342 Chapter 5: Network Protocols
So, the mask must have at least 8 host bits because 2
7
Ч
128 is not enough and 2
8
Ч
256 is more
than enough for numbering 200 hosts in a subnet. The mask must have at least 7 subnet bits,
likewise, because 2
7
is the smallest power of 2 that is larger than 100, which is the required
number of subnets. The first 16 bits in the mask must be binary 1 because a Class B network
(172.16.0.0) is used. Figure 5-50 diagrams the possibilities.
Figure 5-50
Subnet Mask Options for Scenario 5-2
The only bit position in which a decision can be made is the 24th bit, shown with an x in Figure
5-50. That leaves two mask possibilities: 255.255.254.0 and 255.255.255.0. This sample shows
the 255.255.254.0 mask because 255.255.255.0 is more intuitive.
Answers to Task 2 for Scenario 5-2
To choose a mask and pick enough subnets to use for the original topology illustrated in Figure
5-44, a review of the longer binary algorithm and shortcut algorithm for deriving subnet
numbers is required. To review, subnet numbers have the network number binary value in the
network portion of the subnet numbers and have all binary 0s in the host bits. The bits that vary
from subnet to subnet are the subnet bits--in other words, you are numbering different subnets
in the subnet field.
Valid subnets with mask 255.255.254.0 are as follows:
172.16.0.0 (zero subnet)
172.16.2.0
172.16.4.0
172.16.6.0
.
.
.
172.16.252.0
172.16.254.0 (broadcast subnet)
1111111111111111
Network Bits
1111111 X
Minimum
Subnet
Bits
00000000
Minimum
Host
Bits
ch05.fm Page 342 Monday, March 20, 2000 5:06 PM