asked on subnetting ccna
I understand subnetting the long way, but that wont help on the exam. So I need to be able to get it quick for the exam. I understand the powers of 2 chart but I only know part of it and need some help and advice with the rest just to make sure I'm getting it right. Especially on the way they ask the questions or using cidr or not for th esubnet mask.
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 64 32 16 8 4 2 1
128
192
224
240
248
252
254
255
I understand (I think) this so far - How many subnets and hosts can this IP give you? 172.20.15.0 /23?
So a /23 uses 7 extra bits in the mask for the network, so looking at the chart 2^7=128, so they could have 128 subnets (we only minus 2 for hosts from what I learned). So, using 7 bits for the network we have 9 bits leftover for hosts, we look at the chart 2^9 is 512-2=510, then 510 is the number of hosts we can have.
Is that correct? How do I find the subnet mask based on the chart above? Is it if I used 7 bits for the network then I would count 7 on the column on the left and that is 254. So the subet mask for this /23 address is 255.255.254.0, correct?
Also, what subnet does this IP reside on 172.20.15.5 /16? This is where I need help since I dont know how to find the subnet increment.
Is this where I take the /23 as previously and see it uses 7 bits for the network and on the chart that would be increment of 128.
172.20.0.0
172.20.128.0
172.20.256.0
So it must live between the .0 and .128 networks and the range of usable hosts would be
172.20.0.1
172.20.127.254 (I might have messed that up?)
Also, if my logic is correct on this, is that I need to know for the exams?
How in the world can I use that chart for class A addresses with /20 bits used in the SM, seems impossible? How can it be done?
Thanks.
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 64 32 16 8 4 2 1
128
192
224
240
248
252
254
255
I understand (I think) this so far - How many subnets and hosts can this IP give you? 172.20.15.0 /23?
So a /23 uses 7 extra bits in the mask for the network, so looking at the chart 2^7=128, so they could have 128 subnets (we only minus 2 for hosts from what I learned). So, using 7 bits for the network we have 9 bits leftover for hosts, we look at the chart 2^9 is 512-2=510, then 510 is the number of hosts we can have.
Is that correct? How do I find the subnet mask based on the chart above? Is it if I used 7 bits for the network then I would count 7 on the column on the left and that is 254. So the subet mask for this /23 address is 255.255.254.0, correct?
Also, what subnet does this IP reside on 172.20.15.5 /16? This is where I need help since I dont know how to find the subnet increment.
Is this where I take the /23 as previously and see it uses 7 bits for the network and on the chart that would be increment of 128.
172.20.0.0
172.20.128.0
172.20.256.0
So it must live between the .0 and .128 networks and the range of usable hosts would be
172.20.0.1
172.20.127.254 (I might have messed that up?)
Also, if my logic is correct on this, is that I need to know for the exams?
How in the world can I use that chart for class A addresses with /20 bits used in the SM, seems impossible? How can it be done?
Thanks.
Switches / HubsCiscoNetworking
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Cyclops3590
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Simple put... yes... your logic is correct. That's how I figure out everything quickly in my head.
Not all the logic is correct.
==========================
Is this where I take the /23 as previously and see it uses 7 bits for the network and on the chart that would be increment of 128.
172.20.0.0
172.20.128.0
172.20.256.0
So it must live between the .0 and .128 networks and the range of usable hosts would be
172.20.0.1
172.20.127.254 (I might have messed that up?)
==========================
a /23 means the subnet mask goes into the third octet. You are using 7 subnet bits in that octet you are correct. leaving 1 host bit in the third octet. But by using my steps, you'll notice the block size is thus 2 ^1 = 2. So the network addresses are not 172.20.0.0 and 172.20.128.0.
Rather, they are
172.20.0.0
172.20.2.0
172.20.4.0
etc.
everything else is correct though.
==========================
Is this where I take the /23 as previously and see it uses 7 bits for the network and on the chart that would be increment of 128.
172.20.0.0
172.20.128.0
172.20.256.0
So it must live between the .0 and .128 networks and the range of usable hosts would be
172.20.0.1
172.20.127.254 (I might have messed that up?)
==========================
a /23 means the subnet mask goes into the third octet. You are using 7 subnet bits in that octet you are correct. leaving 1 host bit in the third octet. But by using my steps, you'll notice the block size is thus 2 ^1 = 2. So the network addresses are not 172.20.0.0 and 172.20.128.0.
Rather, they are
172.20.0.0
172.20.2.0
172.20.4.0
etc.
everything else is correct though.
Good catch, you are correct.
ASKER
Really good explanation, I'll have to spend some time practicing to get it down quick, thank you.
By the way, this process is effectively the exact same as described in Todd Lammle's CCNA book by Sybex. While I thought there was better technical explanation on most technologies in Wendell Odom's Cisco Press CCNA books, Lammle's I felt was by far the bible reference as to subnetting. I'm not sure what study materials you have but those are the two I used along with GNS and I passed with ease on the first try. I spent a lot of time practicing subnetting though to make sure I could do any of them in my head in under 15 seconds like I mentioned. The reason I say 15 seconds is because there are a lot of questions that require that skill but they don't ask for those answers. By doing it quickly it allows more time for the simulated lab portions which can take some time and you should take time as their worth so much of the exam score.
Good Luck.
Good Luck.
So the answer is 172.20.0.1-172.20.255.254 (where .0 is the network number and .255 is the broadcast).
A class A IP address with a higher mask than /8 would be treated similarly. How many bits of the third octet are you using? Answer - 4. Stay within the octet to determine your count just be sure to put the corresponding mask in the right octet and you'll get the right answer. So, the mask is 255.255.240.0.