IP Network Address Binary Equivalent

TABLE 2 . 1 1 Summary Example
IP Network Address Binary Equivalent
172.16.100.0 10101100.0001000.01100100.0
172.16.101.0 10101100.0001000.01100101.0
172.16.102.0 10101100.0001000.01100110.0
172.16.103.0 10101100.0001000.01100111.0
172.16.104.0 10101100.0001000.01101000.0
172.16.105.0 10101100.0001000.01101001.0
172.16.106.0 10101100.0001000.01101010.0
2. Examine the table to determine the maximum number of bits (starting from the left) that
all of the addresses have in common. (Where they are lined up, we boldfaced them to make
them easier for you to see.) The number of common bits is the prefix length for the summarized
address (/20).
In this example, we can see from the table that all of the addresses have the first 20 bits in
common. The decimal equivalent of these first 20 bits is 172.16.96.0. So, we can write our
new summarized address as 172.16.96.0/20. If we were to later add a network 172.16.98.0,
it would need to be behind the router summarizing this address space. If we didn’t, it could
cause problems.
Another way that works more quickly, once you get it down, is to avoid binary and do just
about everything in decimal, as outlined in the following steps:
1. Count the number of networks you’re trying to summarize, calling the first octet that
changes value among the networks the interesting octet. Make sure you include any networks
that are skipped, just in case you’re not dealing with a continuous run of networks.
Often, the quickest way to accomplish this step is to subtract the value of the interesting
octet of the first network from that of the last network and add 1 to the result, just to make
sure both the first and last networks are counted.
2. Recalling the series of numbers that stems from the powers of 2 (0, 2, 4, 8, and so on), see
if there is a value that matches the number of networks you are trying to summarize. Otherwise,
choose the next highest block size. In this case, we are trying to summarize seven subnets,
so we need to look at the block size of 8.
3. Starting with 0, count up by the block size until you reach or exceed the starting value in
the interesting octet. Alternatively, divide the starting value of the interesting octet by the
block size and ignore the remainder. Then, multiply the result by the block size. In this case,
we get (100 div 8) × 8 = 96, where div is the operand for integer division, which ignores any
remainder from the operation. This result is the closest block boundary in the interesting
octet without going over.
4. Confirm that starting with the boundary from the previous step will include all networks
that we need to summarize. In our case, a block size of 8 starting with 96 will
extend only to a value of 103 in the interesting octet, not enough to cover the end of
our run of networks. Simply increasing the block size to the next power of two, or 16,
in this case, will always cover our needs. The thing we must confirm is that 96 is still
a power of 16, as it was for 8. In this case it is, but if our starting point from step 3 had
been 88 or 104, we would have had to adjust down to 80 or 96, respectively, to accommodate
a block size of 16.
5. Subtract the block size from our magic number, 256, and get 240, in this case. Combine this
information with the result of step 3, and we have our summary address and mask. The 96
contributes to the value of the interesting octet of the summary address, 172.16.96.0, and
the 240 is the value of the same octet in the mask, 255.255.240.0, or /20.
Okay, this is confusing, we know. This is why we’re going to give you three more examples.
The preceding five steps of the “faster” method may take a while to read, but once you get them
down, the process of finding an answer without breaking out the binary is incredibly efficient.