Advanced Subnetting
Example 1:
Y
our ISP has assigned you a Class C network address of 198.47.212.0. You have 3 networks in your
company with the largest containing 134 hosts.
You need to figure out if you can subnet this. If it can
be subnetted you need
to calculate the new subnet mask and specify the available network numbers?
You need to determine how
many
bits you have available to work with. Since this is a standard Class C
network the last octet is available so you have 8 bits to work with.
1.
You n
eed to determine how many bits will be needed to address the largest number of hosts
that may exist on a network. This can easily be done by converting the largest number of hosts
to binary and counting the bits needed.
Largest number of hosts addresses n
eeded in
decimal
134
Largest number of hosts addresses converted to
binary
10000110
Number of bits needed to address hosts
8
2.
You need to determine how many bits will be needed to address the required number of
networks. This can easily be done by con
verting the required number of networks to binary and
counting the bits needed.
Network addresses needed in decimal
3
Network addresses
needed
converted to binary
11
Number of bits needed to address
networks
2
3.
Add the total number of bits needed toget
her and determine if a solution is possible. In this
case 8 + 2 = 10 so 10 bits are needed. We only have 8 available so this problem cannot be
solved.
Example
2
:
Your ISP has assigned you a Class C network address of 198.47.212.0. You have 3 networks i
n your
company with the largest containing 51 hosts. You need to figure out if you can subnet this. If it can be
subnetted you need to calculate the new subnet mask and specify the available network numbers?
1.
You need to determine how many bits you have a
vailable to work with.
Since this is a standard
Class C network the last octet is available so you have 8 bits to work with.
2.
You need to determine how many bits will be needed to address the largest number of hosts
that may exist on a network. This can
easily be done by converting the largest number of hosts
to binary and counting the bits needed.
Largest number of hosts addresses needed in
decimal
51
Largest number of hosts addresses converted to
binary
110011
Number of bits needed to address hosts
6
3.
You need to determine how many bits will be needed to address the required number of
networks. This can easily be done by converting the required number of networks to binary and
counting the bits needed.
Network addresses needed in decimal
3
Network
addresses needed converted to binary
11
Number of bits needed to address networks
2
4.
Add the total number of bits needed together and determine if a solution is possible. In this
case 6 + 2 = 8 so 8 bits are needed. We have 8 available so this proble
m can be solved.
5.
You must then determine the new subnet mask. To do this we take the number of bits needed
to address network from the left side of the available bits. Convert the value to decimal and you
have the subnet mask.
Subnet mask in binary
1100
0000
Subnet mask converted to decimal
192
Entire new subnet mask
–
佲楧楮慬⁰汵猠sh攠
捨cng敳eyou m慤攠瑯 瑨攠to獴spo牴ron
255⸲55⸲55⸱.2
㘮
Finally you need to determine the network addresses available. This can be done by counting
up in binary of the b
its you used for the network and converting the result to decimal.
Network number without
host bits (Both in decimal
and in binary)
Network number with
host bits (in binary)
Network number
Full network number
00 b
–
0
0000 0000
0
198⸴7⸲.2⸰/26
01
–
1
0100 0000
㘴
198⸴7⸲.2⸶4/26
10
–
2
1000 0000
128
198⸴7⸲.2⸱28/26
11
–
3
1100 0000
192
198⸴7⸲.2⸱92/26
Example 3:
Your ISP has assigned you a Class
B
network address of 160.13.0.0. You have 11 networks in your
company with the large
st containing 98 hosts. You need to figure out if you can subnet this. If it can be
subnetted you need to calculate the new subnet mask and specify the available network numbers?
1.
You need to determine how many bits you have available to work with.
Since
this is a standard
Class B network the last
two
octet
s
are
available so you have 16 bits to work with.
2.
You need to determine how many bits will be needed to address the largest number of hosts
that may exist on a network. This can easily be done by convertin
g the largest number of hosts
to binary and counting the bits needed.
Largest number of hosts addresses needed in
decimal
98
Largest number of hosts addresses converted to
binary
1100010
Number of bits needed to address hosts
7
3.
You need to determine h
ow many bits will be needed to address the required number of
networks. This can easily be done by converting the required number of networks to binary and
counting the bits needed.
Network addresses needed in decimal
11
Network addresses needed converte
d to binary
1011
Number of bits needed to address networks
4
4.
Add the total number of bits needed together and determine if a solution is possible. In this
case
7
+
4
=
11
so
11
bits are needed. We have
16
available so this problem can be solved.
5.
You
must then determine the new subnet mask. To do this we take the number of bits needed
to address network from the left side of the available bits. Convert the value to decimal and you
have the subnet mask.
Subnet mask in binary
11
11
0000
Subnet mask con
verted to decimal
240
Entire new subnet mask
–
佲楧楮慬⁰汵猠sh攠
捨cng敳eyou m慤攠瑯 瑨攠to獴spo牴ron
255⸲55.
㈴2
.
0
6.
Finally you need to determine the network addresses available. This can be done by counting
up in binary of the bits you used for the
network and converting the result to decimal.
Network number without
host bits (Both in decimal
and in binary)
Network number with
host bits (in binary)
Network number
Full network number
0
00
0
b
–
0
0000 0000
0
160⸱3⸰⸰
⼲
0
〰
01
–
1
0001
0000
ㄶ
1
60⸱3⸱..0/20
〰
10
–
2
001
0 0000
㌲
160⸱3⸳..0/20
〰
11
–
3
0011
0000
㐸
160⸱3⸴..0/20
0100
–
4
0100
0000
㘴
160⸱3⸶..0/20
0101
–
5
0101 0000
㠰
160⸱3⸸..0/20
0110
–
6
0110
0000
㤶
160⸱3⸹..0/20
0111
–
7
0111 0000
112
160.
13⸱.2⸰/20
1000
–
8
1000
0000
128
160⸱3⸱.8⸰/20
1001
–
9
1001
0000
144
160⸱3⸱.4⸰/20
1010
–
10
1010 0000
160
160⸱3⸱.0⸰/20
1011
–
11
1011
0000
176
160⸱3⸱.6⸰/20
1100
–
12
1100
0000
192
160⸱3⸱.2⸰/20
1101
–
13
1101 0
〰0
208
160⸱3⸲.8⸰/20
1110
–
14
1110
0000
224
160⸱3⸲.4⸰/20
1111
–
15
1111 0000
240
160⸱3⸲.0⸰/20
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