Advanced Subnetting Example 1: Your ISP has assigned you a ...

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23 Οκτ 2013 (πριν από 4 χρόνια και 15 μέρες)

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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




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



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⸰⸰

0


01⁢


1⁤

0001

0000



1
60⸱3⸱..0/20


10⁢


2⁤

001
0 0000



160⸱3⸳..0/20


11⁢


3⁤

0011
0000



160⸱3⸴..0/20

0100⁢


4⁤

0100
0000



160⸱3⸶..0/20

0101⁢


5⁤

0101 0000



160⸱3⸸..0/20

0110⁢


6⁤

0110
0000



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