As we mentioned in last week's
article, having a good understanding of TCP/IP is critical to being able to
administer a Windows 2000 network. In particular, because Windows 2000 sites
are based on subnets, you need to know how TCP/IP subnetting works and how to
subnet your network with an extended, custom subnet mask. But, you will need to
know subnetting for more than just configuring Windows 2000 sites. Knowledge of
subnetting is important for configuring Windows 2000 to provide routing and
remote-access services, such as Virtual Private Networks (VPNs).
Last week, we established a foundation for discussing the calculation of subnet
masks by talking about binary numbers and a logical operation called ANDing.
This week, we are going to finish our two-part series by showing you an easy
technique for calculating subnets. The method I describe is one that I have
used successfully in the classroom for a number of years to teach subnetting.
There are a number of different ways to calculate subnets. If you already know
how to subnet, you may find the method I describe differs from the one you
know. You should use whatever method you are comfortable with to calculate
subnets. However, you should find that this method does have simplicity to
recommend it, unlike a number of the other methods you could use to calculate
subnets.
Let's take a look at an example of two hosts on a Class B network trying to
communicate with one another. Let's assume the IP address of the source host is
172.16.32.1/16 and the destination host address is 172.16.64.1/16. Because we
are using the default subnet mask of /16 or 255.255.0.0, 172.16.32.1/16 and
172.16.64.1.16 are on the same network. As long as there is no change in the
network portion of the address, the two hosts are on the same network. The
network portion of the address is determined by the subnet mask of 255.255.0.0,
which tells the TCP/IP stack that the first two octets, 172.16.0.0, represent
the network portion of the address.
However, what if we need more than one network, perhaps because we have remote
locations or we have more hosts than we can place on a single cable? It makes
sense to sub-divide the network into smaller networks. What if we needed to
create 6 subnets from our larger, single network? In a case like this, we would
need to extend the default subnet mask by borrowing some of the bits from the
host portion of the IP address. To create at least six subnets from the
172.16.0.0/16 network, we need to borrow at least 3 bits. So, our subnet mask
would be 255.255.224.0. In binary, the subnet mask would look like this:
11111111.11111111.11100000.00000000
As we learned in last week's article, the TCP/IP stack will AND the source and
the destination IP address with the subnet mask and compare the results. If the
results of the ANDing are the same, the two hosts are on the same network. If
the results of the ANDing are different, the two hosts are on different
networks.
Let's do the ANDing for 172.16.32.1/19 and 172.16.64.1/19. We can ignore the
first two octets, since they are identical for both addresses. The octet of
interest is the 3rd octet.
32 224 32
00100000 AND 11100000 = 00100000
64 224 64
01000000 AND 11100000 = 01000000
Given a subnet mask of 255.255.224.0, the two hosts are on different networks.
One way you can think about subnet masks is this: any time there is a change in
the bits used to represent the network portion of the address, you have a
separate network. In the case of a custom subnet mask of 224, we create at
least 6 networks. (I say at least because the actual number depends on the
hardware or the software. This is a rather technical issue that you needn't
concern yourself with right now. For the time being, we are going to assume
that neither the network nor the host portion of the address can be all 0's or
all 1's, as per the original 1985 standard for subnetting, RFC 950.)
Here are the possible network IDs in the 3rd octet for our 255.255.224 subnet
mask:
00000000 0 (Normally not allowed according to RFC950)
00100000 32 172.16.32.0/19
01000000 64 172.16.64.0/19
01100000 96 172.16.96.0/19
10000000 128 172.16.128.0/19
10100000 160 172.16.160.0/19
11000000 192 172.16.192.0/19
11100000 224 (Normally not allowed according to RFC950)
Remember, we are dealing only with the 1st 3 left-most bit positions. A subnet
mask of 224 "masks" off these bits to represent the network. For any two IP
addresses, if the values in these bit positions change, the...
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