4/14/2023 0 Comments Cisco subnet mask table![]() Network ID Last Usable IP First Usable IP Subnet Mask Broadcast IP 255.255.240.0 172.16.0.1 Step 6: Calculate the Broadcast IP The Broadcast IP will always be the last IP address of a network. In IPv4, the maximum number in an octet is 255 before it will increment the preceding octet. When the hour hand rolls over the minute hand starts over at 0). ![]() ![]() Remember we always add to the 4th octet and as that fills up it will roll over to the 3rd octet (like how the minute hand on a clock increments until the hour hand rolls over. Now add 1 to the fourth octet and we get 172.16.0.1. Our network ID for this network is 172.16.0.0. The First Usable IP is always the IP address if you add 1 to the network ID. The subnet space we have to work with is 172.16.0.0 255.255.0.0.172.16.0.0 is the first IP address available so it is our network ID First Last Usable Usable Broadcast IP IP IP Network ID Subnet Mask 255.255.240.0 172.16.0.0 Step 5: Calculate the First Usable IP. Note: The 4th octet is because we only add up positions that have a 1 under them and the 4th octet only has O's under it's binary positions First Network | Usable ID IP Last Usable IP Broadcast IP Subnet Mask 255.255.240.0 Step 4: Calculate the Network ID The network ID is always the first IP address. We'll use it in a few steps 3rd Octet 4th Octet 4096 2048 1024 512256 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 128 64 32 16 8 421 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 4 1 Step 3: Now add up all the numbers with a 1 under it Positions 128 through 16 have a 1 under it so if we add all of those numbers together our subnet mask is 255.255.240.0. In this network, the magic number is 16 as it was the last number with a 1 under it. Hint: Remember the smallest number with a 1 under it is the Magic Number. The Green is the host portion of the IP address. Yellow is the network portion of the IP address. 4096 is bigger than 3000 so it gets a 1 under it but 2048 is smaller so it gets a 0. This has been done for you and is marked in Yellow. From left to right, put 1's in the table under the binary position until the host requirement is bigger than the number of hosts supported. I stopped at 4096 since we do not need to go larger for this lab but you can fill in the rest of the table on your own if desired. As you can see it just continues to double the number of hosts for each binary position. ![]() Do not confuse it with the binary positional numbers. This new row is the number of hosts that the binary position will support and it is only there for us to easily see the number of hosts, kinda like a cheatsheet for subnetting the 3rd octet. Also notice that the table has an extra row added to it on the top. The table includes the binary positions of the 3rd and 4th octet. Step 2: Size the Subnet Mask to meet the requirement We use the Binary Positional table below to size it correctly. R2 to R3 point to point network (2 hosts) Hint: Subnet the networks in the above order Step 1: Identify the largest required network In this lab, the largest host network is 3000 hosts R1 to R2 point to point network (2 hosts) e. Host requirements are located on the PT network diagram The required networks to subnet are: a.
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