THANK YOU so much. Oh my gosh. Learning this in school right now and I've been wanting to rip my hair out for weeks. This video clicked it into my brain. Awesome. Finally.
how you find the maximum number of valid subnets and usable hosts per subnet that you can get from the network 172.21.0.0/27 using this calclation plz do you have any trick?
In this case, it will be 10.4.77.96 255.255.255.240 = /28, which means you need 28 binary bits for the subnet mask. 24 + 4. The lowest bit value is 16 ( 4th position ) so 16 multiplied by 6 = 96 which is closer to 99. Therefore subnet id is: 10.4.77.96 with 1st host 10.4.77.97 and last host: 10.4.77.110
what if the host IP is 10.55.66.77 /25 ? what is the subnet ID ? going through the process, you'll find out 256-128 = 128, how do I match the closest to 77 if 128 is above 77 ? Thanks
Hi, actually the technique in the video might not apply 100% of the time. I found an easier way to go around it. Let me explain: IP: 10.55.66.77 /25 Mask: 255.255.255.128 128 in binary is: 10000000. Now, to find out if there are multiple subnet IDs, we have to find the increment. The increment is the lowest bit in the Mask, which is 128. Therefore : Subnet ID: 10.55.66.0 Mask:255.255.255.128 WC: 0.0.0.127 ( finding the wildcard helps you determine what is the broadcast address ) Broadcast: 10.55.66.127 1st host: 10.55.66.1 last host: 10..55.66.126 Resulting in 126 hosts / subnet Going further, the 2nd subnet id is: 10.55.66.128, which means the 1st host is 10.55.66.129 & last :10.55.66.254 The 3rd subnet id is: 10.55.67.0 with 1st host: 10.55.67.1 & last: 10.55.67.126, and so on until the last subnet available 10.255.255.128, having in total 33.554.432 subnets ( 2 to the power of 25) with 126 hosts/subnet
Catalin Chira can you please paste a link where I can learn your method from I started off good on following it but than got lost so please send me a link if you can thanks
Hi, I have no link, it is a combination of multiple ways, after reading, searching online...and eventually it worked for me very well. Some people are taught a different way. Where did you stuck maybe I can help you
felt like i was learning survival cisco with that intro
THANK YOU! great explanation...exactly what i need. now i am ready to find my subnet ID.
thankyou so much i was stuck on this for a while.
Great instruction Wendell! I am a follower!
mind = blown! I feel like I'm in genius mode right now, but I still got a long way to go! I am for sure subscribing to this channel.
You helped me more than my professor did! Thank you!
Much easier than the cake method I was introduced to. Thanks.
THANK YOU so much. Oh my gosh. Learning this in school right now and I've been wanting to rip my hair out for weeks. This video clicked it into my brain. Awesome. Finally.
weeks? we covered this shit in 2 days and it literally passed over my head
Thank you so much, this video is amazing!
how you find the maximum number of valid subnets and usable hosts per subnet that you can get from the network 172.21.0.0/27 using this calclation plz do you have any trick?
This guide is just the best i ever seen
so simple
i actually understand this already, but i lost and stuck in chapter 20
thnks for making this video xd
I have only seen examples where the third octet is neither 255 or 0. Is it the exact same process if the fourth octet is neither 255 or 0?
I got it! Thank you.
Thank You!
You are the best!
Great Stuff. Thank you!
THANK YOU SO MUCH
isnt 10.4 class A ? so doesnt it should be 255.0.0.0 ? pleas answer
Emprator Z Not if you are using variable length subnet masks :)
thank you sir
thank you !
how do I find the usable range for 172.16.85.0/21 Help!!!
and ............... you saved my life .
Thank you !
amazing video ))) thank you very much indeed ))))
Great
Thank you! My professor didn't want to teach it this way. HAHAHA!
thanks man
thanks you helped
thx for your solution to the problem , i prefer this over binary
but what if you only have the subnet mask like 255.255.255.240 what would the subnet id be??
In this case, it will be 10.4.77.96 255.255.255.240 = /28, which means you need 28 binary bits for the subnet mask. 24 + 4. The lowest bit value is 16 ( 4th position ) so 16 multiplied by 6 = 96 which is closer to 99. Therefore subnet id is: 10.4.77.96 with 1st host 10.4.77.97 and last host: 10.4.77.110
@@TheDjexcessive how is 16 the lowest bit when 240 in binary is = 11110000
Or do you look at the host bits only?
what if ip of 3rd octat is 0 then?
The subnet ID will be 10.4.0.0 Broadcast 10.4.15.255, 1st h 10.4.0.1, Last h 10.4.15.254
Wendell may the good LORD bless you
that sounds to easy is it really that simple
ive got your book
what if the host IP is 10.55.66.77 /25 ? what is the subnet ID ? going through the process, you'll find out 256-128 = 128, how do I match the closest to 77 if 128 is above 77 ? Thanks
Catalin Chira I try to do yours and I'm stuck too
Hi, actually the technique in the video might not apply 100% of the time. I found an easier way to go around it. Let me explain:
IP: 10.55.66.77 /25
Mask: 255.255.255.128
128 in binary is: 10000000. Now, to find out if there are multiple subnet IDs, we have to find the increment. The increment is the lowest bit in the Mask, which is 128. Therefore :
Subnet ID: 10.55.66.0
Mask:255.255.255.128
WC: 0.0.0.127 ( finding the wildcard helps you determine what is the broadcast address )
Broadcast: 10.55.66.127
1st host: 10.55.66.1
last host: 10..55.66.126
Resulting in 126 hosts / subnet
Going further, the 2nd subnet id is: 10.55.66.128, which means the 1st host is 10.55.66.129 & last :10.55.66.254
The 3rd subnet id is: 10.55.67.0 with 1st host: 10.55.67.1 & last: 10.55.67.126, and so on until the last subnet available 10.255.255.128, having in total 33.554.432 subnets ( 2 to the power of 25) with 126 hosts/subnet
Catalin Chira can you please paste a link where I can learn your method from I started off good on following it but than got lost so please send me a link if you can thanks
Hi, I have no link, it is a combination of multiple ways, after reading, searching online...and eventually it worked for me very well. Some people are taught a different way. Where did you stuck maybe I can help you
Catalin Chira how did you get 10.55.66.0? and your second subnet fourth octet is the same as the mask fourth octet
are you wendall odom?
thank you . you save my ass :)
no help
Binary is easier