This video tutorial deserves praise. It's so clear and easy to understand especially with the method that is shown step by step throughout. Very nice! :)
After days and 4hrs in a row I can't understand what even the IP subletting is thanks man every IP subletting can communicate with his IP that's make sense 100/200
In order to get no:of host address for an ip 2n - 2 In your example 2 power of 5 and - 2 = 32-2 == 30 So we need to take 5 bits from right to left of the octate For no:of subnets 2n 2 power of 3 = 8 So in order to get 8 subnets we need to take 3 bits from left to right of the octate
00:08 Learn how to subnet a network based on host requirements. 01:42 Subnetting a Class C IP address into 30 hosts per subnet 03:23 Understanding subnet mask and subnet generators. 04:55 Representation of binary number 30 requires 5 bits. 06:30 Identified subnet generator and octet position. 08:07 Generate new subnet mask and start subnet generation. 09:41 Understanding subnetting and IP address range calculation 11:21 Understanding network and broadcast addresses Crafted by Merlin AI.
Sir when will you upload part 2 of subnetting🙃 I have an exam in a day and I was relying on your lectures completely. Sir pls upload soon😭🙏 btw u are God sent😍😍❤❤
After subnetting the class C IP address 216.21.5.0/24 into a range of IP addresses, all the new subnets have 255.255.255.224 or /27 subnet masks. How can they fall into different networks despite having the same subnet mask which is the /27 subnet mask? Is there someone who can help me with this doubt, please!!
The ans is 255.255.255.224 / 27 because there are 27 times 1 in the binary representation of the subnet mask that's why in classless notation there is 27
Hello sir pls solve this question " An Organization is granted 130.56.0.0/16 .The administrator wants to create 1024 subnets. a)Find First and Last address of first subnet. b) Find first and last addresses of the last subnet. This Question was ask in UGC NET EXAM. pls help.
To solve this question, we first need to determine how many bits will be used for subnetting. Since the administrator wants to create 1024 subnets, we need at least 10 bits (2^10 = 1024). Now, let's break down the steps to find the first and last addresses of the first subnet and the last subnet: 1. **Determine the subnet mask:** Since we're starting with a /16 network (130.56.0.0/16), and we need 10 bits for subnetting, the new subnet mask will be /26 (16 bits for the original network + 10 bits for subnetting = 26 bits total). 2. **Calculate the subnet size:** With a /26 subnet mask, each subnet will have 2^(32-26) = 64 IP addresses. However, we'll have to skip two addresses in each subnet (network address and broadcast address), leaving us with 62 usable IP addresses per subnet. 3. **Find the first and last addresses of the first subnet:** - First address: The first subnet will start at 130.56.0.0 (network address) + 1 = 130.56.0.1 - Last address: The last usable address in the first subnet will be 130.56.0.0 (network address) + 62 = 130.56.0.62 4. **Find the first and last addresses of the last subnet:** - To find the last subnet, we need to determine the network address of the last subnet and then calculate the first and last addresses accordingly. - Since there are 1024 subnets and each subnet has 64 addresses, the last subnet will start at 130.56.0.0 (network address of the original network) + 1024 * 64 (addresses per subnet) = 130.56.64.0 - First address of the last subnet: 130.56.64.0 (network address) + 1 = 130.56.64.1 - Last address of the last subnet: 130.56.64.0 (network address) + 62 = 130.56.64.62 So, the answers are: - First subnet: First address = 130.56.0.1, Last address = 130.56.0.62 - Last subnet: First address = 130.56.64.1, Last address = 130.56.64.62
there are some predefined ranges if the starting range of a network is from 0-127 that means it belongs to Class A same goes for the other ones 128-191 (Class B) 192- 223 (Class C) 224- 239 (Class D) 240- 255 (Class E)
4:46, you said there must be 5 bits. but 16+8+4+2 = 30, then why that 0 is necessary, as you said 5 bits are must, is there any firm reason for that necessity ?
Thanks!
Sir.... This... Is very useful for many students.... You did a great job sir... God bless you sir.. And thank you❤
This video tutorial deserves praise. It's so clear and easy to understand especially with the method that is shown step by step throughout. Very nice! :)
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After days and 4hrs in a row I can't understand what even the IP subletting is thanks man every IP subletting can communicate with his IP that's make sense 100/200
Wow this is the best content!!!! So easily understood the whole subnetting theory
Awesome content with simple explanation. Lots of love dear❣❣
In order to get no:of host address for an ip 2n - 2 In your example
2 power of 5 and - 2 = 32-2 == 30
So we need to take 5 bits from right to left of the octate
For no:of subnets
2n
2 power of 3 = 8
So in order to get 8 subnets we need to take 3 bits from left to right of the octate
00:08 Learn how to subnet a network based on host requirements.
01:42 Subnetting a Class C IP address into 30 hosts per subnet
03:23 Understanding subnet mask and subnet generators.
04:55 Representation of binary number 30 requires 5 bits.
06:30 Identified subnet generator and octet position.
08:07 Generate new subnet mask and start subnet generation.
09:41 Understanding subnetting and IP address range calculation
11:21 Understanding network and broadcast addresses
Crafted by Merlin AI.
thank you may god bless you.i dont know how to thank for your explanations.
One of the favourite ❤️❤️❤️ channel... 🌿🌿🌿🌿
Complex topics with simple explanation 🖱️💯😊😊❤️
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han bhai aaj hai 😂😂😂😂😂😂
@@DhruvParekh_ 😂👍
Great video sir! Clearly understood the topic😍
Balu ser 💥💥
Thank you for your video. It is very indeed very clear and easy to understand!
I actually watched this twice and trust me the 2nd time it was different (😯)
Thank you so much sir, this video is really helpful for my tomorrow's exam🙏❤
I failed in my exam again. Now total failed paper is 18. I tried by unable to pass. your video r a great relief.
Really loved this one 🤩,
Kal mera etp hai cse307 ka
Aur abhi yeh dekh rha hu as a last topic for revision
the best explanation on the world !!!!
two extra addresses are used as network address and broadcast address , that is why every subnet has 32 adresses
Clean and clear explanation
Thank you so much❤❤❤❤
Thanks, very good explanation, it makes sense, now I understand it clearly.
Thanks a lot...Your explanation makes it easy to understand
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Thanks for the amazing lecture💗💗
Superb. Very easy to understand
nicely explained and easy to understand the subnetting concept...thnak you
Very good explanation
crystal clear... thankyou so much sir
wonderful, your videos are awesome to understand. Thank you.
Thank you very much sir 🙏
clear and concise.
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Well done👍
Good explanation
fantastic.
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Clear explanation... Thank you!!
Please upload more videos on subnetting
Great explanation , thank you ❤
Awesome and simple video to understand subnetting ❤❤
Great lecture ❤
very good
Saar very good video very good whole playlist thanku with your help i was able to give my 3rd sem Computer network exam❤
having an exam in 5h, thank you sir
Superb explanations !! Hats off and THANK YOU
Superb
Can we use this trick for classless addressing
god level teaching skills
Lol....
Teaching skills are very good but this sentence is wrong
Sir when will you upload part 2 of subnetting🙃 I have an exam in a day and I was relying on your lectures completely. Sir pls upload soon😭🙏 btw u are God sent😍😍❤❤
Soja bro
great sir geo
Wow . Awesome explanation
My Lecturer😊😊
Wonder Full
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After subnetting the class C IP address 216.21.5.0/24 into a range of IP addresses, all the new subnets have 255.255.255.224 or /27 subnet masks. How can they fall into different networks despite having the same subnet mask which is the /27 subnet mask? Is there someone who can help me with this doubt, please!!
The ans is 255.255.255.224 / 27 because there are 27 times 1 in the binary representation of the subnet mask that's why in classless notation there is 27
Hello sir pls solve this question " An Organization is granted 130.56.0.0/16 .The administrator wants to create 1024 subnets. a)Find First and Last address of first subnet. b) Find first and last addresses of the last subnet.
This Question was ask in UGC NET EXAM. pls help.
a)136.56.0-130.56.7.0
b)136.56.248-136.56.255.0
To solve this question, we first need to determine how many bits will be used for subnetting. Since the administrator wants to create 1024 subnets, we need at least 10 bits (2^10 = 1024).
Now, let's break down the steps to find the first and last addresses of the first subnet and the last subnet:
1. **Determine the subnet mask:**
Since we're starting with a /16 network (130.56.0.0/16), and we need 10 bits for subnetting, the new subnet mask will be /26 (16 bits for the original network + 10 bits for subnetting = 26 bits total).
2. **Calculate the subnet size:**
With a /26 subnet mask, each subnet will have 2^(32-26) = 64 IP addresses. However, we'll have to skip two addresses in each subnet (network address and broadcast address), leaving us with 62 usable IP addresses per subnet.
3. **Find the first and last addresses of the first subnet:**
- First address: The first subnet will start at 130.56.0.0 (network address) + 1 = 130.56.0.1
- Last address: The last usable address in the first subnet will be 130.56.0.0 (network address) + 62 = 130.56.0.62
4. **Find the first and last addresses of the last subnet:**
- To find the last subnet, we need to determine the network address of the last subnet and then calculate the first and last addresses accordingly.
- Since there are 1024 subnets and each subnet has 64 addresses, the last subnet will start at 130.56.0.0 (network address of the original network) + 1024 * 64 (addresses per subnet) = 130.56.64.0
- First address of the last subnet: 130.56.64.0 (network address) + 1 = 130.56.64.1
- Last address of the last subnet: 130.56.64.0 (network address) + 62 = 130.56.64.62
So, the answers are:
- First subnet: First address = 130.56.0.1, Last address = 130.56.0.62
- Last subnet: First address = 130.56.64.1, Last address = 130.56.64.62
Thank uuuu
nice
Plz upload power electronics videos
if the binary number of "total host number is start with zero", then can we generate subnet's ? please tell me sir!
Tou are telling the classless subnetting but you are using the classes of ip.
Anybody answer
Sir plz can you give homework question on this video too 🙏
The reservation process is from left to right i guess. And now here it is from right to left
IF anyone is aware of this topic,I have one doubt EX:- For my company i need three branch and in each branch i need to have 50 hosts ? Can you !.
How did you get 224?
I understand the 27 but not the 224
Oh never mind I got it, thank you
Shouldn't the subnet generator be 64?
Why 5 bits specifically?
5 bits are the least number of bits that represent 30 in binary
For a class c IP address 4 octets will be there ..but in your example only 3 octets are filled.216.215.0 ..why so?
It is 216.21.5.0
❤️❤️❤️❤️
Why is there 224 in the 4th octet of the new subnet mask?
value of 11100000 is 128+64+32=224 for the 4th octet
how do you identify that it is class C- type
there are some predefined ranges
if the starting range of a network is from 0-127 that means it belongs to Class A
same goes for the other ones
128-191 (Class B)
192- 223 (Class C)
224- 239 (Class D)
240- 255 (Class E)
Why didn't I watch this before my exam? Because I didn't have time. Or did I?
To tu exam bad ku dekh Raha he
@@ketansharma1411to tu exam ke baad usko reply kyu kr rha h?
On a class B network, how many subnets are available with a subnet mask of 248? write if anybody no the answer with explanation
216.215.0 kaha se aaya?
💖🙏
Wooow
Behave your self
😘
bhai mujhe kuch jhaat nhi samja
4:46, you said there must be 5 bits. but 16+8+4+2 = 30, then why that 0 is necessary, as you said 5 bits are must, is there any firm reason for that necessity ?
Huh. 30 is 11110, which is 5 bits
Do you know @nesoacademy that I am from JHAATU COLLEGE?😢