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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