I came here randomly searching subnetting explanations. Looked this video and the previous one through, and finally I decided to do the whole series from the beginning. The best thing a teacher can hope for is when students ask for more. Great work, Jeremy!
I passed my CCNA today, and i thank you jeremy for being the best resource to pass it, thank you for your efforts and gifting us with this amazing FREE content.
1./23 prefix length 2.172.21.96.0/20 3. 192.168.91.127/26 4. 172.16.64.0 - 172.16.127.255 5. 64 subnets, because of the 6 bits borrowed from the 3rd octate. keep the good work up bro..thanks
in the 5th its all maths. because we have used prefix /23 before for 500 hosts,logicly, we need to take the step back, and have 500*2=1000. originally we have /16 on our big network, /22-/16=6. 2 in the power of 6 is 64.
500 hosts. So 2^H -2 >=500, where H is is number of host bits. So we need 9 host bits to have 500 addresses. So out of total 32 bits, we are using 9 hosts bits so remaining network bits will be 32-9 =23. Hence /23. This is how I calculated.
I wanted to seriously thank you for this. This is by far the best introduction I've had to subnetting, especially understanding subnetting versus just solving problems. Thank you for being so practical in your approach.
You are a blessing, believe me when I say that subnetting by far is one of the toughest things to get a grasp on, at least for me. I have watched countless videos from different instructors and you have made it click for me. Truly appreciated all you do.
Incredibly helpful course! I'm binge watching it (currently on a 14th video) and just wanted to say thank you very much! I'm not planning on taking CCNA exam, I just wanted to learn basics of Networking and your course helped me so SO SO much. Looking forward for your videos and other courses if you're planning them. Best networking teacher!
Aloha.......brah...iv been watching neils course and others....lets just say yours are by far, above and beyond anything thats out there ....im grateful for ur time and effort...MAHALO
This video really had me thinking and using my brain. Learning subnetting is difficult but I get better each time I study. Love that I can go back and watch videos. I found it helpful for me to add tags to the Anki cards corresponding with the video they came from. I have found that this process of tagging allows me to identify the concepts I am struggling. To identify the video I need to watch I browse the deck in Anki and sort by the reviews field. The more reviews there are the less I know that card and can re-watch the day associated with it. Has been incredibly helpful so far.
This is incredible, I am used to doing it the long way and had no idea all it take is converting bits instead of multiplying and writing out each range. Thank you!
Thank you for this content. I've learned a trick to finding which subnet a host address belongs to. For the example you used, I divide the subnet ID by the binary value 217/8=27.125 (we don't care about the decimal value, only the whole number). Then I multiply the whole number by the binary value 27*8=216. Then I replace the subnet ID from the problem with the new subnet ID to get the network range and its home subnetwork. Shout out to Keith Barker for the shortcut.
does not work if .17 /27. We have 3 borrowed bits , where the last network bit is 32, so subnet ids go .0 .32 0.64 etc. .17 octate in binary is 0 0 0 1 0 0 0 1, change network bits (5 last bits in this case) to 0 and you get 0 in the whole octate. So .17 /27 belongs to .0 /27. BUT if you use your trick and 17/8=2.125 and then 2x8=16. We have subnet id increment of 32, so .16 /27 cannot be a subnet, because it is within the ip range of .0/27. Nor .16 /27 can be a subnet, because it does not add up in binary calculations. Where did i mess up ? :P
Hi Jeremy. Thanks for the lesson. 1. /23 2. 172.21.96.0/20 3.192.168.91.127/26 4. using a /18 subnet 172.16.64.0 - 172.16.127.255 5. 64 subnets the last question I've to take a glance at the last chart in your slide :DDDDD don't know if it's good
@@bluesparrowlove6846 Late response, but for anyone who might be wondering: Since each subnet has 64 addresses, including broadcast + network (2^6), the network address of the second subnet (192.168.91.64) counts as an address in the subnet, if that makes sense. Another way to look at it is how .0 to .63 in subnet 1 is counted as 64 addresses because the network address, 192.168.91.0 is counted the first address. Hopefully that makes sense! It took me quite a while to understand and I also got this question wrong initially.
I have Network+ but I still didn't feel very confident with subnetting. Watching these videos and sleeping on it a few nights made it all click. Thank you so much!
Hi, the quiz is always helpful. My answers are Q1. The prefix we should use is 23. 172.30.0.0/23 first subnet Network address Q2. The host 172.21.11.201/20 is belongs to subnet address: 172.21.96.0/20 Q3. The broadcast address of 192.168.91.78/26 is 192.168.91.127/26 Q4. The 2nd subnet address is: 172.16.64.0/18 & Broadcast address is 172.16.127.255/18 Q5. We can able to make 64 subnets.
In Q3, how did you get 192.168.91.127/26? Since prefix is /26, shouldn't we set all host bits after 26th bit to 1 to get the broadcast address? For me, I got 192.168.91.63, please help
Did anyone else get ring bells in their heads at 10:48 when you realized the trick you can use to get the subnet of the host address ? so easy and simple this men has been God Sent to teach us all !
I already understood the basics of binary arithmetic as I have some programming background but this course is really improving my ability to quickly think in binary
Suggesting we change the variables in the formula to avoid confusion 2 ^ y where y = borrowed network bits borrowed from right to left 2 ^ x - 2 where x = borrowed hosts bits borrowed from left to right
2:16 The Borrowed Bit. For those of you struggling to understand the concept of the "borrowed bit," just convert 192.168.1.0/26 from decimal to binary: 11000000.10101000.00000001.**00**000000 Then, you can see that you are using a /26 subnet, so the first two zeros in the fourth octet still belong to the network portion. This means that those two zeros can be used because they are the last bits of the network portion. idk if it is the correct way but... it helped me to understand. :) hope it helps.
I love the subnetting trick. I had no problems doing the quiz because there was only 4 possible combinations of bits but if I had a /29 subnet or something like that I think i'd have trouble keeping track of it all without that trick.
for review purposes :D 0:42 Things we'll cover 1:16 Answer to the last video's QUIZ 5:20 Subnetting Trick (LSB :D) 6:40 Execise 8:07 Number of Subnets Formula 10:50 Execise - Identify the Subnet 12:51 Class C Subnets/Hosts Chart 14:20 Subnetting Class B 20:59 Class B Subnets/Hosts Chart 21:42 QUIZ ___________________________________________ 1. /23 2. 172.21.96.0/20 3. 192.168.91.127/26 4. 172.16.64.0/18 172.16.127.255/18 5. 64 subnet
Hi Jeremy, I would like to thank you very much for your great lessons. I have learned so much from your lessons. Thanks a million for giving them for free.
I am the very worst at subnet masking but I'll give the quiz a shot! :>) Thanks Jeremy!! Question #1: /23 Question #2: 172.21.96.0 /20 Question #3: 192.168.91.127 /26 Question #4: 172.16.64.0 /18 Question #5: 62
qn 5 : is 64 subnets..because it is 1000 hosts needed, so here means 2^10-2, which means the host bits should be 10, and the number of borrowed bits from the 3rd octate is 6 bits, which makes the 64 subnets (2^6)..
Cannot really thank you enough for this great explanation. As said in french :" Chapeau a vous". I'm following both yourself and Imran Rafei and he started subnetting at a very early stage - day 3. I was still uncomfortable with subnetting, especially class A and B even after watching 2 or 3 times. At that time, i then tried following Astrit Krisniki who made things a little clearer. But the way you resolve the problem, especially finding network id & broadcast id for any given network is really painless now. For it would be very cumbersome if we tried to addup the value to get to the next block address, especially with class B & A.👍
Hi Jeremy, Thanks for this simple way to find out the subnets and the host range. I was stuck in this video and looked elsewhere but came back later to your video and I understood it quite easily. You are excellent. Thanks. God bless. Hope all is fine in Japan.
most awaited course ,really really helpfull jeremy's sir, its very very interesting ,at present situation i can't donate you, but surely I will. so I humble request you to please complete this course , Its really helpfull for Us. your way of teaching very lucent also. great course| Thank You very much
They don't give us calculators on the exam so one thing I highly recommend memorizing is the values of adding powers of two. Memorize 128+64+32+16 and all combinations thereof. It's not hard to calculate but you cannot get stuck crunching binary on the CCNA when you barely have any time to spend on a question.
Q.1 172.30.0.0/23 Q.2 172.21.96.0/20 Q.3 192.168.91.127/26 Q.4 172.16.64.0/18 network 172.16.127.255/18 Broadcast Q.5 126 subnets. Don't know if the answer is right , just posting the finded answer. Thanks for the great lecture ❤
12:02 Another way to solve the subnet is to: 1. map out each value of the octet: 128,64,32,16,8,4,2,1 . 2. subtract the prefix length (/29 in the example) with the closest multiple of 8 (24 in this scenario) 29-24 = 5 , This is the index for the octet value list we made earlier. 3. using the index and octet list, we find that 8 is that value of the last network bit. 4. The closest multiple of 8 to the interesting octet value (219) is 216. Mentally you could do 219 - 160 = 59. 59-56 = 3. 219-3 =216
Thank you Jeremy for making me easily understand subnetting. Really Cool!!! Here are answers to the questions. Quiz#1. Prefix length is a /23. Quiz#2. Subnet ID for the host 172.21.111.201/ 20 is 172.21.96.0/20. Quiz#3. Broadcast address is 192.168.91.127/26 Quiz#4. For the second subnet, Network address is 172.16.64.0/18 and broadcast address is 171.16.127.255/18. Quiz#5. We can make 64 subnets (2 to the power of 6).
it take me much time and little bit confusion but i solved it 1. 172.30.0.0/23 22:15 2. 172.21.96.0/20 22:25 3. 192.168.91.127/26 22:42 4. 172.16.64.0 - 172.16.127.255 22:57 5. 64 subnet 23:07
Example at 18:14, the subnetting trick to find all the possible subnets (where you keep adding decimal value of the last borrowed bit) doesn't seem to work. Decimal value of the last borrowed bit is 128, so if you add it,you get 256, 384, 512... Any comments?
Jeremy you are amazingly organized to create these videos. A great salute to your effort. Just found a little typo and wanted to point it out to all. @ 21.09 for /21 the no of hosts should be 2048-2=2046. So 2044 would be a typo.
Hello, to find next subnet address of every 500 subnet in the 18:11 example. How can I do it? I do'nt get it at all... I tried to follow the trick of converting the last network bit to decimal in the last octet, so it gives me 128, it means that I would have 500 with a difference of 128 between subnet 1 , 2 , 3 and so on? I got a little confused. Sorry for my english, I'm not a native speaker.
I could be wrong but I think first few addresses are .. 172.22.0.1 through 172.22.0.127, second subnet.. 172.22.0.128 through 172.22.0.254. Heres where it gets easy.. third subnet. Third subnet, 172.22.1.1 through 172.22.1.127, fourth subnet 172.22.1.128 through 172.22.1.255. Fifth subnet ..172.22.2.1 through 172.22.2.127 and sixth..172.22.2.128 through. 255. You have a whole octet to use (256 addresses) doubled as you use only half of the available 256 users in last octet = 512 networks. I think LOL.
what will be the first and second subnet of the problem at min 19:00 regarding up 172.22.0.0/25 borrowing 9 bits=512 subnets but I have question in how many host per subnet will be as well as first and second subnets?
Ok so when we are trying to work out the Network/Subnets we are stealing bits from left to right and while we are working out for a hosts we are stealing bits and we count from right to left...
At 19:49 There are 32 available subnets. You gave us the exact subnet ID, but out of the 32 available subnets, which subnet number 1-32 is 172.25.216.0 ? Is there a fast way to figure that out? I believe it's subnet number 27, because 216/8 is 27... And the hosts per network is 8, right?
In 17:50 how the second subnet becomes 172.16.1.0 ? and 172.16.2.0... please explain I am confused and stuck in this lesson and trying to review more and more but still I haven't get it.
Can someone please explain the answer between 17:50 and 18:09? as I understand the next subnet is 1 higher than the broadcast address of the previous subnet, but with these answers this is not possible because if I Set all the bits in the host portion to 1 it would not add up?
At 11:43, I am a little confused. If this 192.168.5.57 is a class C address with and the network bits are constant and cannot be changed, why mention them when you are trying to find out what range the IP address belongs to?
Hello all i have a question if any one can ans thanks in advance.At 18:12 on the example it is clear to me what prefix we are using but a bit confusion to me is how we will make the first and last address because now we are using the last 2 octets .....if any once can please a bit explain....Thanks and Regards
The process is the same. The first usable address is always the network address +1 (for example, network address 172.16.0.0/16, first usable address 172.16.0.1) The last usable address is always the broadcast address -1 (for example, broadcast address 172.16.255.255/16, last usable address 172.16.255.254)
Hi Jeremy, I'm currently working through the video and am looking at the table shown at 21:22. For the /32 prefix length I thought the number of hosts was 2046 and not 2044? Wanted to check to see if the formula was correct. Thanks for the great vids!
Thanks for simplifying the class C network address, still trying to understand the calculations and to memorise all of the subnets and host numbers, very useful video I recommend everyone who is studing CCNA or Network+
geez- this could eat some time on the CCNA exam! I eventually landed on the right answer. Used right technique, but slopping moving my 1s and 0s around.
At 18:30 if we borrow 9 bits it becomes /25 mask right? If yes, that means only 2 possible networks...but we need 500...am I missing sth? Thank you very much.
Q1- /23 (need to borrow 7 bits from host bits to get 128 subnets which is close to 100 subnets requirements, will get 510 hosts which cover 500 hosts requirements) Q2- 172.21.111.201 is belong to 172.21.96.0/20 subnet Q3- Broadcast Address : 192.168.91.127/26 Q4- Network Address : 172.16.64.0/18 and Broadcast address : 172.16.127.255/18 Q5- 64 Subnets Thanks for sharing Sir
Hi Jeremy another great video, thanks so much for all the hours of work you put into these videos. I have a question, if anyone could help clear this up. At 11:44 why is the answer the 32, I thought in a /27 subnet there are 3 borrowed bits? Shouldn't it be 128 + 64 +32 ? thanks
First I would like to appreciate you for you hard work and enable us to study with a free course, thank a lot. The answers are: Q1- sol = 172.30.0.0/23 Q2 - sol = 172.121.96.0/20 Q3- sol = broadcast add = 192.168.91.127 Q4- sol= a- 172.16.63.0 /16 b- network add = 172.16.64.0 /16 c- Broadcast add =172.16.128.0 /16 Q5- sol = 172.30.0.0/22 I hope it's all correct. Best regards...
one thing that keeps tripping me up is when i turn all the host bits to 1 for the broadcast I keep wanting to use the 2 to the power of the number of bits as the network value which keeps landing me 1 off
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
ERRATA: At 3:45, the range of subnet 2 is 192.168.1.64 - 192.168.1.127
At 21:00, for a /21 prefix length the number of hosts should be 2046, NOT 2044.
Thanks!
Thanks for pointing it out to me!
@@JeremysITLab pin it please :]
@@LarrySoliman Oops, I thought I did. Thanks!
CANT SEEM TO FIGURE OUT THE BORROWED BIT THING. SO LOST
I came here randomly searching subnetting explanations. Looked this video and the previous one through, and finally I decided to do the whole series from the beginning. The best thing a teacher can hope for is when students ask for more. Great work, Jeremy!
I passed my CCNA today, and i thank you jeremy for being the best resource to pass it, thank you for your efforts and gifting us with this amazing FREE content.
What did you do to prepare besides this course ? Im at day 14 and i'm hoping this course can get as close as i can to a CCNA, thanks !
please tell me if you study from other resources and if so can you share it with mw thank you
@@hanaelsadig7511 you have to lab lab lab lab lab, and more lab
@@hanaelsadig7511 the only way to truly understand this stuff, is to apply it and verify it (why did i do this, what does it do)
You should tip him
1./23 prefix length
2.172.21.96.0/20
3. 192.168.91.127/26
4. 172.16.64.0 - 172.16.127.255
5. 64 subnets, because of the 6 bits borrowed from the 3rd octate.
keep the good work up bro..thanks
in the 5th its all maths. because we have used prefix /23 before for 500 hosts,logicly, we need to take the step back, and have 500*2=1000. originally we have /16 on our big network, /22-/16=6. 2 in the power of 6 is 64.
@@Anton2452r2n can u tell me how u got/23 when 500 hosts are asked in Q1?
500 hosts. So 2^H -2 >=500, where H is is number of host bits. So we need 9 host bits to have 500 addresses. So out of total 32 bits, we are using 9 hosts bits so remaining network bits will be 32-9 =23. Hence /23. This is how I calculated.
@@SkMariamFathima atleast 500 hosts
I wanted to seriously thank you for this. This is by far the best introduction I've had to subnetting, especially understanding subnetting versus just solving problems. Thank you for being so practical in your approach.
Thanks Evan, glad to hear that :)
I'm so grateful for this opportunity to learn CCNA at my own pace and on my own time. Thank you for making this course available to everyone.
You are a blessing, believe me when I say that subnetting by far is one of the toughest things to get a grasp on, at least for me. I have watched countless videos from different instructors and you have made it click for me. Truly appreciated all you do.
AGREED!!
Yes but once you get the hang of it it seems simple. Subnetting IPv6 networks is a blast! :)
Incredibly helpful course! I'm binge watching it (currently on a 14th video) and just wanted to say thank you very much! I'm not planning on taking CCNA exam, I just wanted to learn basics of Networking and your course helped me so SO SO much. Looking forward for your videos and other courses if you're planning them. Best networking teacher!
Thanks so much!
Did you pass?
@@muffinbutton1484 they're not planning to take the exam.
Aloha.......brah...iv been watching neils course and others....lets just say yours are by far, above and beyond anything thats out there ....im grateful for ur time and effort...MAHALO
I finally starting to get it. Holy moly. It took rewatching you and Neil Anderson multiple times and this time making better notes.
Finally, the long awaited video is in...
Thank you very much.
Thank you!
This video really had me thinking and using my brain. Learning subnetting is difficult but I get better each time I study. Love that I can go back and watch videos. I found it helpful for me to add tags to the Anki cards corresponding with the video they came from. I have found that this process of tagging allows me to identify the concepts I am struggling. To identify the video I need to watch I browse the deck in Anki and sort by the reviews field. The more reviews there are the less I know that card and can re-watch the day associated with it. Has been incredibly helpful so far.
Thanks James :)
Hi James, thats is such a good idea, where / how do you add tags? My brain hurts after this too haha
I am not giving CCNA exam but I am loving these videos as You are explaining it with so much of dedication. Thank you sir
Thanks Bikash :)
This is incredible, I am used to doing it the long way and had no idea all it take is converting bits instead of multiplying and writing out each range. Thank you!
Thank you for this content. I've learned a trick to finding which subnet a host address belongs to. For the example you used, I divide the subnet ID by the binary value 217/8=27.125 (we don't care about the decimal value, only the whole number). Then I multiply the whole number by the binary value 27*8=216. Then I replace the subnet ID from the problem with the new subnet ID to get the network range and its home subnetwork. Shout out to Keith Barker for the shortcut.
Nice idea. I am just wondering how did you arrive at the binary value of 8.
@@vaitheeshiyer 8 is the block size, or number of unique hosts (0 - 7)
does not work if .17 /27. We have 3 borrowed bits , where the last network bit is 32, so subnet ids go .0 .32 0.64 etc. .17 octate in binary is 0 0 0 1 0 0 0 1, change network bits (5 last bits in this case) to 0 and you get 0 in the whole octate. So .17 /27 belongs to .0 /27. BUT if you use your trick and 17/8=2.125 and then 2x8=16. We have subnet id increment of 32, so .16 /27 cannot be a subnet, because it is within the ip range of .0/27. Nor .16 /27 can be a subnet, because it does not add up in binary calculations.
Where did i mess up ? :P
after watching these subnetting videos for the past week on repeat. im finally starting to understand. yay!
Hi Jeremy. Thanks for the lesson.
1. /23
2. 172.21.96.0/20
3.192.168.91.127/26
4. using a /18 subnet 172.16.64.0 - 172.16.127.255
5. 64 subnets
the last question I've to take a glance at the last chart in your slide :DDDDD don't know if it's good
Perfect answers! 👍
@@JeremysITLab Hey Jeremy, On quiz 3, How is it .127 in the last octet. I get all the part till finding that 192.168.91.128 is the subnet.
@@bluesparrowlove6846 Late response, but for anyone who might be wondering:
Since each subnet has 64 addresses, including broadcast + network (2^6), the network address of the second subnet (192.168.91.64) counts as an address in the subnet, if that makes sense.
Another way to look at it is how .0 to .63 in subnet 1 is counted as 64 addresses because the network address, 192.168.91.0 is counted the first address.
Hopefully that makes sense! It took me quite a while to understand and I also got this question wrong initially.
After your videos. i feel i'm an expert in subnettings, you made it so easy to understand and resolve all solutions, thank you so much
Thanks for the series. Foe once, in all my years of training and working in IT, this was explained easily and now seems less daunting.
Thank you, I'm glad to hear that :)
I have Network+ but I still didn't feel very confident with subnetting. Watching these videos and sleeping on it a few nights made it all click. Thank you so much!
1) 255.255.254.0
2) 172.21.111.201/20 belongs to 172.21.96.0/20 subnet
3) 192.168.91.64 subnet and 192.168.91.127 broadcast
4) subnet is 172.16.64.0 and broadcast 172.16.127.255
5) 172.30.0.0/22 = 64 subnets
Thanks for your excellent videos.
Thanks J, this requires constant practice, helpful walkthroughs. Always like your guidelines
Thanks Glenn!
Hi, the quiz is always helpful. My answers are
Q1. The prefix we should use is 23. 172.30.0.0/23 first subnet Network
address
Q2. The host 172.21.11.201/20 is belongs to subnet address: 172.21.96.0/20
Q3. The broadcast address of 192.168.91.78/26 is 192.168.91.127/26
Q4. The 2nd subnet address is: 172.16.64.0/18 & Broadcast
address is 172.16.127.255/18
Q5. We can able to make 64 subnets.
Q.5 answer should be 16 subnets
In Q3, how did you get 192.168.91.127/26?
Since prefix is /26, shouldn't we set all host bits after 26th bit to 1 to get the broadcast address?
For me, I got 192.168.91.63, please help
I would like to say, "THANK YOU VERY MUCH"! You have a great way of explaining the process. By far, the best video out there
today.
Did anyone else get ring bells in their heads at 10:48 when you realized the trick you can use to get the subnet of the host address ? so easy and simple this men has been God Sent to teach us all !
I already understood the basics of binary arithmetic as I have some programming background but this course is really improving my ability to quickly think in binary
You kick butt!!! Just found this series and so grateful that it's online. Thanks Jeremy!
Suggesting we change the variables in the formula to avoid confusion
2 ^ y
where y = borrowed network bits
borrowed from right to left
2 ^ x - 2
where x = borrowed hosts bits
borrowed from left to right
2:16 The Borrowed Bit.
For those of you struggling to understand the concept of the "borrowed bit," just convert 192.168.1.0/26 from decimal to binary:
11000000.10101000.00000001.**00**000000
Then, you can see that you are using a /26 subnet, so the first two zeros in the fourth octet still belong to the network portion. This means that those two zeros can be used because they are the last bits of the network portion.
idk if it is the correct way but... it helped me to understand. :) hope it helps.
I love the subnetting trick. I had no problems doing the quiz because there was only 4 possible combinations of bits but if I had a /29 subnet or something like that I think i'd have trouble keeping track of it all without that trick.
for review purposes :D
0:42 Things we'll cover
1:16 Answer to the last video's QUIZ
5:20 Subnetting Trick (LSB :D)
6:40 Execise
8:07 Number of Subnets Formula
10:50 Execise - Identify the Subnet
12:51 Class C Subnets/Hosts Chart
14:20 Subnetting Class B
20:59 Class B Subnets/Hosts Chart
21:42 QUIZ
___________________________________________
1. /23
2. 172.21.96.0/20
3. 192.168.91.127/26
4. 172.16.64.0/18 172.16.127.255/18
5. 64 subnet
@@speedysui master this first, then watch practical networking's subnetting series
I think 2 is 104.0
Hi Jeremy,
I would like to thank you very much for your great lessons. I have learned so much from your lessons.
Thanks a million for giving them for free.
Thank you, I'm glad they're helpful!
Thank you once again Jeremy.
All of us aspiring network engineers appreciate your work!!
I am the very worst at subnet masking but I'll give the quiz a shot! :>) Thanks Jeremy!!
Question #1: /23
Question #2: 172.21.96.0 /20
Question #3: 192.168.91.127 /26
Question #4: 172.16.64.0 /18
Question #5: 62
Jack N5Io dude same here i am struggling with subnetting big time
qn 5 : is 64 subnets..because it is 1000 hosts needed, so here means 2^10-2, which means the host bits should be 10, and the number of borrowed bits from the 3rd octate is 6 bits, which makes the 64 subnets (2^6)..
11:56
Quick and dirty maths:
Mask determination: /29 = 29/8= 3,625 octets = 255.255.255.(0,625*8 =5=0b11111=)248.
Network size determination: 256-248=8
--
Find subnet fast:
Convert host block to binary:
Divide by 2. Round down. Example 219/2=109,5 > 109.
1
Cannot really thank you enough for this great explanation. As said in french :" Chapeau a vous".
I'm following both yourself and Imran Rafei and he started subnetting at a very early stage - day 3. I was still uncomfortable with subnetting, especially class A and B even after watching 2 or 3 times.
At that time, i then tried following Astrit Krisniki who made things a little clearer.
But the way you resolve the problem, especially finding network id & broadcast id for any given network is really painless now. For it would be very cumbersome if we tried to addup the value to get to the next block address, especially with class B & A.👍
Thanks for your comment, I'm glad to hear it :)
your explanation is very clear brother.
This channel is getting a spot in my bookmarks… definitely
That trick with the last bit borrowed from the host portion is pretty neat lol. Thank you for the great explanation!
Such a great course! Thanks for putting this together! You're a very good teacher!
I love your way of deepening the topic understanding through practice problems,quizzes and the like. Thanks a lot!
Hi Jeremy, Thanks for this simple way to find out the subnets and the host range. I was stuck in this video and looked elsewhere but came back later to your video and I understood it quite easily. You are excellent. Thanks. God bless. Hope all is fine in Japan.
Thanks Sanjay, all is fine here, I hope the same for you ;)
most awaited course ,really really helpfull jeremy's sir, its very very interesting ,at present situation i can't donate you, but surely I will. so I humble request you to please complete this course , Its really helpfull for Us. your way of teaching very lucent also. great course| Thank You very much
Why do we turn borrowed bit from 0 to 1 when finding broadcast address of block of addresses?
Can somebody explain the answer to the question which is mentioned at 18:13
Question 1) 172.30.0.0/23
Question 2) 172.21.96.0/20
Question 3) 192.168.91.127/26
Question 4) 172.16.64.0 - 172.16.127.255
Question 5) (/18) 64 subnets
Thanks for all you doing, it's amazing!
They don't give us calculators on the exam so one thing I highly recommend memorizing is the values of adding powers of two. Memorize 128+64+32+16 and all combinations thereof. It's not hard to calculate but you cannot get stuck crunching binary on the CCNA when you barely have any time to spend on a question.
Thank You Jeremy, really Thank You... Respect to the other videos I saw, You are really the best!!!!!!!
Thanks, Pietro :)
The Goat!
thank you for this explanation of subnetting. it was clear. for class b number of hosts, i still use 2^n-2; where n = 16 - number of borrowed bit/s
Q.1 172.30.0.0/23
Q.2 172.21.96.0/20
Q.3 192.168.91.127/26
Q.4 172.16.64.0/18 network
172.16.127.255/18 Broadcast
Q.5 126 subnets.
Don't know if the answer is right , just posting the finded answer. Thanks for the great lecture ❤
12:02
Another way to solve the subnet is to:
1. map out each value of the octet: 128,64,32,16,8,4,2,1 .
2. subtract the prefix length (/29 in the example) with the closest multiple of 8 (24 in this scenario) 29-24 = 5 , This is the index for the octet value list we made earlier.
3. using the index and octet list, we find that 8 is that value of the last network bit.
4. The closest multiple of 8 to the interesting octet value (219) is 216. Mentally you could do 219 - 160 = 59. 59-56 = 3. 219-3 =216
@11:00 do we need to know how to calculate these things without pen & paper for the exam?
Thank you Jeremy for making me easily understand subnetting. Really Cool!!!
Here are answers to the questions.
Quiz#1. Prefix length is a /23.
Quiz#2. Subnet ID for the host 172.21.111.201/ 20 is 172.21.96.0/20.
Quiz#3. Broadcast address is 192.168.91.127/26
Quiz#4. For the second subnet, Network address is 172.16.64.0/18 and broadcast address is 171.16.127.255/18.
Quiz#5. We can make 64 subnets (2 to the power of 6).
it take me much time and little bit confusion but i solved it
1. 172.30.0.0/23 22:15
2. 172.21.96.0/20 22:25
3. 192.168.91.127/26 22:42
4. 172.16.64.0 - 172.16.127.255 22:57
5. 64 subnet 23:07
Example at 18:14, the subnetting trick to find all the possible subnets (where you keep adding decimal value of the last borrowed bit) doesn't seem to work. Decimal value of the last borrowed bit is 128, so if you add it,you get 256, 384, 512... Any comments?
Jeremy you are amazingly organized to create these videos. A great salute to your effort. Just found a little typo and wanted to point it out to all. @ 21.09 for /21 the no of hosts should be 2048-2=2046. So 2044 would be a typo.
He pinned a comment 3 years ago pointing out this mistake so no need at all.
Thank you for another great video with clear, concise explanations!
nice very nice i watch a lot of subnetting and ur video is the best direct to the point u also included the point to point network.
Thank you, I'm glad you like it :)
First time to know how to find the network address by turning the host bits to zero...I used to count till reach the network that may has this host .🙌
Thank you so much sir for this valuble explain ❤❤ now I can understand this . Wish you all the best . Again Thankyou so much😍😍
I'm the man who is going to be king of the network admins
Hello, to find next subnet address of every 500 subnet in the 18:11 example. How can I do it? I do'nt get it at all...
I tried to follow the trick of converting the last network bit to decimal in the last octet, so it gives me 128, it means that I would have 500 with a difference of 128 between subnet 1 , 2 , 3 and so on?
I got a little confused. Sorry for my english, I'm not a native speaker.
I could be wrong but I think first few addresses are .. 172.22.0.1 through 172.22.0.127, second subnet.. 172.22.0.128 through 172.22.0.254. Heres where it gets easy.. third subnet.
Third subnet, 172.22.1.1 through 172.22.1.127, fourth subnet 172.22.1.128 through 172.22.1.255.
Fifth subnet ..172.22.2.1 through 172.22.2.127 and sixth..172.22.2.128 through. 255. You have a whole octet to use (256 addresses) doubled as you use only half of the available 256 users in last octet = 512 networks. I think LOL.
what will be the first and second subnet of the problem at min 19:00 regarding up 172.22.0.0/25 borrowing 9 bits=512 subnets but I have question in how many host per subnet will be as well as first and second subnets?
Well, there are 7 host bits...so you can calculate how many hosts will be in each subnet!
@@JeremysITLab so the first network will be 172.22.0.0 to 172.22.0.127 and second subnet 172.22.0.128 to 172.22.0.254 ?
Sir , we don't need any tricks
Your direct method is far more better....👌👌👌
Thank you ;)
Jeremy, thank you for making this content.
Yes, for part 3, I noticed that I had made a mistake. ;)
Thanks
Ok so when we are trying to work out the Network/Subnets we are stealing bits from left to right and while we are working out for a hosts we are stealing bits and we count from right to left...
At 19:49 There are 32 available subnets. You gave us the exact subnet ID, but out of the 32 available subnets, which subnet number 1-32 is 172.25.216.0 ? Is there a fast way to figure that out? I believe it's subnet number 27, because 216/8 is 27... And the hosts per network is 8, right?
In 17:50 how the second subnet becomes 172.16.1.0 ? and 172.16.2.0... please explain I am confused and stuck in this lesson and trying to review more and more but still I haven't get it.
The first subnet is 172.16.0.0/23. To get the second subnet, change the 'smallest' bit to 1. That makes 172.16.2.0/23.
Can someone please explain the answer between 17:50 and 18:09? as I understand the next subnet is 1 higher than the broadcast address of the previous subnet, but with these answers this is not possible because if I Set all the bits in the host portion to 1 it would not add up?
if u set host portion to all 1's, then it will be equal to 255. this is the broadcast address. next address will be the network address of next subnet
Got a question. At 12:28 all that process I need to do fast in my head (since there are no papers on ccna exam) ?
day two, and I am able to answer this question in 10 - 15 sec... it is possible... last day it didn't seem so.
why do even collage degrees and paid coursed are needed now?
Dude you should make your own website where CCNA, CCNP and CCIE courses are available. Everyone will buy.
I'd love to! But first I have to make the CCNP and CCIE courses ;)
At 11:43, I am a little confused. If this 192.168.5.57 is a class C address with and the network bits are constant and cannot be changed, why mention them when you are trying to find out what range the IP address belongs to?
You just made it so easy for me.. Thank you so much.
Thank you for watching!
hey what are the subnets for the 500 separate subnets? do i just add 64?
Hello all i have a question if any one can ans thanks in advance.At 18:12 on the example it is clear to me what prefix we are using but a bit confusion to me is how we will make the first and last address because now we are using the last 2 octets .....if any once can please a bit explain....Thanks and Regards
The process is the same. The first usable address is always the network address +1 (for example, network address 172.16.0.0/16, first usable address 172.16.0.1)
The last usable address is always the broadcast address -1 (for example, broadcast address 172.16.255.255/16, last usable address 172.16.255.254)
@@JeremysITLab Thanks you indeed Sir.......Thank u once again :)
Hi Jeremy, I'm currently working through the video and am looking at the table shown at 21:22. For the /32 prefix length I thought the number of hosts was 2046 and not 2044? Wanted to check to see if the formula was correct. Thanks for the great vids!
Do you mean /21? You're correct, that should be 2046!
Thanks for simplifying the class C network address, still trying to understand the calculations and to memorise all of the subnets and host numbers, very useful video I recommend everyone who is studing CCNA or Network+
geez- this could eat some time on the CCNA exam!
I eventually landed on the right answer. Used right technique, but slopping moving my 1s and 0s around.
With some practice you'll get much more efficient at it!
becoming a subnetting-ninja after this.
man! what great and polished work you have done, thank you very much!!
number 1 is /24,
number 2 is 172.21.96.2,
number 3 is 192.168.91.127,
number 4 is 172.16.64.0 to 172.16.127.255,
number 5 is 64
Watch day 15's video to check your answers ;)
Thank you so much for lovely clear lectures Subnetting (Part 1 & 2)
You are SOOO GOOOOOOD at what you do, Thank you for everything you're doing and then some!
Thanks for putting this very informative videos out. Is there more videos to come
Why did we change the borrowed host bit to 1 at 3:22?
At 18:30 if we borrow 9 bits it becomes /25 mask right? If yes, that means only 2 possible networks...but we need 500...am I missing sth? Thank you very much.
9 borrowed bits means we have 9 bits to change freely and make new network addresses (new subnets)! 2^9= 512 subnets
@@JeremysITLab Got it. Thanks a lot Jeremy.
Q1- /23 (need to borrow 7 bits from host bits to get 128 subnets which is close to 100 subnets requirements, will get 510 hosts which cover 500 hosts requirements)
Q2- 172.21.111.201 is belong to 172.21.96.0/20 subnet
Q3- Broadcast Address : 192.168.91.127/26
Q4- Network Address : 172.16.64.0/18 and
Broadcast address : 172.16.127.255/18
Q5- 64 Subnets
Thanks for sharing Sir
best subnetting explanation ever
Thank you, Saif!
From these videos, I am confident to lie on my CV having CCNA certificate😀, thanks Jeremy for this incredible talent.
Hi Jeremy another great video, thanks so much for all the hours of work you put into these videos. I have a question, if anyone could help clear this up. At 11:44 why is the answer the 32, I thought in a /27 subnet there are 3 borrowed bits? Shouldn't it be 128 + 64 +32 ? thanks
Hi Philip! The 128 and 64 bits are both 0. Only the 32 bit is set to 1, so the value is 32.
First I would like to appreciate you for you hard work and enable us to study with a free course, thank a lot.
The answers are:
Q1- sol = 172.30.0.0/23
Q2 - sol = 172.121.96.0/20
Q3- sol = broadcast add = 192.168.91.127
Q4- sol=
a- 172.16.63.0 /16
b- network add = 172.16.64.0 /16
c- Broadcast add =172.16.128.0 /16
Q5- sol = 172.30.0.0/22
I hope it's all correct.
Best regards...
Quiz 1 /23, Quiz 2 Subnet 96 , Quiz 63, Quiz 4 Network is 16 and broadcast 31 and Quiz 5 64 subnets
It's getting hard to grasp! Thanks by the way:)
q1: /23
q2: 172.21.96.0/20
q3: 192.168.91.127/26
q4: NA-172.16.64.0/18 BA-172.16.127.255/18
q5: 64 subnets
Nice, check the next video for the answers ;)
Hello Jermy, what you mean Split the network in half? in /25 13:10
one thing that keeps tripping me up is when i turn all the host bits to 1 for the broadcast I keep wanting to use the 2 to the power of the number of bits as the network value which keeps landing me 1 off
No worries, just keep practicing and you'll get used to it ;)
Great job Jeremy, keep up the good work.