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Formulating Subnet Information Quicker
Hello,
I am going for my Network+ N10-006 test this next week and I am having some issues with some subnetting. I need to find faster ways to obtain the following details with IP addresses. I can work them out in the long run, taking a lot of time with converting dotted decimal to binary then comparing the subnet mask to the IP address to find network etc. etc. But the real problem is in my test I will be under the time clock with NO calculator. I have done quite a bit of these and not seeming to get faster at them so I am hoping some of you have some good tricks to help speed it up. I have searched the web too but most of them require excel and calculators which I wont be able to use. Info I need to determine quickly is below.
A random IP, 10.150.174.20/17 (If you know a quicker way to convert to binary instead of lining up the formula and plugging in 1/0 in the following: 128 64 32 16 8 4 2 1, that would also be super helpful!)
Network ID
Broadcast
First Valid IP
Last Valid IP
One other thing that is kind of escaping me that I cannot fully grasp yet. If my number of subnets are quite high, like in a /27 that allows me 32 diferent subnet ranges, with 30 valid addresses in each one minus my ID and Broadcast. When you add those up, it goes above the .254 range for the octet and has to increase the octet prior, correct? If so, is there an easier way to calculate these ranges from ID to Broadcast easier then setting down and writing each range out individually? I also experienced that at some point I will hit mid range at .254 and cannot go higher so does that range just lose the rest of the valid IP's or does it carry over with the octet increase?
I know this is a lot of beginner questions to you guys probably but at 300.00 for my cert test I cannot afford to not know why I am answering something haha. Thanks so much in advance!
I am going for my Network+ N10-006 test this next week and I am having some issues with some subnetting. I need to find faster ways to obtain the following details with IP addresses. I can work them out in the long run, taking a lot of time with converting dotted decimal to binary then comparing the subnet mask to the IP address to find network etc. etc. But the real problem is in my test I will be under the time clock with NO calculator. I have done quite a bit of these and not seeming to get faster at them so I am hoping some of you have some good tricks to help speed it up. I have searched the web too but most of them require excel and calculators which I wont be able to use. Info I need to determine quickly is below.
A random IP, 10.150.174.20/17 (If you know a quicker way to convert to binary instead of lining up the formula and plugging in 1/0 in the following: 128 64 32 16 8 4 2 1, that would also be super helpful!)
Network ID
Broadcast
First Valid IP
Last Valid IP
One other thing that is kind of escaping me that I cannot fully grasp yet. If my number of subnets are quite high, like in a /27 that allows me 32 diferent subnet ranges, with 30 valid addresses in each one minus my ID and Broadcast. When you add those up, it goes above the .254 range for the octet and has to increase the octet prior, correct? If so, is there an easier way to calculate these ranges from ID to Broadcast easier then setting down and writing each range out individually? I also experienced that at some point I will hit mid range at .254 and cannot go higher so does that range just lose the rest of the valid IP's or does it carry over with the octet increase?
I know this is a lot of beginner questions to you guys probably but at 300.00 for my cert test I cannot afford to not know why I am answering something haha. Thanks so much in advance!
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As a CCSI for about the last 20 years, I've heard every trick and shortcut for subnetting. The ONLY thing that works for certain is practice, practice, practice.
Just go to a site that keeps hitting you with subnetting questions and work them out. After a few hundred, you will become faster at it.
google "practice subnetting questions" and you'll find a bunch.
Just go to a site that keeps hitting you with subnetting questions and work them out. After a few hundred, you will become faster at it.
google "practice subnetting questions" and you'll find a bunch.
@Hemil - The explanation provided was great, thanks! I do have one bit of confusion tho that I cannot follow along. I am quoting you on one section that I have not seen details on in any other study material I found to support why you would do this. Can you elaborate?
"Since I knew the value of my last octet which was equal to 128 I proceed to sum up 128 and gives me a total of 256.
128 is lower than 174 therefore I had to sum up the same value to see if I can get closer to such value. because the fact I couldn't get closer. the value remains as 128 because it belongs to the range of the network 128+128=256."
1. What tells you that you had to sum up the value to see if you can get closer to such value. Is there some experienced trick in trying to sum up to meet the 174?
2. I assume you chose the 174 because it was the third octet, which is the same as where the 1=Network meets the 0=Host on the subnet mask. Also, what values would you be using to try to reach that number from 128 to 174? The predetermined bit value for each in the 128 64 32 16 8 4 2 1, for example 128+64, 128+32, 128+16 etc?
3. Lastly, what happens if you could find a value that gets closer to 174, what change would that make?
@Don I appreciate you taking the time to share that information with me. I have been hitting the Examcompass website for those. Like I said before I can do it the old fashioned paper and pencil way but it takes me a long time to check then double check and triple check my work which I am afraid may cripple me on the test since I only get a crappy marker and dry eraser board lol.
"Since I knew the value of my last octet which was equal to 128 I proceed to sum up 128 and gives me a total of 256.
128 is lower than 174 therefore I had to sum up the same value to see if I can get closer to such value. because the fact I couldn't get closer. the value remains as 128 because it belongs to the range of the network 128+128=256."
1. What tells you that you had to sum up the value to see if you can get closer to such value. Is there some experienced trick in trying to sum up to meet the 174?
2. I assume you chose the 174 because it was the third octet, which is the same as where the 1=Network meets the 0=Host on the subnet mask. Also, what values would you be using to try to reach that number from 128 to 174? The predetermined bit value for each in the 128 64 32 16 8 4 2 1, for example 128+64, 128+32, 128+16 etc?
3. Lastly, what happens if you could find a value that gets closer to 174, what change would that make?
@Don I appreciate you taking the time to share that information with me. I have been hitting the Examcompass website for those. Like I said before I can do it the old fashioned paper and pencil way but it takes me a long time to check then double check and triple check my work which I am afraid may cripple me on the test since I only get a crappy marker and dry eraser board lol.
1. What tells you that you had to sum up the value to see if you can get closer to such value. Is there some experienced trick in trying to sum up to meet the 174?
10.150.174. ( let's make it simple.)
The last octet /17 is equal to 128 as you know (11111111.11111111.(1)=128
This is my tricky way to deal with is. Once you know the value like of the last octet you need to use the number and sum up until you get closer to the network. Let me give you an example
192.168.154.0/26
This is my binary ------128 (64) 32 16 4 2 1-----
In this network I am working with the last octet (11111111.11111111.1111111
knowing that 64 it's my magic number I will have to do this
****Sorry to update this, I have made another mistake, I apology since it's really late in the east coast**
Network 192.168.154.0/26
Since we are working with the last octet, the network start with 0, but your magic number is 64 thefore you will have to start from 0 to 64. Like in my example below
Network= 192.168.154.0
First IP= 192.168.154.1
Last IP= 192.168.154.62
Brodca= 192.168.154.63
Next network =192.168.154.64
Another Example:
Binary 128 64 32 16 (8) 4 2 1
In this network I am working with the last octet (11111111.11111111.1111111
172.10.94.0/29 we are looking for 5 usable IP addresses.
We are working with the last value but the last octet start with zero therefore your magic # is 8
Network= 172.10.94.0
First IP= 172.10.94.1
Last IP= 172.10.94.6
Brodca= 172.10.94.7
Next network it's 172.10.94.8
I think you follow me right?
I have made some mistake that I've fixed.. I didn't check right my values but it's fine now
@Hemil Okay, I am understanding much better now. The only lasting question with this new information is, in the two examples I see how you sum the number with itself to get as close as you can to the last octet in network. What I do not understand is what you do with that information.
I.E 64+64=128, 8+8+8+8+8+8+8+8+8+8+8= 88. What do you do with this information. I do not see the 128 or 88 applying in either of the two examples. You say 88 is the best match to create the network, but how does it do that? Or how does the 128 match with the subnet range?
I.E 64+64=128, 8+8+8+8+8+8+8+8+8+8+8= 88. What do you do with this information. I do not see the 128 or 88 applying in either of the two examples. You say 88 is the best match to create the network, but how does it do that? Or how does the 128 match with the subnet range?
I.E 64+64=128, 8+8+8+8+8+8+8+8+8+8+8= 88. What do you do with this information. I do not see the 128 or 88 applying in either of the two examples. You say 88 is the best match to create the network, but how does it do that? Or how does the 128 match with the subnet range?
Give me few minutes you will understand it in a bit I'm writing you more samples.
Here is another example
Subnet this network (192.168.8.40 /29)
Binary 128 64 32 16 (8) 4 2 1
Our network finish in /29 11111111.11111111.11111111
Network 192.168.8.40 --You see this network (40) you need to use your magic number (8) to sum itself up until it reach the closest value.
Example
8+8+8+8+8=40 ( 40 it's in the range or equal to it's value, therefore it will be use as the start point for subnetting.)
8+8+8+8+8+8=48 ( 48 is greater than 40 therefore cannot be use to subnet this network.
Network 192.168.8.40
First IP 192.168.8.41
Last IP 192.168.8.46
Broad 192.168.8.47
Next network 192.168.8.48/29
Subnet this network (192.168.8.40 /29)
Binary 128 64 32 16 (8) 4 2 1
Our network finish in /29 11111111.11111111.11111111
Network 192.168.8.40 --You see this network (40) you need to use your magic number (8) to sum itself up until it reach the closest value.
Example
8+8+8+8+8=40 ( 40 it's in the range or equal to it's value, therefore it will be use as the start point for subnetting.)
8+8+8+8+8+8=48 ( 48 is greater than 40 therefore cannot be use to subnet this network.
Network 192.168.8.40
First IP 192.168.8.41
Last IP 192.168.8.46
Broad 192.168.8.47
Next network 192.168.8.48/29
Another example
10.72.95.0 /19
Binary 128 64 32 16 8 4 2 1
Our network finish in /19 11111111.11111111.11(1)=32
We will be working with the third octet. 10.72.(95)
32+32=64 ( 64 is less than 95 therefore belongs to the range of IP network)
32+32+32=96 (96 is greater than 95 therefore cannot be able to subnet this network) This is the next network
Network 10.72.64.0
First IP 10.72.64.1
Last IP 10.72.95.254
Broad 10.72.95.255
Next network 10.72.96.0
Does that make any sense to you?
10.72.95.0 /19
Binary 128 64 32 16 8 4 2 1
Our network finish in /19 11111111.11111111.11(1)=32
We will be working with the third octet. 10.72.(95)
32+32=64 ( 64 is less than 95 therefore belongs to the range of IP network)
32+32+32=96 (96 is greater than 95 therefore cannot be able to subnet this network) This is the next network
Network 10.72.64.0
First IP 10.72.64.1
Last IP 10.72.95.254
Broad 10.72.95.255
Next network 10.72.96.0
Does that make any sense to you?
This look right?
Example 1
Subnet this network (192.168.8.39 /30)
Binary 128 64 32 16 8 (4) 2 1
11111111. 11111111. 11111111. 11111(1) = 4
Magic Number = 4
4+4+4+4+4+4+4+4+4=36 (Can use because equal or lower than 39)
4+4+4+4+4+4+4+4+4+4=40 (Cannot use, over 39)
Network: 192.168.8.36
First IP: 192.168.8.37
Last IP: 192.168.8.38
Broadcast: 192.168.8.39
Next network: 192.168.8.40
Example 2 (This one is tricky because im working in the third octate, the others we done was always the fourth.) CHECK MY MATH
Subnet this network (192.164.23.39 /21)
Binary 128 64 32 16 (8) 4 2 1
11111111. 11111111. 1111(1) =8
Magic Number = 8
8+8=16 (Can use because equal or lower than 23)
8+8+8=24 (Cannot use, over 23)
Network: 192.164.23.16
First IP: 192.164.23.17
Last IP: 192.164.23.22 (JUST CORRECTED)
Broadcast: 192.164.23.23
Next network: 192.164.23.24
Example 1
Subnet this network (192.168.8.39 /30)
Binary 128 64 32 16 8 (4) 2 1
11111111. 11111111. 11111111. 11111(1) = 4
Magic Number = 4
4+4+4+4+4+4+4+4+4=36 (Can use because equal or lower than 39)
4+4+4+4+4+4+4+4+4+4=40 (Cannot use, over 39)
Network: 192.168.8.36
First IP: 192.168.8.37
Last IP: 192.168.8.38
Broadcast: 192.168.8.39
Next network: 192.168.8.40
Example 2 (This one is tricky because im working in the third octate, the others we done was always the fourth.) CHECK MY MATH
Subnet this network (192.164.23.39 /21)
Binary 128 64 32 16 (8) 4 2 1
11111111. 11111111. 1111(1) =8
Magic Number = 8
8+8=16 (Can use because equal or lower than 23)
8+8+8=24 (Cannot use, over 23)
Network: 192.164.23.16
First IP: 192.164.23.17
Last IP: 192.164.23.22 (JUST CORRECTED)
Broadcast: 192.164.23.23
Next network: 192.164.23.24
The first one was 100% good
Now, the second one was 50% good
This is what you got
Subnet this network (192.164.23.39 /21) you have /21 and means, you are working with the third octet (23)
Binary 128 64 32 16 (8) 4 2 1 -- Your magic # was good.
11111111. 11111111. 1111(1) =8
Magic Number = 8
8+8=16 (Can use because equal or lower than 23) --good--
8+8+8=24 (Cannot use, over 23) --good--
Since you were working in the third octet you need to replace the 23 by 16 and that's how you should do it.
But you're getting there thou
Network: 192.164.16.0
First IP: 192.164.16.1
Last IP: 192.164.23.254
Broadcast: 192.164.23.255
Next network: 192.164.24.0
Now, the second one was 50% good
This is what you got
Subnet this network (192.164.23.39 /21) you have /21 and means, you are working with the third octet (23)
Binary 128 64 32 16 (8) 4 2 1 -- Your magic # was good.
11111111. 11111111. 1111(1) =8
Magic Number = 8
8+8=16 (Can use because equal or lower than 23) --good--
8+8+8=24 (Cannot use, over 23) --good--
Since you were working in the third octet you need to replace the 23 by 16 and that's how you should do it.
But you're getting there thou
Network: 192.164.16.0
First IP: 192.164.16.1
Last IP: 192.164.23.254
Broadcast: 192.164.23.255
Next network: 192.164.24.0
This one is extreme, but looking better?
Subnet this network (172.172.15.45 /12)
Binary 128 64 32 1(6) 8 4 2 1
11111111. 111(1) = Working in Second Octet
Magic Number = 16
16+16+16+16+16+16+16+16+16
16+16+16+16+16+16+16+16+16
Network: 172.160.0.0
First IP: 172.160.0.1
Last IP: 172.172.255.254
Broadcast: 172.172.255.255
Next network: 172.173.0.0
Max number of hosts = 1,048,574
Max number of networks = 16
Subnet this network (172.172.15.45 /12)
Binary 128 64 32 1(6) 8 4 2 1
11111111. 111(1) = Working in Second Octet
Magic Number = 16
16+16+16+16+16+16+16+16+16
16+16+16+16+16+16+16+16+16
Network: 172.160.0.0
First IP: 172.160.0.1
Last IP: 172.172.255.254
Broadcast: 172.172.255.255
Next network: 172.173.0.0
Max number of hosts = 1,048,574
Max number of networks = 16
haha thanks. I dont think I will see such extreme answers on the test but its good to know how they work past the fourth octet. If you think its getting good enough to study this formula more I can quit keeping you awake fact checking me haha.
Network: 172.160.0.0
First IP: 172.160.0.1
Last IP: 172.176.255.254
Broadcast: 172.176.255.255
Next Network: 172.177.0.0
maybe?
First IP: 172.160.0.1
Last IP: 172.176.255.254
Broadcast: 172.176.255.255
Next Network: 172.177.0.0
maybe?
16+16+16+16+16+16+16+16+16
16+16+16+16+16+16+16+16+16
160 it's your network
176 it's the next network. You have to replace the Last IP and broadcast as follow 172.175.255.254
Broadcast 172.175.255.255
Next nextwork 172.176.0.0
Network: 172.160.0.0
First IP: 172.160.0.1
Last IP: 172.172.255.254 -- You forgot to change the value in the second octed
Broadcast: 172.172.255.255 -- Here as well
Next network: 172.173.0.0
Max number of hosts = 1,048,574
Max number of networks = 16
16+16+16+16+16+16+16+16+16
160 it's your network
176 it's the next network. You have to replace the Last IP and broadcast as follow 172.175.255.254
Broadcast 172.175.255.255
Next nextwork 172.176.0.0
Network: 172.160.0.0
First IP: 172.160.0.1
Last IP: 172.172.255.254 -- You forgot to change the value in the second octed
Broadcast: 172.172.255.255 -- Here as well
Next network: 172.173.0.0
Max number of hosts = 1,048,574
Max number of networks = 16
Yep, I seen it just about the same time you did, and tried to edit the post as you posted yours lol.
Network: 172.160.0.0
First IP: 172.160.0.1
Last IP: 172.175.255.254
Broadcast: 172.175.255.255
Next Network: 172.176.0.0
First IP: 172.160.0.1
Last IP: 172.175.255.254
Broadcast: 172.175.255.255
Next Network: 172.176.0.0
Fingers Crossed!
172.65.25.15 /13
11111111. 1111(1) = 8
8+8+8+8+8+8+8+8=64
8+8+8+8+8+8+8+8+8= 72
Network: 172.64.0.0
First IP: 172.64.0.1
Last IP: 172.71.255.254
Broadcast: 172.71.255.255
Next Network: 172.72.0.0
Max Hosts = 524,286
Max Networks = 32
172.65.25.15 /13
11111111. 1111(1) = 8
8+8+8+8+8+8+8+8=64
8+8+8+8+8+8+8+8+8= 72
Network: 172.64.0.0
First IP: 172.64.0.1
Last IP: 172.71.255.254
Broadcast: 172.71.255.255
Next Network: 172.72.0.0
Max Hosts = 524,286
Max Networks = 32
8+8+8+8+8+8+8+8=64
8+8+8+8+8+8+8+8+8= 72 this is your next network therefore always rest -1
Once you get this values think about this
Network 172.64.0.0 -- Good
First IP 172.64.0.1 -- Good
Last IP 172.71.255.254
Broad 172.71.255.255
Last IP: 172.95.255.254
Broadcast: 172.95.255.255
Next Network: 172.96.0.0
Wrong
8+8+8+8+8+8+8+8+8= 72 this is your next network therefore always rest -1
Once you get this values think about this
Network 172.64.0.0 -- Good
First IP 172.64.0.1 -- Good
Last IP 172.71.255.254
Broad 172.71.255.255
Last IP: 172.95.255.254
Broadcast: 172.95.255.255
Next Network: 172.96.0.0
Wrong
I'm going to sleep, practice and I will check tomorrow Pal, I cant even focus looking on my screen xD
Ok, so I actually had that correctly at first, because I counted up from 64 - 96 on the second octet. Which is a network jump of 32. But then I looked at next network and seen it started at 72 according to my magic number sums, so i changed it.
Ok man, I will do some tomorrow too and maybe post a few of them up for you to check. One from each octet just for fun. Have a great night and thanks again!
10.56.58.92/19
11111111. 11111111. 11(1) = 32
32+32 = 64 (won't work)
Network: 10.56.32.0
First IP: 10.56.32.1
Last IP: 10.56.63.254
Broadcast: 10.56.63.255
Next Network: 10.56.64.0
11111111. 11111111. 11(1) = 32
32+32 = 64 (won't work)
Network: 10.56.32.0
First IP: 10.56.32.1
Last IP: 10.56.63.254
Broadcast: 10.56.63.255
Next Network: 10.56.64.0
There you go.
That one it's good as hell.
Keep practicing and you will kickass. That's my trick when I needed to subnet on my CCNA.
That one it's good as hell.
Keep practicing and you will kickass. That's my trick when I needed to subnet on my CCNA.
Thanks a lot, it was a little confusing but this is definitely what I was looking for. I can do most of these now in my head with a few exceptions, so it will make a huge difference on the exam. Have a great one and hope we get to speak again in the future!
Like I said before I can do it the old fashioned paper and pencil way but it takes me a long time to check then double check and triple check my work
I've taken the CCNA exam at least 20 times (back in the day, we had to get at least a 925). I always work out the subnetting problems in writing. Once you've done it enough, there's no reason to double and triple check. You'll know when you get there because you're always right the first time.
The dry erase boards do suck. When they give you the pen, try it out right there. If the tip is worn down, demand another. After they've been used a hundred or so times they get so soft and wide that your writing is too large.
Hey dude, I have created an article to help people how to easily subnet.
Here is my link
https://www.experts-exchange.com/articles/31249/How-to-subnet.html?notificationFollowed=200737802
Here is my link
https://www.experts-exchange.com/articles/31249/How-to-subnet.html?notificationFollowed=200737802
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I'm going to do my best to help you understand how can you do it quicker, but again require lots of practice.
I need network of this subnet, 10.150.174.20/17
Do you see the /17. That's the number you have to pay attention because it tells you where you network start and finish.
now every IP it's equal to 8. Meaning each number has a value of an octet. Example
128 64 32 16 8 4 2 1 -- This is my binary, you need to memorize it to catch up with the network.
10 = 00001010
150= 10010110
172= 10101100
20= 00010100
Now, let's crank it up. (10.150.174.20 /17)
Since we are working with the /17 network that means that we are going to work with the 3rd octet 174 why? becuase that's where the network takes a place. Example 11111111.11111111.10000000
The subnet mask for this network is 255.255.128.0 why? because if you sum this up 128+64+32+8+4+2+1=255 128+64+32+8+4+2+1=255 128 " the last number 128 it's because there's no more ones therefore I stopped the count.
Remember this 1= to network 0= hosts
So we know the subnet mask is 255.255.128.0 now we need to know the rest:
10.150.174.20 /17
Network ID = 10.150.128.0 -
Broadcast = 10.150.255.255
First Valid IP = 10.150.128.1
Last Valid IP = 10.150.255.254
I am going to break it down for you
Network ID = 10.150.128.0 -- Where the heck did I get 128? you wonder.
Since I knew the value of my last octet which was equal to 128 I proceed to sum up 128 and gives me a total of 256.
128 is lower than 174 therefore I had to sum up the same value to see if I can get closer to such value. because the fact I couldn't get closer. the value remains as 128 because it belongs to the range of the network 128+128=256.
Broadcast = 10.150.255.255 After knowing the total of 128+128=256. we have to deduct -1 to 256 because 256 doesn't belong to the binary values. Giving you a total of 255
Broadcast = 10.150.255.255
After knowing the value of the network and broadcast ip Looking for the first and last host will be a cup cake.
First Valid IP = 10.150.128.1
Last Valid IP = 10.150.255.254
I think this doesn't need an explication.
I hope this helps,