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 ?
