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Chapter 3
IP Addressing
We now have a network of 172.16.10.0 with a subnet mask of
255.255.254.0, which can also be written as 172.16.10.0/23. Again, by
doing the math (2
9
2 = 510, since we have nine host bits), we see that we
now have 510 available IP addresses instead of 254.
Let's work through a VLSM design example, as depicted in Figure 3.2.
F I G U R E 3 . 2
VLSM design example
In our example, we have the following set of requirements for our net-
work addressing:
A server farm requires 300 IP addresses.
A user segment requires 200 IP addresses.
A serial link between two routers requires two IP addresses.
A hub segment interconnecting four routers requires four IP addresses.
We have been assigned the Class B network of 172.16.0.0.
T A B L E 3 . 5
VLSM Example 2
Decimal
255
255
254
0
Binary
11111111
11111111
11111110
00000000
Router Interconnections
(4 Addresses Needed)
Point to Point HDLC Link
(2 Addresses Needed)
Ethernet User Segment
(200 Addresses Needed)
Class B Address: 172.16.0.0
Server Farm
(300 Addresses Needed)
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