I've been working on subnetting all day and in the last 20 minutes Ive learned more from these videos than from any other material.... Nice work brother
The other trick to finding the magic number which tells you the block size (and to avoid going into binary which slows us down) is to figure out the most significant bit (usually the number in the fourth octet which is to the right) and subtracting it from 256. So, let's say we have a mask of 255.255.255.192 (192 is the most significant) so we do 256-192 to end up with the 64 bit block size shown above.
256 is the answer but because we reserve 2 bits for both the network and broadcast address we're left with 254 useable host addresses in that scenario.
Timestamp selected: 7:03 Additional details: The presenter stated 5 zeroes instead of 6 zeroes ; resultant after borrowing 2 bits from 8 will leave you with six bits (ie six zeroes).
Good video. BTW if it hasn't already been mentioned, you can also get the magic number by subtracting the interesting octet subnet mask (that which doesn't fall on a classful boundary) from 256. Per the example: IP = 192.168.1.55/26, Mask = 255.255.255.192. 256-192 = 64 = magic number.
@CambioDRuta Hello and thanks for the feedback! I just looked at the video and you are correct. That was my mistake, I just forgot to add the last 2 in the example. for the Hosts if the last octet is 11000000 then you have 2x2x2x2x2x2 = 64 hosts The magic number tells you that too...
No problem. This has helped me: Within any octet (8 bits between the "dots"), the total possibilities of hosts AND subnets will always total 256. So if you have 2 "borrowed" subnet bits, they give you 4 possibilities (2 to the 2nd power). The remaining 6 host bits give you 64 possibilities (2 to the 6th power). 4 X 64 = 256. It's a good way to check your work and your thinking. Todd Lammle CCNA Study Guide is a great book - Chapter 4 is better than anything I've found on UA-cam.
thanks for these great tutorials! they are really helpful,don't wanna sound like a smartass or something but i think that the number of hosts in a subnet would be 2^x - 2,so in this example it would be 64 - 2 = 62? correct me if i'm wrong
you did a fantastic job. anyone that's just brushing up like i am understands already that you subtract 2 to get the number of valid hosts. and the place value of the last borrowed bit is one less step than subtracting the last octet from 256. thanks for this. i have CCENT coming up and i really needed a refresher.
Sorry, but did you make a mistake during 7:25? You said that 2^5 is 64, you also said that in the last octet we have 2 ones and 5 zeros. But that's wrong, it is 6 zeros? 2^6 is 64. so it's actually 2x2x2x2x2x2 = 64, or have I misunderstood something?
32 hosts - 2 (the first and last host) means 30 available hosts. You can not use the network address (the first) or the broadcast address (the last) :)
Awesome job, other than the notation error in this vid of course. My professor at my local CC tries to pat himself on the back by stating that he can teach subnetting easily, but I beg to differ - this has been a much clearer explanation to me.
Annotations, Annotations, Annotations. If you mess up a video and you cant re-record it then at least take the 4 minutes to put up a descriptive and corrective annotation to notate it. We all make mistakes but laziness is a choice.
I hate to be that guy that tries to correct everything. I find your videos extremly helpful. But you made one error. To find the numbers of host the equation is 2*2*2*2*2*2=64 - 2. because we all know that we cannot use the first number in the subnet(64) because that is the network and we cannot use the last(127) that is the broadcast. So there are only 62 available host. so to simply put it, the equations to find the following Subnetworks - 2(to the power of the number of ones) in this case there are 2 ones = #subnets Hosts - 2(to the power of the number of zeros) in this case there are 6 zeros MINUS 2 = #HOSTs
I know this is an old video and audio isn't the best, but these are the most informative videos I have seen on UA-cam. The Vids with over a million views on the same subject are not nearly as informative IMO.
Thanks for the videos. You explain things well. But, how come the network and broadcast addresses are included in the number of hosts, if they cannot be used for host addresses?
I'm sorry, but I think i'm misunderstanding something. I thought the number of hosts would actually be 62 not 64 because you have to subtract the network address and the broadcast address?
Not to be picky, but you're saying there are 5 host bits. There are 6 shown, and you get 64 options with 6 bits, not 5. This could be really confusing for someone who didn't know where you were going with this. But thanks for the info - the idea of finding the magic number using the last subnet bit is really quick. I learned from a Todd Lammle book that you subtract the subnet mask (192 in your case) from 256 (always use 256) and you get 64. But your method is fast too.
I was going to make this comment as I was confused I had missed something. Reading the comments op advised it was an error in notation so I am back on track. 7:04> op says it is 5 zero's it is actually 6.
Can anybody tell me where did the 2 came from that he’s multiplying? I understand that 2 bits multiply by another 2. Where did that other 2 came from? Im new to networking
Keep up the great work man, i love your tutorials (except one about VLSM where the volume was messed up). Not complaining though, big fan of yours here.
Two more examples to help you see the pattern. If you have 3 borrowed subnet bits, they give you 8 possibilities (2 to the 3rd power, 2X2X2). The remaining 5 host bits give you 32 possibilities (2 to the 5th power, 2X2X2x2X2). 8 subnets X 32 hosts = 256. If you have 4 borrowed subnet bits, they give you 16 possibilities (2 to the 4th power, 2X2X2X2 = 16). The remaining 4 host bits give you the same (2X2X2X2 = 16). 16 subnets X 16 hosts = 256.
Dan, taking the CCNA currently, watching your videos to explain the difficult to understand Cisco text, thanks a milli. That said, I learned to find the majic number by taking the value of the netmask .192 in this case and subtracting it from 255. That gives me 63, then simply add 1 to account for the .0, giving me 64 as my magic number. Do you consider this correct as well?
So basically if you subtract two from the magical number you will get the number of hosts per network. And every time you see a "1" in the last octet put change it to "2" and multiply it. So in your last octet you have "11000000" change it to "22000000" multiply the "2's" so "2x2(000000)" = 4. If it was "11100000" change it to 22200000 which is 2x2x2(00000) or 8 hosts.
@ljay103 i caught that right away maybe professor didn't include 2 itself which was 2 x 1 = 2. he started @ 2x2 = 4. i maybe wrong though. just a noob here.
Mighty KC in his 4th octet (the top that has the two 1s and the six 0s). He has two 1s and the first one is in the place value of 128 the one next to it is 64, (if there was another 1 next to it it would be 32) But to get the magic number it is the last #1 place value in the octet.
Awesome video! Just to clarify the other question ("What network is 192.168.1.55 in?")... I assume that would be 192.168.1.0.... not 192.168.1.64... not 192.168.1.128... nor 192.168.1.192 because the host address is 192.168.1.55 and falls between 192.168.1.0 and 192.168.1.63). Think I just solved my own question :)
62 hosts because, for example, if you are talking about the first network, 192.168.1.0 up to 192.168.1.63 are controlled by that first switch or router (being the beginner that I am, I do not know which device would be used) however, that router/switch needs an address for itself (which would be 192.168.1.0) and a broadcast IP address (which would be 192.168.1.63) therefore only 62 addresses are left for users, or in other words, hosts.
I'm not sure about 256-2 = hosts. I mean you have to know how many bits are used for the host. Let say 5 bits, that will give you 32, subtract 2 and you have 30 hosts per subnet.
Yes it was 2x2x2x2x2x2=64, simple mistake. Either way the magic number shows the block size and number of total hosts, and of course -2 for the network id and broadcast address. I learned an invaluable lesson today almost 12 years later.
nice explanation Sir, but little error der in dis video, dat's nt a big matter ..While defining the host bits i.e 6 host bits are ...So no of hosts woulds 64(2*2*2*2*2*2*) but only 5 2's are define in explanation ..
I'm amazed. U figured out that there were 5 host bits here, and u wrote 2x2x2x2x2 = 64 The answer is correct according to that there are actually 6 host bits which should equal 64, but you say 5 and write 5 and still manage to get it right. hehe
Would be a great lecture if not for the unsynchronized audio/video. Tried refreshing, replaying, rewinding and it still doesn't help. The audio is ahead of the video by a good 10 seconds or so.
I've been working on subnetting all day and in the last 20 minutes Ive learned more from these videos than from any other material.... Nice work brother
The other trick to finding the magic number which tells you the block size (and to avoid going into binary which slows us down) is to figure out the most significant bit (usually the number in the fourth octet which is to the right) and subtracting it from 256. So, let's say we have a mask of 255.255.255.192 (192 is the most significant) so we do 256-192 to end up with the 64 bit block size shown above.
awesome! thanks !
well done Brian =)
So if the mask is 255.255.255.0 then it would be 256 - 0 for the magic number?
256 is the answer but because we reserve 2 bits for both the network and broadcast address we're left with 254 useable host addresses in that scenario.
Brian Pacheco Thanks!
Timestamp selected:
7:03
Additional details:
The presenter stated 5 zeroes instead of 6 zeroes ; resultant after borrowing 2 bits from 8 will leave you with six bits (ie six zeroes).
Dude. It's 6 zeroes in a host part. Not 5. So 2^6=64.
You said and wrote 5. 2x2x2x2x2=32.
It should be 6. 2x2x2x2x2x2=64
I was confused here, thought he was wrong. Always good to check comments to confirm!
why the magic number is that magical. i mean do you know from where is this information? where is it coming from. i still cant understand it
why the network goes up by 64? explanation?
Correct
Hala I thought I was the only one
Good video. BTW if it hasn't already been mentioned, you can also get the magic number by subtracting the interesting octet subnet mask (that which doesn't fall on a classful boundary) from 256. Per the example: IP = 192.168.1.55/26, Mask = 255.255.255.192. 256-192 = 64 = magic number.
Yes, I did this one a long time ago. Maybe I can edit it, split the audio and video tracks and repost.
@CambioDRuta
Hello and thanks for the feedback!
I just looked at the video and you are correct. That was my mistake, I just forgot to add the last 2 in the example.
for the Hosts if the last octet is 11000000 then you have 2x2x2x2x2x2 = 64 hosts The magic number tells you that too...
No problem.
This has helped me: Within any octet (8 bits between the "dots"), the total possibilities of hosts AND subnets will always total 256. So if you have 2 "borrowed" subnet bits, they give you 4 possibilities (2 to the 2nd power). The remaining 6 host bits give you 64 possibilities (2 to the 6th power). 4 X 64 = 256. It's a good way to check your work and your thinking.
Todd Lammle CCNA Study Guide is a great book - Chapter 4 is better than anything I've found on UA-cam.
6th bit is lack in host id that is 2*2*2*2*2*2=64
+Shubham Thakur Yes, I made an error in notation. Good job paying attention. Many people have noted it in the comments.
thanks for these great tutorials! they are really helpful,don't wanna sound like a smartass or something but i think that the number of hosts in a subnet would be 2^x - 2,so in this example it would be 64 - 2 = 62? correct me if i'm wrong
Maybe I can help. What is your question Jose?
You do not have 64 Host. You have 62 valid hosts. 6 bits of hosts = 64 minus the nwtwork address and the broadcast
You have 64 IP addresses but only 62 are usable.
you did a fantastic job. anyone that's just brushing up like i am understands already that you subtract 2 to get the number of valid hosts. and the place value of the last borrowed bit is one less step than subtracting the last octet from 256. thanks for this. i have CCENT coming up and i really needed a refresher.
@ 7:18, would there not be 6 zeros? Am I getting something wrong here? Please help. Thanks.
Got it. I thought that was the case, but wasn't entirely sure. Thanks, bro. On a side note, it's still an excellent video.
yeah. I want to take the ICND 1, but all I have is the CBT Nuggets and this. I think it'll be enough.
Volf Khat Thank you. I was like, what am I doing wrong!!!
Taj Mahal its 6 zero's . 2^6-2. The 1st host is the network and last host broadcast id. I dont know where he keeps getting 5 zero's
Yeah. I think it was a minor oversight during his flow. Simple mistake anyone can make. Thanks for the reply though.
Sorry, but did you make a mistake during 7:25? You said that 2^5 is 64, you also said that in the last octet we have 2 ones and 5 zeros. But that's wrong, it is 6 zeros? 2^6 is 64. so it's actually 2x2x2x2x2x2 = 64, or have I misunderstood something?
32 hosts - 2 (the first and last host) means 30 available hosts. You can not use the network address (the first) or the broadcast address (the last) :)
Awesome job, other than the notation error in this vid of course. My professor at my local CC tries to pat himself on the back by stating that he can teach subnetting easily, but I beg to differ - this has been a much clearer explanation to me.
Damn, these vids are so easy to understand. Dan, you're da man!
These were the videos that finally made it all click for me. Thanks!
I'm curious, at 7:13 you say we have 5 zeros. Why don't we have 6 zeros for the host, since I see 6 zeros up there? I'm a bit lost on that, thank you.
Annotations, Annotations, Annotations. If you mess up a video and you cant re-record it then at least take the 4 minutes to put up a descriptive and corrective annotation to notate it. We all make mistakes but laziness is a choice.
I hate to be that guy that tries to correct everything. I find your videos extremly helpful. But you made one error. To find the numbers of host the equation is 2*2*2*2*2*2=64 - 2. because we all know that we cannot use the first number in the subnet(64) because that is the network and we cannot use the last(127) that is the broadcast. So there are only 62 available host.
so to simply put it, the equations to find the following
Subnetworks - 2(to the power of the number of ones) in this case there are 2 ones = #subnets
Hosts - 2(to the power of the number of zeros) in this case there are 6 zeros MINUS 2 = #HOSTs
good explanation.thanks
This is the best I have seen for subnetting!!!!!!
Hello. I don't understand. in the last part of the video, how could the magic number 63 come? could somebody tell me? Thank you.
The host bits are 2^6 - 2 so we have 2*2*2*2*2*2 = 64 - 2 = 62 valid hosts without the subnetwork address and the broadcast address
I know this is an old video and audio isn't the best, but these are the most informative videos I have seen on UA-cam. The Vids with over a million views on the same subject are not nearly as informative IMO.
Thanks for the videos. You explain things well.
But, how come the network and broadcast addresses are included in the number of hosts, if they cannot be used for host addresses?
I'm sorry, but I think i'm misunderstanding something. I thought the number of hosts would actually be 62 not 64 because you have to subtract the network address and the broadcast address?
Nevermind, the next video makes the distinction between hosts and usable hosts. Thanks!
Where did you get the 192 form in the subnet mask ?
Thanks for the great explanation in this video; I was struggling to understand this for my networking exam but this totally made it click for me!
Not to be picky, but you're saying there are 5 host bits. There are 6 shown, and you get 64 options with 6 bits, not 5. This could be really confusing for someone who didn't know where you were going with this.
But thanks for the info - the idea of finding the magic number using the last subnet bit is really quick. I learned from a Todd Lammle book that you subtract the subnet mask (192 in your case) from 256 (always use 256) and you get 64. But your method is fast too.
At 7:30 its 2^6 = 64 not 2^5 = 32
I was going to make this comment as I was confused I had missed something. Reading the comments op advised it was an error in notation so I am back on track. 7:04> op says it is 5 zero's it is actually 6.
What about /27 ?
There is 3 bits from the last octet, 3x2 is 6 subnetworks? I dont understand what you mean
he made an error around 7:20. It is 2 to the power of 6. making it 64. so 2 to the power of 5 is not 64 rather it is 32.
im wondering if the network part and host part of 0-63 is still usable? because you said you can have 64 hosts in that part..
6 hosts = 128.
thanks alot, this explains subnetting a lot clearer.
Im trying to grasp the formula for networks and hosts... which is which and why must we subtract 2 from it?
ok but how you know that this networks was created and exists or this is hipotetical exitances of the other networks
how do we know that it is a class c net mask that we are going to use????
so befour we only had one network but now we have 4 right ?
with the classless subnet thing ?
I have confusion in getting the borrowed bits, how will I know how many should I be borrowing?
if you know the answer now tell me.
By the Subnet mask, or you can tell by the cider block notation.
Can anybody tell me where did the 2 came from that he’s multiplying? I understand that 2 bits multiply by another 2. Where did that other 2 came from? Im new to networking
Wait. a host can't use the broadcast address so really there are only 31 available hosts on each network correct?
does the magic number works for every class?
Keep up the great work man, i love your tutorials (except one about VLSM where the volume was messed up). Not complaining though, big fan of yours here.
Two more examples to help you see the pattern.
If you have 3 borrowed subnet bits, they give you 8 possibilities (2 to the 3rd power, 2X2X2). The remaining 5 host bits give you 32 possibilities (2 to the 5th power, 2X2X2x2X2). 8 subnets X 32 hosts = 256.
If you have 4 borrowed subnet bits, they give you 16 possibilities (2 to the 4th power, 2X2X2X2 = 16). The remaining 4 host bits give you the same (2X2X2X2 = 16). 16 subnets X 16 hosts = 256.
Thank god for a simple straight forward explanation.
How do you find the magic number of /24?
How many subnets, and how many host on each subnet can there be if the netmask used is: 255.255.255.224 and the network is class B
32 host and 8 nets......30 host to be precise.
LelouchVi Britannia Thanks
Dan, taking the CCNA currently, watching your videos to explain the difficult to understand Cisco text, thanks a milli. That said, I learned to find the majic number by taking the value of the netmask .192 in this case and subtracting it from 255. That gives me 63, then simply add 1 to account for the .0, giving me 64 as my magic number. Do you consider this correct as well?
what an amazing video series, for anyone who want to learn or refresh sub-netting skills ..congrats
the 2 that is subtracted are the network and the broadcast addresses
So basically if you subtract two from the magical number you will get the number of hosts per network. And every time you see a "1" in the last octet put change it to "2" and multiply it. So in your last octet you have "11000000" change it to "22000000" multiply the "2's" so "2x2(000000)" = 4. If it was "11100000" change it to 22200000 which is 2x2x2(00000) or 8 hosts.
These videos are what made it click for me - so thank you.
You are welcome and thank you for the feedback
why magic number is equal to number of hosts any reason behind it???
It is 2^6 not 2^5.
Wondering how you still got the 64
@ljay103 i caught that right away maybe professor didn't include 2 itself which was 2 x 1 = 2. he started @ 2x2 = 4. i maybe wrong though. just a noob here.
can someone tell me where did he get or how did he get the magic number 64?
Mighty KC in his 4th octet (the top that has the two 1s and the six 0s). He has two 1s and the first one is in the place value of 128 the one next to it is 64, (if there was another 1 next to it it would be 32) But to get the magic number it is the last #1 place value in the octet.
I dont understand how you arrived at finding 4 subnetworks.
Dan your amazing! the way You explain it is so SOO AMAZIN :D
Dan, you are the man. It's starting to click.
Awesome video! Just to clarify the other question ("What network is 192.168.1.55 in?")... I assume that would be 192.168.1.0.... not 192.168.1.64... not 192.168.1.128... nor 192.168.1.192 because the host address is 192.168.1.55 and falls between 192.168.1.0 and 192.168.1.63). Think I just solved my own question :)
So 64 hosts for each of the 4 networks?
62 hosts because, for example, if you are talking about the first network, 192.168.1.0 up to 192.168.1.63 are controlled by that first switch or router (being the beginner that I am, I do not know which device would be used) however, that router/switch needs an address for itself (which would be 192.168.1.0) and a broadcast IP address (which would be 192.168.1.63) therefore only 62 addresses are left for users, or in other words, hosts.
What if subnet is10101000 then also the last "1" would be magic number.?
in about the end of the course the sound has been muted .
Dan you are great brother you are helping ma a lot. I Thank you for what you do brother man...keep it up for real. GOD BLESS YOU
Thank you so much! I was having so much trouble understanding these and they're very easy really haha. You're amazing!
I'm not sure about 256-2 = hosts. I mean you have to know how many bits are used for the host. Let say 5 bits, that will give you 32, subtract 2 and you have 30 hosts per subnet.
Yes it was 2x2x2x2x2x2=64, simple mistake. Either way the magic number shows the block size and number of total hosts, and of course -2 for the network id and broadcast address. I learned an invaluable lesson today almost 12 years later.
hey there are six hosts not five!
Dan there are 2 to the 6 hosts you put only 2 to the 5 = 64
So if WE find Magic number Ex: 64
So thé Host avalable IS 64-2 =62
nice explanation Sir, but little error der in dis video, dat's nt a big matter ..While defining the host bits i.e 6 host bits are ...So no of hosts woulds 64(2*2*2*2*2*2*)
but only 5 2's are define in explanation ..
sub net is process use to divide large network into small network.
magic number host id 2*2*2*2*2=64
I'm amazed. U figured out that there were 5 host bits here, and u wrote 2x2x2x2x2 = 64
The answer is correct according to that there are actually 6 host bits which should equal 64, but you say 5 and write 5 and still manage to get it right. hehe
in this the no of host u said is 64
but 2 multiply by 5 times is 32
This has been very helpful. Thank you!
No offense, but I am lost at the 5 zeros adding up to 64. I don't see that at all. 5 hosts should add up to 32?? This throws the logic way off...
Hi, thanks for this tuto. Can you have a subnet with bits not in a row like:
11111111. 11111111. 11111111. 1100 1000 = 128 + 64 + 8 = 255.255.255.200
7:03
"How many zeros do we have? We have 5 zeros..."
You mean, 6 zeros???
I'm confused...
Thanks a ton awesome Master Dan.
Hello sir, there are 6 0s
dont you subtract 2 from the network and host
to get a total of 2 networks and 62 host...........
2 to the 5 th is 32 and two the sixth is 64. 2 + 5 is 7 and you are missing a bit in the last octet.
Insanely helpful video, thanks so much
yeah, that happens to me sometimes. :)
really enjoyed this video. thanks for posting.
And yeah, the number of usable host is 2 in the power of number of 0s minus 2. Thats why 64-2 is gonna be 62
How can you switch the topic from Part 2. ? This gives no sense. You might be right, but not organized
But the formula is 2^n -2 so why 2x2
I like how you teach it .. I understand it ..
owesome. .what a way to teach subnetting
thanks a lot.
Thank you Sir, I really mean it. 🙏
its 6 zeros. 2 raise to the power of 6 is is 64.
this magic number is not may be 2*2*2*2*2=32 then in host id this magic number is 2*2*2*2*2*2=64 shall be 6
thank you very much , very useful .
Would be a great lecture if not for the unsynchronized audio/video. Tried refreshing, replaying, rewinding and it still doesn't help. The audio is ahead of the video by a good 10 seconds or so.
Really confusing... Last time I checked, 2^5=32
And why not simply, as Todd Lammle suggest, subtract your node from 256
256 - 192 = 64 Easy....
Dat wordjoke at 4:48 quite a "bit"
I see what you did there :D
These are 6 zeros hence 2 to the sixt is 64