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Once you have mathematically enabled the bits that will give you the number of networks you are looking for, the next step is from right to left to turn off any bits that are not being used up to the last bit you have enabled Once you have done that, you count the number of bits you have worked with from right to left In the example, you will simply need to disable the first bit (far right side), and then count from right to left the number of bits that you have worked with In the example, there are two bits that have been manipulated, so the number of bits that you need to create two networks is two You can see the work in the following table:
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8 128 7 64 6 32 5 16 4 8 3 4 2 2 1 1 1 0
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You can mathematically verify that you have the correct number of bits you want to work with by using this formula: 2masked bits 2 = number of networks It was decided that two bits were needed to mask the subnet mask to do the subnetting; in checking your work by filling in the formula, you should have 22 2 = 2 networks Having two networks was your goal, so you are on track! Now that you know that two bits must be masked to create the two subnets, the next step is to look at your IP range of 10000 and ask yourself, What is the default subnet mask of this address You should come up with 255000, because it is a class A address Once you know the default subnet mask, the next step is to break the subnet mask down into binary A default class A subnet mask in binary looks like the following:
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255 11111111 0 00000000 0 00000000 0 00000000
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Once the subnet mask has been written out in binary, this will be the work area to calculate the new addresses of the two subnets The first thing you need to do is steal, or mask, two bits from the host ID portion of the subnet mask and make them network bits by setting them to a 1 state Remember, the reason for taking
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two bits is because of your calculation earlier Because the last three octets of this subnet mask represent the host bits of a class A address, work from left to right and mask two additional bits to get the new binary table that follows:
Decimal binary
255 11111111 0 11000000 0 00000000 0 00000000
Notice that the two bits that are now enabled are in the second octet Also notice that you always work from left to right, except for that little cheat method you saw earlier for calculating the number of bits required to mask That cheat is just to get the number of bits to work with but once you actually work with the bits, you always work from left to right The next thing you want to do is convert all those octets to decimal values again; then you will have the new subnet mask of the two networks (subnets) we are building If you convert the new binary value in the preceding table of 11111111110000000000000000000000, you should get 25519200 as the new subnet mask of your two new subnets Write that number down because it will be needed later The next step is to calculate the IP ranges for the two different networks, but before that, you should be aware that there are five pieces of information you should know about each network when your calculations are over These five pieces of information are
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