Pick two network address from the list below. Everybody must pick a unique network address and number of subnet. Please reply to this conference with your selection. Don’t select a network address that somebody else already selected. For each of them, I need the following information: The new subnet mask after the subnetting The following information for the first FOUR subnets: Subnet’s network address Subnet’s broadcast address Subnet’s range of available IP addresses The complete calculations on how you get to the answers. This is very important!! Your goal is to subnet them with as little subnet as possible but still meeting the requirement. In other word, maximize the number of hosts that is available for each subnet. These are the two I need help with. 1. Subnet 188.8.131.52/16 into 34 subnets 2. Subnet 184.108.40.206/8 into 210 subnets
Need it tonight please!
Thanks, XXXXX XXXXX like to wait but I do have a deadline to meet as well and will appreciate if you could find someone before it reaches.
Hi David, Can you please include the calculations as to how you arrived at the subnetting address net work address, subnet broadcast address and subnet's range of available IP addresses.
No, there is no particular method. Any method you use is fine, I just need the calculations included as a reference.
Please click the link below to download the modified document:
David I was hoping you show the calculations in figures instead of words. I attached this example for you to see.
I convert 220.127.116.11 into binary numbers:
199 . 0 . 0 . 0 11000111 . 00000000 . 00000000 . 00000000
Next, the "/8" at the end of the network means that the subnet mask consists of eight 1's followed by 0's. We know that subnet mask is 32 bits long, so there are eight 1's and twenty-four 0's:
11111111 . 00000000 . 00000000 . 00000000
Now, line up both binary numbers:
11000111 . 00000000 . 00000000 . 00000000 11111111 . 00000000 . 00000000 . 00000000
The part of the IP address where the subnet mask is 1 are called the "network" bits and the other is called the "host" bits.
To subnet a network means that we are going to "borrow" some bits from the "host" bits and use it as a "network" bits.
To calculate how many bits we need to "borrow", we need to find "which power of 2 is at least equal to or greater to the number of subnet needed".
Since we need to have 100 subnets, 2 to the power of 7 will fit our needs (2^7 = 128 and 128 is the smallest power of 2 that is greater than the number of our subnets).
So, we need to borrow 7 bits from the host bits and use it as network bits.
now we have the following:
11000111 . 00000000 . 00000000 . 00000000 -> IP Address 11111111 . 11111110 . 00000000 . 00000000 -> New Subnet Mask.
Note that the seven 0's are now converted into 1's.
So, to answer the first question.. what is the new subnet mask, we just convert the subnet mask from binary back to decimal number:
11111111 . 11111110 . 00000000 . 00000000
255 . 254 . 0 . 0
Now, for the first four subnets, we need to find the network address, broadcast address, and range of available IP addressed.
We know that the first 8 bits are the "network" bits, and the last 17 bits are the "host" bits. The 7 bits that we borrowed from the "host" bits and now used as part of the new "network" bits are called "subnet" bits.
Each of the 100 subnets that we will now have, we have a unique "subnet" bits assigned to them.. The easiest way is to just assign them 1, 2, 3, 4, 5, 6, 7, ... and so on.
So, the first subnet, will have the subnet bits that is equal to 1. Now, remember, this "1" is in decimal notation and will need to be converted into a 7 bits binary format: 0000001
So, for the first subnet, the first 15 bits will look like this:
11000111 . 0000001_ . ________ . ________
Now, some new rules.. - Network Address is one where the "host" bits are all 0's - Broadcast Address is one where the "host" bits are all 1's - The range of available IP addresses are the IP addresses between the two numbers above
So, the network address for the first subnet is:
11000111 . 00000010 . 00000000 . 00000000 or 18.104.22.168
and the broadcast address is:
11000111 . 00000011 . 11111111 . 11111111 or 22.214.171.124
So, if the network address is 126.96.36.199 and broadcast address is 188.8.131.52, then the available IP addresses must be: 184.108.40.206 until 220.127.116.11.
Now, repeat this for the other three subnet by specifying 2, 3, and 4 for the "subnet" bits, then report back what the network address, broadcast address, and range of available IP addresses for each of the three subnets.
Hello Daivid. I need your help with a question and will like to ask how can I add it as an attachment and send it to you.Please let me know when you get this.
Hello David,I don't know if you got my last message but I have a question that I need help with and was wondering if you are online to look at it for me. It is too large to add here, so if you can please send me a way I could