# Addressing Activities Network Classes 6.5.3.1 Classful Subnetting

```Addressing Activities
 6.5.4
Calculating network address from host IP and
 6.5.5
Calculating the number of Hosts (note you only
– Examples
 6.5.6
Network Fundamentals – Chapter 6
 6.5.7, 6.5.8
Packet Tracer
 6.5.3.2 to 6.5.3.5 VLSM example
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Network Classes
Class First octet
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6.5.3.1 Classful Subnetting
Network and Default subnet
Host bits
Hosts per
network
N.H.H.H /8
16,777,214
A
1 to 126
255.0.0.0
B
128 to 191 N.N.H.H /16 255.255.0.0
C
192 to 223 N.N.N.H /24 255.255.255.0
65,534
 Require 7 subnets, i.e. borrow 3 bits,
 Subnetmask: 24 + 3 bits = /27
2^3 = 8 subnets
which is 255.255.255.224
254
 This leaves 30 – 27 = 5 bits for host bits, 2^5 – 2 = 30 hosts
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Classful Subnetting
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Classful Subnetting
 A class A, B or C network is used with a custom
 The same mask is used on all interfaces making
the subnets all the same size
 What subnet mask should be
used?
 How many hosts can each
subnet connect?
 This is inefficient use of addresses
6
not used, limiting future growth
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 The number of bits borrowed is determined by
1. Number of subnets required
2. Max number of hosts on a subnet
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Subnetting a Subnet
 Using more than one subnet mask on a
network is called Variable Length Subnet
 Use one of the subnets for the WAN links
say 192.168.20.192 /27
 Require 2 host IPs per subnet
 VLSM is a more efficient way of assigning
2^2 – 2 = 2, so use /30 mask
 With VLSM the subnets do not have to be all
the same size, so fewer addresses are wasted
 Borrow 3 bits gives 2^3 = 8 subnets
 In particular, a WAN link only requires a subnet
with 2 addresses, so use /30 subnets for these
 Subnets 4, 5 and 7 are now available for
192.168.20.192/30
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Subnetting a Subnet
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 VLSM is a more efficient way of assigning
addresses than classful subnetting, but it can
only be used with modern routing protocols
 Determine how many addresses required for
each subnet,
number of hosts + router interface
 Work out largest subnet first, WAN links last
 How many bits must be borrowed for the
subnet?
Number of hosts = 2 ^ (no. of bits) - 2
 This determines the size of the subnet
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6.5.3.2 Scenario
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6.5.3.2 Scenario
 This scenario has the following requirements:
 Design an addressing scheme using the
 Classful subnetting will not work. Why not?
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2
Num of
Host
Subnet
Prefix
A
IP’s
58
bits
6
192.168.15.0
192.168.15.63
/26
B
26
5
192.168.15.64
192.168.15.95
/27
C
10
4
192.168.15.96
192.168.15.111
/28
D
10
4
192.168.15.112
192.168.15.127
/28
W1
2
2
192.168.15.128
192.168.15.131
/30
W2
2
2
192.168.15.132
192.168.15.135
/30
W3
2
2
192.168.15.136
192.168.15.139
/30 13
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