.
So for your example, 192.168.15.230/30 i know its going to be a multiple of 4 based on the subnet mask 256-252 = 4
i also know that the IP is 230, and that 240 is a multiple of 2 value, so i simple subtract 12 from that to get 228. The reason i subtract 12 is that 8 would give me 232 which is more than 230, so subtract another 4 makes it 228.
for me having to work out without a calculator what 230/4 is, takes more time than me subtracting 12 from 240.
As i said though, it's really up to you and what you can understand or learn or what seems to make sense best for you. And remember, Practice makes perfect. You are correct in the being able to do the subnet calculations in 30 seconds or less though.
I found these websites great for practicing
Thanks everyone for your replies.
@Zerihun: Thanks for that screenshot. I am following Odom's book and that mehod (magic number) was the one I tried at first but didn't quite get comfortable with it.
Hence, I started looking around and found the below method:
which was easy to understand and quickly to calculate in my head (Just using those boundaries and the current mask in / notation).
@Sara: Thanks for the time spent in your posts. I see what you are saying. I know that certain divisions are difficult without pen and paper and I might spend valuable time in them. The thing is, since the method I am using doesn't really consider the decimal notation of the mask, I didn't want to add a new step that looks at that notation.
I am not sure if I am making sense. But just tring to find a reliable way that gives me the right values all the time.
In any case, may just be that I need to keep practising. I have only started a month ago and thought I would have managed to do all the calculations easily in my mind now. What I am also doing is spending time daily practising separate for the other material for the icnd1 exam.
As already mentioned by @Raymond, the easiest way to find the network ID of any IP Address is by following these steps:
1. Find the IP address class (Very much necessary, because you could be having a class A address with /24 mask)
2. Subtract subnet mask bits from 256 (for each octet if more than 1 octet is being subnetted), and that will be your jump size for that particular octet
3. Turn all the bits on for that subnet and you have broadcast address.
You may try this course of CCNA, where this topic along with others has been covered in detail.
Another way, as long as you remember some key numbers, you can work up from there. For instance, the basic ones for subnetting, 0, 128, 192, 224, 240, 248 (you need to memorize these anyway)
So if you're looking for 231, you know that 224 is a good starting point.
Or just memorize the 64s: 0, 64, 128, 192.
Then the 32s: 32, 64, 96, 128, 160, 192, 224, 256
No need to memorize 4, 8, or 16 multiples, you can just count up from the closest 32 multiple.
For example, you're trying to find 192.168.0.70 /30 network and broadcast. Start at 64 and keep adding 4 until you get there.
Hope that helps.
All the fellas here have given you excellent options for doing subnetting, I use most of them sometimes. However I recommend you as well to perform the AND operation between the subnet mask (in the octet that its different of 255, 252 this case) and the number of interest in the IP address.
So, in your example you have the IP 192.168.15.30 with a mask of 255.255.255.252, so our interest values are: 30 and 252, convert them into binary and perform the AND operation.
30 - 0001 1110
252 - 1111 1100
Performing AND operation
28 - 0001 1100
So that result indicates that 192.168.15.28 is the Subnet ID
Hope this helps as well.
****, I didn't know that the support in these forums was so good.
I think I'll experiment with a few of those methods and see how I feel. I actually get good results when just focus on the actual problem and am not worried about how much time I am spending (is it below/over 20/30 seconds?). It sounds silly but that is taken me away from the actual problem I am trying to resolve?
Gregory's explanation might work for me as it is kind of similar of what I am doing now. If you memorise the 16,32 and 64 times table (along with the subnet mask decimal table) you have a few starting point.
I was also doing some examples last night. If someone is quick enough to divide the "interesting" octect by the size (being this 2,4 or 8 as the most "feasible" divisions to do), you may get an accurate result. If the division is not exact, you may just substract the remainder which gives you the subnet id.
Anyway, I'll keep working on it and will let you know how I get on.
Thanks again to everyone.
Everybody was in the same boat. when they got to subnetting they were befuddled. At the school they told me a lot students give up because of subnetting. I thought I would never get it but I kept practicing. realize you are only dealing with 4 octets with 8 bits in each. there are only a certain amount of valid masks to deal with. thats when it started to click