Transcript Ex. 02 - IP Subnet

Exercise: IPv4 subnetting
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Task 1
Given is an IP network with address
194.141.0.0:
Divide this network into 8 subnets.
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Solution
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This network is class C because the
starting bits are 110.
In binary form:
11000010. 10001101.00000000.00000000
The first 3 octets represent the network
address. The fourth octet is for subnet and host
address.
We need a subnet mask with 3 bits ones,
because 8 = 23 :
11111111.11111111.11111111.11100000
In decimal form: 255.255.255.224
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Comments
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We use this mask in the corresponding field of
the IP configuration.
The available subnets have the following
addresses:
194.141.0.0
194.141.0.1
194.141.0.2
194.141.0.3
194.141.0.4
194.141.0.5
194.141.0.6
194.141.0.7
Can not be
used.
In fact we have
6 real subnets.
Hosts with address 0 or
255 can not be used.
Therefor each subnet
contains up to
256/8 - 2 = 30 hosts.
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Task 2
Given is an IP network with address 162.251.0.0:
Divide this network into 32 subnets.
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Solution
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The subnet mask is
11111111.11111111.11111000.00000000
in decimal: 255.255.248.0
30 subnets with up to 2046 hosts each.
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Task 3
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Given is a subnet mask: 255.254.0.0
Which class is the network?
How many real subnets?
How many real hosts may contain each subnet?
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Solution
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This is a subnet mask for network class A.
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There are 27 – 2 = 126 real subnets.
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Each subnet may has up to
217 – 2 = 131070 real hosts.
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