Subnetting Part 2 - Subnetting for Hosts

[@tahersadeghi6773]

The way you describe the subject is great. It would be very nice to see more of your videos on the subnetting topic. Thanks

[@bloemkaaskop123]

Very usefull indeed. showing the numbers on the screen and having it explained along with it makes this very clear to follow thank you.

[@AlexslaneyHD]

Thank you very much for the video, it has really helped me get a firm understanding of subnetting.

[@AlIslam02]

Thank you very much!

[@stevenstclair6704]

thank you so much this was just what i needed!

[@manarahmed8574]

Thank you so much man you are amazing

[@stuwieFF]

ty for this, hope it helps with my Cisco exam :)

[@moya6647]

oh my God thank you soooooo much !!!

[@nsmaepa]

Wow Thank you so much zixster... May you please make /upload a video whereby you are subtting IPV6 addresses.

[@jnaz000]

I think there's no need to subnet IPV6 cause it's ip are sooooo many.

[@mubashems2141]

the best

[@malindabarron8165]

Thanks!

[@kevinm38]

0:54 ... only if we are discussing classful addresses.

[@vinodjaiswal7625]

For example 2: There will be 8 subnets subnet1:192.168.10.0/27-192.168.10.31/27 Subnet2:192.168.10.32/27-192.168.10.63/27 Subnet3:192.168.10.64/27-192.168.10.95/27 Subnet4:192.168.10.96/27-192.168.10.127/27 Subnet5:192.168.10.128/27-192.168.10.159/27 Subnet6:192.168.10.160/27-192.168.10.191/27 Subnet7:192.168.10.192/27-192.168.10.223/27 Subnet8:192.168.10.224/27-192.168.10.255/27 Default mask:255.255.255.224 or /27

[@raghavsood6156]

Please tell me why are 6 bits reserved for your host? When you say 16+32+2 = 50 then why are you refering to the rest of the 6 bits? shouldn't you only choose 3 bits with place value equivelent to 50?

[@A1cheeze]

def just help me with my homework.

[@G_Rapids]

Do you teach at GRCC? because you sound like my Security+ teacher lol.

[@ChrisJayCayton]

very helpful thank you god bless :)

[@ianchapman4619]

thanks zixster

[@swiss7800]

what do you do if subnet class is b but the required host is 24...

[@zixster]

Answer will be the same if the starting network class is A, B, or C. the number 24 is represented as 11000, so you would need to use 5 total bits to represent at least 24 hosts, the remaining 3 bits would be converted to 1's. so the resulting mask would be 255.255.255.224. The 224 in binary is 11100000, so three 1's for the network and 5 0's for the hosts.

[@fun_world9079]

good for students

[@saunders2112]

This isn't exactly right. Try an example where you have a class C address and need 31 hosts. If you do it this way, you end up with a range of address where the networks have 32 host addresses and when you take of the first (for the network address) and the last (for the broadcast address), you end up with 30 usable addresses. To use this method and have it always work out, you have to add 1 to the number of hosts you want to find before figuring out how many host bits you need. So, if you are trying to find the subnet mask that will give you 31 hosts, find how many bits you would need for 32 (31+1). I believe this problem will occur if you are looking for 1, 3, 7, 15, 31, 63, or 127 hosts per network.

[@zixster]

Adam Saunders That is technically correct. An easier way to say it though is if you are looking for a number of hosts and the end result you get is a single contiguous string of ones. In those odd-ball cases you'll be short. But you'll also figure out pretty quickly when working through the numbers. Realistically, when you use a method like this and see how it works you stop working it out on paper and just do it in your head. If someone asks the question "I need a network that can accommodate 45 hosts" you just need to look at the options. If I go down to 224, that's give me 32-2 hosts (256-224=32, then 32-2=30) so I would need to go to 192 (265-192=64, then 64-2=62) so the subnet I would need is 255.255.255.192 to allow for those 45 hosts. However, to get to that point you need to understand where the numbers come from.

[@saunders2112]

zixster Sounds good. Thanks for the response.

[@raghavsood6156]

Why are you switching the first 2 bits of the last octet to "1" and only the place value under 16 and 32 + 3 = 50. NOT THE OTHER BITS. So why do you say the total number of ones that make the number 50 are 6 bits??? NO THEY ARE ACTUALLY 3 BITS meaning, 16, 32 + 2 not all 6 bits. How could that be?

[@zixster]

Raghav Sood Because if you look at how the number 50 is represented in binary ( you can put it in a calculator and convert from dec to bin and see for yourself) it uses all 6 bits. You don't just dismiss the 0's they are part of representing 50 in binary