In a /19, the line is going to be after 3 bits on the third octet. If we subnet it into/19 networks, how many subnets do we have, and how many hosts per subnet? In our example, we've been allocated a Class A 60.0.0.0/8. This time we're going to do a Class A on the third octet. We know that the next block starts at 80, so the broadcast address must be 60.15.10.79, and the valid hosts would be 60.15.10.65 to 60.15.10.78. Now, we know that the address blocks go up in multiples of 16 until we get to the closest subnet. So if the number we're subnetting at is 240, then 240 taken away from 256 is going to be 16. Then what we do with the magic number is we take that number away from 256. So 4 bits is going to be 128 plus 64 is 192, plus 32 is 224, plus 16 is going to be 240. A /28 is going to use the first 4 bits in the last octet. Especially if you were given a subnet mask in dotted decimal notation rather than in slash notation.Įven if you have been given a slash notation, you can convert it to dotted decimal first. In this is way, you can do it quite quickly in your head. And the range of valid host addresses falls between the network address and the broadcast address, 60.15.10.65 up to 60.15.10.78.Īnother way you can do it is by using the magic number. Our broadcast address here must be 60.15.10.79. The network address is going to be 60.15.10.64 and if I add 16 to that, the next network address will be 60.15.10.80. Therefore, the network address is going to go up in multiples of 16. 2 to the power of 20 works out a little over one million subnets.įor the IP address, 60.15.10.75/28, what are the network address, the broadcast address, and the range of valid IP addresses?įor this example, the line is after the 16 when we draw it out. We have 20 bits for the network addresses which is the difference between the default /8 and the /28 that we're using. So, that's 2 to the power of 4 equals 16 minus 2 and it gives us 14 hosts per network. We put the line in after 16, and we can see that we've got 4 bits for our host addressing. We were given a dotted decimal 255.255.255.240, which is the same as a subnet mask of /28. If we apply with the subnet mask 255.255.255.240, how many subnets do we have, and how many hosts per network? In this example, we'll do a Class A where we're going to subnet on the fourth octet. Then, we add 8 to the 136 to get the 144 and we know that that is the next address block. We find which block of 8 is closest to that, which is 136. In that example, our address was 135.15.10.138. 256 minus 248 gives you 8, and now you know that the network addresses are going to go up in blocks of 8. So 248 in this case, you take that away from 256. What you do with the magic number is you take the value in the octet that is being subnetted. So a /29, if we wrote that out in dotted decimal notation it's 255.255.255.248. This one is very handy if you've been given the subnet mask in dotted decimal notation rather than with a slash. You'll see this being cited in quite a few places on the internet. So that's 135.15.10.137 up to 135.15.10.142.Īnother popular way of calculating the network address, the broadcast address, and the host addresses is by using the magic number method. The valid addresses for our hosts fall between the network address and the broadcast address. Now, if the next network address is 144, then the broadcast address is 135.15.10.143. The next network address would be 135.15.10.144. The line is after the 8, so we add 8 to 136 and that is 144. We add those two together which is equal to 136. ![]() In the top right, we've got ones under 128 and 8. We're putting a line after the /29 and the line is going in after the 8. We've got the IP address and the subnet mask. 5 plus 8 gives us 13 bits and 2 to the power of 13 is going to allow a total of 8,192 subnets.įor the IP address, 135.15.10.138/29, what would be the network address, the broadcast address, and the range of valid IP addresses? It would give us 5 bits but since it is Class B, we've got those extra 8 bits. If it was a Class C, we would only have 5 bits for the network address because 29 minus 24 has a difference of 5. In a Class B /16 range we're going to have 13 bits for the network address. So, a /29 will give us 6 available hosts per network even if we're using a Class A, B, or C. This will allow us 6 hosts per network because 2 to the power of 3 is equal to 8 minus 2 gives us 6 hosts. We've got 32 bits in the address, 32 minus 29 gives us our 3 bits. ![]() If we subnet that into /29 subnets, we're going to have 3 bits for host addressing. For our first example, we've been allocated a Class B network with an IP address of 135.15.0.0/16.
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