3 Ways To Calculate Network And Broadcast Address

Subnet masks can be 0, 128, 192, 224, 240, 248, 252, 254 and 255. The number of bits used for subnetting (n) to their corresponding subnet mask is as follows: 0=0, 128=1, 192=2, 224=3, 240=4, 248=5, 252=6, 254=7, and 255=8. Subnet mask 255 is default, so it’ll not be considered for subnet masking. For example: Let’s assume the IP address is 210. 1. 1. 100 and Subnet mask is 255. 255. 255. 224. The total bits= Tb = 8. The number of bits used for subnetting for subnet mask 224 is 3.

Using the example above, n=3. The number of bits left for host is (m) = 8 - 3 = 5. 5 is the number of bits you have left to host.

Using the example above, n=3. The number of bits left for host is (m) = 8 - 3 = 5. 5 is the number of bits you have left to host.

In our example, the number of subnets is 2n = 23 = 8. 8 is the total number of subnets.

In our example, the value of last bit used for subnet masking is Δ = 25 = 32. The value of the last bit used is 32.

In our example, the value of last bit used for subnet masking is Δ = 25 = 32. The value of the last bit used is 32.

The 8 subnets (as calculated in previous step) are shown above. Each of them has 32 addresses.

Our example IP address 210. 1. 1. 100 falls in the 210. 1. 1. 96 - 210. 1. 1. 127 subnet (see the previous step table). So 210. 1. 1. 96 is network address and 210. 1. 1. 127 is broadcast address.

Example: If the bit-length prefix is 27, then write it as 8 + 8 + 8 + 3 . Example: If bit-length prefix is 12, then write it as 8 + 4 + 0 + 0 . Example: Default bit-length prefix is 32, then write it as 8 + 8 + 8 + 8.

Using another example, the IP address is 170. 1. 0. 0/26 . Using above table, you can write the bit-length prefix 26 as 8+8+8+2. Using the chart above, this converts to 225. 225. 225. 192. Now the IP address is 170. 1. 0. 0 and subnet mask in quad-dotted decimal format is 255. 255. 255. 192 .

For subnet mask 255 is default, so it’ll not consider for subnet masking. From the previous step, you got IP address = 170. 1. 0. 0 and Sub-net mask = 255. 255. 255. 192 Total bits = Tb = 8 Number of bits used for subnetting = n. As the subnet mask = 192, its corresponding number of bits used for Subnetting is 2 from above table.

In our example, the number of bits used for subnetting (n) is 2. So the number of bits left for host is m = 8 - 2 = 6. The total bits left for the host is 6.

In our example, the number of bits used for subnetting (n) is 2. So the number of bits left for host is m = 8 - 2 = 6. The total bits left for the host is 6.

In our example, the number of subnets = 22 = 4. The total number of subnets is 4.

In our example, the value of last bit used for subnet masking = Δ = 26 = 64. The value of the last bit used for subnet masking is 64.

In our example, the last value used for subnet masking is 64. This produces 4 subnets with 64 addresses.

Our example IP address is 170. 1. 0. 0. So 170. 1. 0. 0 is network address and 170. 1. 0. 63 is broadcast address.