VLSM SUBNETTING TUTORIAL | How to do VLSM and configure in Cisco Router with live example
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- čas přidán 17. 04. 2018
- VLSM SUBNETTING TUTORIAL | How to do VLSM and configure in Cisco Router with a live example
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#subnetting #vlsm #ccna - Věda a technologie
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Thank you for the final question , i learned a lot during to find the answer. Well what i found is : Question was for R4 network , so before it we have R3 wich is 192.168.5.192/28 in this case its broadcast id should be 192.168.5.207. The R4 network must to start after this , wich is 192.168.5.208 ( Network ID ) , the best possible host number was " 15 " for requested hosts , that means Broadcast ID is 192.168.5.223 . Host Ip range would be then 192.168.5.208 -> 192.168.5.222 . Since we borrowed last 4 bits from last octet , there is 8+8+8+4 left and that gives us the prefix /28 CIDR. And ip was also Class C ( yes VLSM is classless but.. ) so the remaining "1"'s ( 128+64+32+16 ) telling us that the Sub Mask is 255.255.255.240. i hope thats correct.
Hi Evren Ersoy,
You are partially correct. The possible host was 16 that is 2 power 4. The first address is always network and the last address is a broadcast address. I can see your network id and broadcast id are correct but the host range starts from 192.168.5.209-192.168.5.222.
Thank you
Before your video, I knew nothing but in the end, I understand everything. Thank you and keep it up
Thanks for your feedback.
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I was having difficulty understanding this concept but you made it look simple. thanks a lot
You are a life saver. Thank you 😊
Kya explain kiya gajab....
Thank you so much! I passed my test watching this video!
Nice tutorial. I noticed you count from 2^1 instead of 2^0.
2^0 is included in the 8 bits. 2^12 is 13 bits making 5 host bits on the second octate from left.
Sir lots of thank you for making video on IP address and I can understand the problem it because of you thank you sir
Wow ...now the VLSM is very easy to me...Thank you boss....great job....continue....
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Thank you sir your video very helpful for me♥️
I can't thankq enough Bro❤️
You really taught well .
well done and clear sir!!!!!
i subscribed you only because of this video
Great Video!!! Thank you for doing this!! Appreciate it.
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Thank you 😊
Great Tutorial. So easy to understand. -Thank you, God bless you
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I love u bro .....thanks for this video
This helped a lot.
One of fantastic tutorial for understanding VLSM concept... Keep it up bro...
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Thank for teaching
You’re the reason I passed my college course thanks bro
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Bhai kaise
Hey dude, thanks for the video!
This has really helped me out in my understanding.
Update: I got all of the hosts parts correct on my test!
Thanks for the help dude!
I did mess up the WAN part though, because I did /29 as the subnet mask, instead of /30.
I am so dumb XD
Update 28/02/2021
Thanks you dude for this video man. It really helped me put all of the things I learnt in all of the weeks of classes and all together.
You really helped me out and I got a B at a 76% mark in the overall subject.
The low score is because I did /29 as the subnet for the WAN instead of /30 subnet, this was a dumb mistake on my part XD
Bro your explanation is superb, you are helping me a lot. :) :) :)
Thank you for watching my video.
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Good work gentleman. Just like, cake knowledge is sweet when shared.
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Network Id is 192.168.5.208, IP Range is from 192.168.5.209 to 192.18.5.222, Broadcast Id is 192.168.5.223, CIDR value is /28 and the Subnetmask is 255.255.255.240,
My Name is Damian. Pls let me know whether my answer is correct or wrong,
I have watched couple of videos of yours and they are awesome. keep up the good work Sir,
Excellent Damian,
Feeling glad that I am able to clear the VLSM concept.
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@@networkingplus where can I found full answer
Hi. How about this answer: Network id is 192.168.5.48, subnetmask is 255.255.255.240, IP range 192.168.5.49 - 192.168.5.62, Broadcast id is 192.168.5.63...... Could you take a look and let me know if that answer is right as well?
thanku you soo much it was very helpfull to me
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Man tank you i was near to loosing my mind of not understanding, thank you very much...
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Asanti sana! That means 'thank you very much in Swahili' For this nice tutorial...
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the beautiful birdsong in the background makes this video feel better to watch haha
Thanks for your feedback. I live in village and this birds always come to me while recording videos. 😀
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Nice to hear
Thanks for your feedback. I will try to make one.
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smart and specially said
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Really nice...…….. good work.
I always struggle to assign this to topology.
Could you please make a video and assign this ip networks to topology
Thanks for your feedback. I will try to make.
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Your videos always super. Please provide vlan topic and configuration.
Thanks for your feedback.
I will work on this topic.
TECHNICAL TECH thanks for you reply but please provide vlan quickly.
One Mistake in the tutorial is that 2^12=4096 not 4086 Anyway great Tutorial
Network 4: Network Id: 192.168.5.208 Broadcast Id: 192.168.5.223 Subnet mask: 255.255.255.240 CIDR: /28
superb
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is there any one for response?
Given a network of 159.82.145.227 /22, subnet the network into
atleast 4 networks of 64 Addresses, 2 networks of 128 Addresses and 2 networks of at least 256 Addresses.
i watched your video but still cannot the solution for these question. Please help me out
please make a vedio on surprise test for this vedio or give explanation plz
Try to solve it and let me know.
I will try to clear your doubt.
Hi Sir thanks for tutorial I really appreciate it. but I got little confused why you put 6 usable IP address for just 3 routers? As far as I know every router just need 1 IP address. Thanks
Thanks for your feedback.
On every router, we have used two ports (S0/1 and S0/2). So for three routers, we need 6 IP Addresses. Hope it is clear.
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can u please do a vlsm on class B and C? ILL BE GREATFULL
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Please solve this..
consider an ip address 157.110.88.0 we need 30 subnet in such a way that each subnet may host upto 64 host.
1.how many unuseable subnet here
2. subnet mask
3.default mask
4.what will be the address of 47th host of 25th subnet??
the ip adresses written in red at the end of the video are wrong
Hi. How about this answer: Network id is 192.168.5.48, subnetmask is 255.255.255.240, IP range 192.168.5.49 - 192.168.5.62, Broadcast id is 192.168.5.63...... Could you take a look and let me know if that answer is right as well?
At 5:18, the 12th bit value would be 4096 and not 4086...
Thanks for your feedback. Its a typing mistake.
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Is my math wrong? I calculate 2 to the power of 12 to be 4096?
CAN YOU HELP ME
Why the bit borrow started from fourth octate why not third octate please tell me
you have to check the requirement of ip address.
5.193 - 5.208 ? Is it correct?
Can you make video on supernetting
You explain very well
If possible make a video on it
Thanks for your feedback. I will try.
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can someone helped me why it turned 15?
The part at 6:22 is where it is at.
Basically, when you are trying to find the network side, you will do addition of the 2^n values.
For example:
In this octect, we have 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1, all of which are 2^n results up to n = 8.
The IP 172.31.0.0 has 4 octects.
The 172 and 31 are the network octects and the two 0 values are the host octects as we are in class B.
When we are calculating the host, we calculate it from right to left. To obtain the hosts, we had to use the 2^n, and with that, we have had to use up 12 bits from the right side.
To obtain the 15, what we are doing is counting from the right side to the left side with the values of 2^n for all that we require to obtain the hosts.
So the following would happen:
8+4+2+1
Adding that is equal to 15, and hence is how we get to 15.
I hope this makes sense :D
Sir how u assigned final ip address to router 1 and router 3, router 2 and router 3 I think it's starts with 172.31.20.132/30 for router 1 and 3, 172.31.20.136/30 for router 2 and 3
I didn't understand your question. can you explain in brief.
@@networkingplus sir 5th n/w Ip range is 172.31.20.132 to 172.31.20.135 and 6th n/w Ip range is 172.31.20.136 to 172.231.20.139 so ip address for router 1 and 3 is 172.31.20.132/30 and router 2 and 3 is 172.31.20.136 but u have return in video 172.31.20.31/30 for router 1 and 3 and 172.31.20.135/30 for router 2 and 3
Thanks for the correction. It is a typing mistake.
woow
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isn't it 4096? 2^12?
yes
hi there,im reetesh ,after long time..nice to visit u again .
the answer is : usuable ip range 192.168.5.209 to 192.168.5.222 and n/w id is 192.168.5.208,broadcast ip is 192.168.5.223
Thanks brother.
Network 4 id 192.168.5.208, RANGE 192.168.5.209-192.168.5.222 broadcast 192.168.5.223 cidr /28 subnet 255.255.255.240
Very Good.
12:50 if we add 2+4-8+16+33+64+128 then we will get 256 not 127
This is not wrong.
There is 7 bits remaining in the host section that is 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 127
Hey bro u have a minute mistake it will start from 2 to the power 0 to 2 to the power 7 but u have started from 2 to the power 1 and we add all those no. It will come more than 255 that isn't possible so please change it ......thanks 😊😊
Edit:- a minute mistake led to all your solution get wrong so please delete this video or edit it I will suggest to delete as a minute mistake made all your video wrong
Thanks for your feedback but the video is right.
The number mentioned in the grey box are the value of binary bits. and the number I have mentioned in the green box are to find the number hosts. as we know that the formula for finding the number of host is 2^n (n= number of bits reserve for the host) e.g. If there is 7 bits reserved for the host = 2^7=128.
I will recommend you to watch my subnetting video first. Here is the link: czcams.com/video/q7wNcYliJ1Q/video.html
I also noted the same mistake 2^12 is 4096, not 4086 thus the 172.13.15 is wrong,I thought the broadcast id was supposed to be 172.13.16 instead
14 host want 14 IP address
The question is to find the subnet mask, CIDR, Network ID, Broadcast ID.
Still confuse , hurts lol
srm se ho
@@anshchandrakar9257 I think this is an English forum mate
what is your doubt, Let me know. I will try to help you.
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