Subnet Examples

Subnet Examples

In the following examples, determine if the address/subnet pair is legal. If it is legal, determine the network number and the range of host addresses for that network. Also determine for the mask, the number of available networks and available hosts per network.

1.   IP address = 193.144.233.130

Subnet mask 5 255.255.255.192

For a Class C address, we only need to look at the last octet of the address and the mask.

130 = 1000 0010

192 = 1100 0000

This is a legal pair because neither the subnet nor the host is all zeros or all ones.

Network equals 193.144.233.128 because the mask selects the upper two bits of the address (130) and the rest of the bits are set to zero to identify the network.

Range of hosts = 193.144.233.129—193.144.233.190.

The host portion (last six bits) can have values ranging from 000001 to 111110 (remember they can’t be all zeros or all ones). Add in the subnet portion, which is the upper two bits of the address (in this case, 1 0), and you have 10 000001 to 10 111110 for the host addresses.

From Table 2-6, the number of available networks is 2 and the number of hosts is 62.

2.   IP address = 156.26.30.60

Subnet Mask = 255.255.255.0

This is relatively easy because the entire third octet is used for the subnet and the entire fourth octet is used for the host. This is a legal pair because neither the subnet nor the host is all zeros or all ones.

Network = 156.26.30.0

Range of hosts = 156.26.30.1—156.26.30.254

From Table 2-5, the number of networks is 254 and the number of hosts is 254.

3.   IP address = 199.200.201.50

Mask = 255.255.255.128

This is illegal because the subnet mask only borrows 1 bit from the host and that bit has to be either zero or one.

4.   IP address = 191.200.201.50

Mask = 255.255.255.128

This is a legal pair because the address is Class B and we are borrowing 9 bits from the host portion.

Network = 191.200.201.0

Range of hosts = 191.200.201.1—191.200.201.126

From Table 2-5, the number of networks is 510 and the number of hosts is 126.

Subnetting can be viewed as creating a three-part hierarchical address. The network portion of the address can be found by applying the standard subnet mask to the IP address (refer to Figure 2-3). The subnet is determined from the bits “borrowed” from the host portion and the host number is simply those bits that are left over. For an example, we will examine the Class B address/mask pair

144.223.0.0/255.255.255.0

and determine the network number, the subnetwork numbers, and the range of host numbers. The network number is found by applying the standard Class B 16-bit subnet mask, which yields the network

144.223.0.0

The subnet is the entire third octet, so the 254 subnets are

144.223.1.0

144.223.2.0

.

.

.

144.223.254.0

and the range of hosts for each subnet is 1 to 254. Now let’s try a bit more complicated example. Consider the address/mask pair

144.223.0.0/255.255.255.224

The network number is still 144.223.0.0. The subnet mask borrows 11 bits from the host portion of the address. The first 8 bits borrowed include the entire third octet, which has a value of 0 to 255. The 3 bits borrowed from the third octet have the values

00000000=0

00100000=32

01000000=64

01100000=96

10000000=128

10100000=160

11000000=192

11100000=224

Why are the values 0 (all zeros) and 255 (all ones) for the third octet, and 0 (all zeros) and 224 (all ones) from the fourth octet included? The third octet can be 0 if the 3 bits in the fourth octet are not zero. The third octet can also be all ones if the 3 bits in the fourth octet are not all ones. The 3 bits in the fourth octet can be all zeros if the third octet is not all zeros, and the 3 bits from the fourth octet can be all ones if the third octet is not all ones. In other words, the 11 subnet bits cannot be all zeros or all ones. Therefore, the range of subnet numbers is

144.223.0.32

144.223.0.64

.

.

.

144.223.0.224

144.223.1.0

144.223.1.32

.

.

.

144.223.255.0

.

.

.

144.223.255.192

Determining the range of host addresses for each subnet requires more effort. The bit pattern for the fourth octet of network 144.223.0.32 is

001 hhhhh

where hhhhh represents the host number, which cannot be all zeros or all ones. Therefore, the first legal host number is 00001, making the fourth octet

00100001 = 33

so the first host address is

144.223.0.33

and the last legal host bit pattern for the fourth octet is

00111110 = 62

which gives the range of hosts’ addresses for the first subnet as

144.223.0.33—144.223.0.62

The broadcast address for each subnet is found by setting all the bits in the host portion to 1. The broadcast address for subnet 144.223.0.32 is determined by setting the last 5 bits of the fourth octet to 1 yielding

00111111 = 63

Putting it all together gives us the broadcast address

144.223.0.63

5.   Determine all the subnet numbers for the address/mask pair 193.128.55.0/255.255.255.240. Also determine the range of host addresses and the broadcast address for the fourth subnet.

Network

Hosts

193.128.55.0

1—14 (If IP subnet-zero is used)

193.128.55.16

17—30

193.128.55.32

33—46

193.128.55.48

49—62, Broadcast address = 193.128.55.63

193.128.55.64

65—78

193.128.55.80

81—94

193.128.55.96

97—110

193.128.55.112

113—126

193.128.55.128

129—142

193.128.55.144

145—158

193.128.55.160

161—174

193.128.55.176

177—190

193.128.55.192

193—206

193.128.55.208

209—222

193.128.55.224

225—238

193.128.55.240

241—254