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# Todd Lammle - page 14 / 54

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Practice Example #3C: 255.255.255.240 (/28) Let’s practice on another one: 192.168.10.0 = Network address 255.255.255.240 = Subnet mask Subnets? 240 is 11110000 in binary. 24 = 16. Hosts? 4 host bits, or 24 – 2 = 14.

• Valid subnets? 256 – 240 = 16. Start at 0. 0 + 16 = 16. 16 + 16 = 32. 32 + 16 = 48. 48 + 16 =

• 64.

64 + 16 = 80. 80 + 16 = 96. 96 + 16 = 112. 112 + 16 = 128. 128 + 16 = 144. 144 + 16 =

• 160.

160 + 16 = 176. 176 + 16 = 192. 192 + 16 = 208. 208 + 16 = 224. 224 + 16 = 240.

• Valid hosts?

To answer questions 4 and 5, check out the following table. It gives you the subnets, valid hosts, and broadcast addresses for each subnet. First, find the address of each subnet using the block size (increment). Second, find the broadcast address of each subnet increment (it’s always the number right before the next valid subnet), then just fill in the host addresses. The following table shows the available subnets, hosts, and broadcast addresses provided from a Class C 255.255.255.240 mask.

0

16 32 48

64

1

17 33 49

65

Subnet First host Last host

14 30 46 62 78

80 96 112 128 144 160 176 192 208 224 240 81 97 113 129 145 161 177 193 209 225 241 94 110 126 142 158 174 190 206 222 238 254

# Broadcast 15 31 47 63

79

95 111 127 143 159 175 191 207 223 239 255

Cisco has figured out the most people cannot count in sixteens and therefore have a hard time finding valid subnets, hosts, and broadcast addresses with the Class C 255.255.255.240 mask. You’d be wise to study this mask.

Practice Example #4C: 255.255.255.248 (/29) Let’s keep practicing: 192.168.10.0 = Network address

• 255.255.255.248

• Subnets? 248 in binary = 11111000. 25 = 32.

• Hosts? 23 – 2 = 6.

• Valid subnets? 256 – 248 = 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120,

128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, and 248.